https://www.astro-init.top/api.php?action=feedcontributions&feedformat=atom&user=%E6%9B%B4%E6%96%B9%E6%9B%B4%E6%AD%A3%E7%9A%84%E7%89%A9%E7%90%86astro-init - 用户贡献 [zh-cn]2026-04-29T14:56:43Z用户贡献MediaWiki 1.32.2https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15402019年BAAO第1题-帕克太阳探测器2019-10-07T12:29:23Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>{{内容需要完善}}<br />
<br />
==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
[Third mark is for eliminating ''b'' and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)] <br />
<br />
'''e. [5 mks]''' With the perihelion distance we can find the eccentricity, <br />
<br />
$$r_{peri}=a(1-e)\therefore\ e=1-\frac{r_{peri}}{a}=1-\frac{9.86\times6.96\times10^8}{5.79\times10^{10}}=0.882$$ [''1 mks'']<br />
<br />
Using the formula just derived we can find ''E'' for ''r'' = 0.25 au, <br />
<br />
$$E=cos^{-1}(\frac{1}{e}(1-\frac{r}{a}))=cos^{-1}(\frac{1}{0.882}(1-\frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}))=1.16\rm\ rad(=66.4^{\circ})$$ [''1 mks'']<br />
<br />
Using the formula given we can find M, <br />
<br />
$$M=E-e\ sin\ E=1.16-0.882\times sin\ 1.16=0.351\rm\ rad(=20.1^{\circ})$$ [''1 mks'']<br />
<br />
Allowing for both sides of the ellipse, $$\Delta M=0.703\rm\ rad(=40.3^{\circ})$$ [''1 mks'']<br />
<br />
Movement through the circular orbit has constant angular velocity, meaning<br />
<br />
$$\frac{\Delta M}{\Delta t}=\frac{2\pi}{T}\ \therefore\ \Delta t=\frac{T\Delta M}{2\pi}=\frac{88\times0.703}{2\pi}=9.84\rm\ days(=9\ days\ 20\ hours\ 12\ mins)$$ [''1 mks'']<br />
<br />
[If they forget the factor of two (so $$\Delta t = 4.92\rm\ days$$) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part ''b''. Must be given in days for the final mark]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15392019年BAAO第1题-帕克太阳探测器2019-10-07T12:28:40Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>{{内容需要完善}}<br />
<br />
==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
[Third mark is for eliminating ''b'' and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)] <br />
<br />
'''e. [5 mks]''' With the perihelion distance we can find the eccentricity, <br />
<br />
$$r_{peri}=a(1-e)\therefore\ e=1-\frac{r_{peri}}{a}=1-\frac{9.86\times6.96\times10^8}{5.79\times10^{10}}=0.882$$ [''1 mks'']<br />
<br />
Using the formula just derived we can find ''E'' for ''r'' = 0.25 au, <br />
<br />
$$E=cos^{-1}(\frac{1}{e}(1-\frac{r}{a}))=cos^{-1}(\frac{1}{0.882}(1-\frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}))=1.16\rm\ rad(=66.4\circ)$$ [''1 mks'']<br />
<br />
Using the formula given we can find M, <br />
<br />
$$M=E-e\ sin\ E=1.16-0.882\times sin\ 1.16=0.351\rm\ rad(=20.1\circ)$$ [''1 mks'']<br />
<br />
Allowing for both sides of the ellipse, $$\Delta M=0.703\rm\ rad(=40.3\circ)$$ [''1 mks'']<br />
<br />
Movement through the circular orbit has constant angular velocity, meaning<br />
<br />
$$\frac{\Delta M}{\Delta t}=\frac{2\pi}{T}\ \therefore\ \Delta t=\frac{T\Delta M}{2\pi}=\frac{88\times0.703}{2\pi}=9.84\rm\ days(=9\ days\ 20\ hours\ 12\ mins)$$ [''1 mks'']<br />
<br />
[If they forget the factor of two (so $$\Delta t = 4.92\rm\ days$$) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part ''b''. Must be given in days for the final mark]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15382019年BAAO第1题-帕克太阳探测器2019-10-07T12:27:32Z<p>更方更正的物理：</p>
<hr />
<div>{{内容需要完善}}<br />
<br />
==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
[Third mark is for eliminating ''b'' and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)] <br />
<br />
'''e. [5 mks]''' With the perihelion distance we can find the eccentricity, <br />
<br />
$$r_{peri}=a(1-e)\therefore\ e=1-\frac{r_{peri}}{a}=1-\frac{9.86\times6.96\times10^8}{5.79\times10^{10}}=0.882$$ [''1 mks'']<br />
<br />
Using the formula just derived we can find ''E'' for ''r'' = 0.25 au, <br />
<br />
$$E=cos^{-1}(\frac{1}{e}(1-\frac{r}{a}))=cos^{-1}(\frac{1}{0.882}(1-\frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}))=1.16\rm\ rad(=66.4\degree)$$ [''1 mks'']<br />
<br />
Using the formula given we can find M, <br />
<br />
$$M=E-e\ sin\ E=1.16-0.882\times sin\ 1.16=0.351\rm\ rad(=20.1\degree)$$ [''1 mks'']<br />
<br />
Allowing for both sides of the ellipse, $$\Delta M=0.703\rm\ rad(=40.3\degree)$$ [''1 mks'']<br />
<br />
Movement through the circular orbit has constant angular velocity, meaning<br />
<br />
$$\frac{\Delta M}{\Delta t}=\frac{2\pi}{T}\ \therefore\ \Delta t=\frac{T\Delta M}{2\pi}=\frac{88\times0.703}{2\pi}=9.84\rm\ days(=9\ days\ 20\ hours\ 12\ mins)$$ [''1 mks'']<br />
<br />
[If they forget the factor of two (so $$\Delta t = 4.92\rm\ days$$) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part ''b''. Must be given in days for the final mark]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15372019年BAAO第1题-帕克太阳探测器2019-10-07T12:27:02Z<p>更方更正的物理：</p>
<hr />
<div>[[需要完善]]<br />
<br />
==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
[Third mark is for eliminating ''b'' and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)] <br />
<br />
'''e. [5 mks]''' With the perihelion distance we can find the eccentricity, <br />
<br />
$$r_{peri}=a(1-e)\therefore\ e=1-\frac{r_{peri}}{a}=1-\frac{9.86\times6.96\times10^8}{5.79\times10^{10}}=0.882$$ [''1 mks'']<br />
<br />
Using the formula just derived we can find ''E'' for ''r'' = 0.25 au, <br />
<br />
$$E=cos^{-1}(\frac{1}{e}(1-\frac{r}{a}))=cos^{-1}(\frac{1}{0.882}(1-\frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}))=1.16\rm\ rad(=66.4\degree)$$ [''1 mks'']<br />
<br />
Using the formula given we can find M, <br />
<br />
$$M=E-e\ sin\ E=1.16-0.882\times sin\ 1.16=0.351\rm\ rad(=20.1\degree)$$ [''1 mks'']<br />
<br />
Allowing for both sides of the ellipse, $$\Delta M=0.703\rm\ rad(=40.3\degree)$$ [''1 mks'']<br />
<br />
Movement through the circular orbit has constant angular velocity, meaning<br />
<br />
$$\frac{\Delta M}{\Delta t}=\frac{2\pi}{T}\ \therefore\ \Delta t=\frac{T\Delta M}{2\pi}=\frac{88\times0.703}{2\pi}=9.84\rm\ days(=9\ days\ 20\ hours\ 12\ mins)$$ [''1 mks'']<br />
<br />
[If they forget the factor of two (so $$\Delta t = 4.92\rm\ days$$) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part ''b''. Must be given in days for the final mark]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15362019年BAAO第1题-帕克太阳探测器2019-10-07T12:26:39Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
[Third mark is for eliminating ''b'' and writing all expressions in terms of only one trigonometric function. A reasonable attempt at a derivation (allowing alternative methods) must be present to get the marks for this question (simply writing the answer only scores 1 mark)] <br />
<br />
'''e. [5 mks]''' With the perihelion distance we can find the eccentricity, <br />
<br />
$$r_{peri}=a(1-e)\therefore\ e=1-\frac{r_{peri}}{a}=1-\frac{9.86\times6.96\times10^8}{5.79\times10^{10}}=0.882$$ [''1 mks'']<br />
<br />
Using the formula just derived we can find ''E'' for ''r'' = 0.25 au, <br />
<br />
$$E=cos^{-1}(\frac{1}{e}(1-\frac{r}{a}))=cos^{-1}(\frac{1}{0.882}(1-\frac{0.25\times1.50\times10^{11}}{5.79\times10^{10}}))=1.16\rm\ rad(=66.4\degree)$$ [''1 mks'']<br />
<br />
Using the formula given we can find M, <br />
<br />
$$M=E-e\ sin\ E=1.16-0.882\times sin\ 1.16=0.351\rm\ rad(=20.1\degree)$$ [''1 mks'']<br />
<br />
Allowing for both sides of the ellipse, $$\Delta M=0.703\rm\ rad(=40.3\degree)$$ [''1 mks'']<br />
<br />
Movement through the circular orbit has constant angular velocity, meaning<br />
<br />
$$\frac{\Delta M}{\Delta t}=\frac{2\pi}{T}\ \therefore\ \Delta t=\frac{T\Delta M}{2\pi}=\frac{88\times0.703}{2\pi}=9.84\rm\ days(=9\ days\ 20\ hours\ 12\ mins)$$ [''1 mks'']<br />
<br />
[If they forget the factor of two (so $$\Delta t = 4.92\rm\ days$$) allow 4 marks. Allow ±1 hour on the final answer to account for intermediate rounding errors. Allow full ecf for using their value of the semi-major axis from part ''b''. Must be given in days for the final mark]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15352019年BAAO第1题-帕克太阳探测器2019-10-07T12:10:56Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
Using Pythagoras' theorem, <br />
<br />
('''SP''')<sup>2</sup>=('''Pd''')<sup>2</sup>+('''Sd''')<sup>2</sup><br />
<br />
$$\therefore\ r^2=(b\ sin\ E)^2+(a\ cos\ E-ae)^2$$ [''1 mks'']<br />
<br />
$$ r^2=a^2(1-e^2)\times(1-cos^2\ E)+a^2(cos^2\ E-2e\ cos\ E+e^2)$$ [''1 mks'']<br />
<br />
$$r^2=a^2-2a^2e\ cos\ E+a^2e^2\ cos^2\ E=a^2(1-e\ cos\ E)^2$$<br />
<br />
$$\therefore\ r=a(1-e\ cos\ E)$$ [''1 mks'']<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15342019年BAAO第1题-帕克太阳探测器2019-10-07T12:00:52Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
'''SP''' = ''r''<br />
<br />
'''cS''' = ''ae''<br />
<br />
'''cd''' = ''U''<br />
<br />
'''Pd''' = ''V''<br />
<br />
therefore '''Sd''' = '''cd''' – '''cs''' = ''U'' – ''ae''<br />
<br />
For an ellipse, the general equation is $$\frac{U^2}{a^2}+\frac{V^2}{b^2}=1$$, so the points (U,V) on the ellipse are <br />
<br />
$$cos\ E=\frac{U}{a}$$ and $$sin\ E=\frac{V}{b}$$ [''1 mks'']<br />
<br />
We are also given that $$e=\sqrt{1-\frac{b^2}{a^2}}\ \therefore\ b^2=a^2(1-e^2)$$<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15332019年BAAO第1题-帕克太阳探测器2019-10-07T11:43:47Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
'''d. [4 mks]''' In '''Figure 3''',<br />
<br />
'''cx''' = '''cz''' = ''a''<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15322019年BAAO第1题-帕克太阳探测器2019-10-07T11:40:17Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
'''c. [7 mks]''' Using a similar method to the previous question to work out the semi-major axis and perihelion speed for the first orbit, <br />
<br />
$$a_{initial}=8.27\times10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v_{initial}=95.3\rm \ km\ s^{-1}$$ [''1 mks'']<br />
<br />
The total energy is the sum of the potential and kinetic energies, <br />
<br />
$$E_{tot}=E_p+E_k$$ and $$E_p=-\frac{GM}{r}$$<br />
<br />
At the first perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{35.7\times6.96\times10^8}+\frac{1}{2}\times685\times(95279)^2$$<br />
$$E_{tot}=-3.66\times10^{12}+3.11\times10^{12}=-5.50\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
At the final perihelion, <br />
<br />
$$E_{tot}=-\frac{6.67\times10^{-11}\times1.99\times10^{30}\times685}{9.86\times6.96\times10^8}+\frac{1}{2}\times685\times(190766)^2$$<br />
$$E_{tot}=-1.32\times10^{13}+1.25\times10^{13}=-7.85\times10^{11}\rm\ J$$ [Allow 1 mark if only have a correct E<sub>p</sub> or E<sub>k</sub> term] [''2 mks'']<br />
<br />
$$\therefore \Delta E_{tot}=2.35\times10^{11}\rm\ J$$ [Ignore a minus sign] [''1 mks'']<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15312019年BAAO第1题-帕克太阳探测器2019-10-07T11:21:58Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$ [''1 mks'']<br />
<br />
$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}=\sqrt{6.67\times10^{-11}\times1.99\times10^{30}(\frac{2}{9.86\times6.96\times10^8}-\frac{1}{5.79\times10^{10}})}$$ [''1 mks'']<br />
<br />
$$v=(190776\rm \ m\ s^{-1})=191\rm km\ s^{-1}$$ [Must be in km s<sup>-1</sup>] [''1 mks'']<br />
<br />
[This will mean that the Parker Solar Probe will become the fastest spacecraft (relative to the Sun) ever flown – at this speed you could travel from New York to Tokyo in less than a minute!]<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15302019年BAAO第1题-帕克太阳探测器2019-10-07T11:11:58Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[4]{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[4]{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[3]{\frac{GM}{4\pi^2}T^2}=\sqrt[3]{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15292019年BAAO第1题-帕克太阳探测器2019-10-07T11:10:55Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i) [2 mks]''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] <br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii) [3 mks]''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}=2\sigma T^4$$<br />
<br />
$$\therefore T=\sqrt[n]{4}{\frac{L_{odot}}{8\pi \sigma r^2}}$$ [''1 mks'']<br />
<br />
$$T=\sqrt[n]{4}{\frac{3.85\times10^{26}}{8\pi \times 5.67\times10^{-8}\times (9.86\times6.96\times10^8)^2}}$$ [''1 mks'']<br />
<br />
$$T=1550\rm\ K$$ [''1 mks'']<br />
<br />
'''b. [4 mks]''' Using Kepler’s third law to find the semi-major axis of the final orbit, <br />
<br />
$$T^2=\frac{4\pi^2}{GM}a^3$$ <br />
<br />
$$\therefore a=\sqrt[n]{3}{\frac{GM}{4\pi^2}T^2}=\sqrt[n]{3}{\frac{6.67\times10^{-11}\times1.99\times10^{30}}{4\pi^2}\times(88\times24\times3600)^2} $$ [''1 mks'']<br />
<br />
$$a=5.79\times 10^{10}\rm \ m$$<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15282019年BAAO第1题-帕克太阳探测器2019-10-07T08:40:40Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i)''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] ['''total 2 mks''']<br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii)''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15272019年BAAO第1题-帕克太阳探测器2019-10-07T08:30:51Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i)''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] ['''total 2 mks''']<br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii)''' For thermal balance, power absorbed must equal power emitted, so<br />
$$\frac{L_{odot}}{4\pi r^2}\times A_{abs}=\sigma A_{emit}T^4$$<br />
But since $$A_{emit}=2\times A_{abs}$$ then<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15262019年BAAO第1题-帕克太阳探测器2019-10-07T08:27:21Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i)''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] ['''total 2 mks''']<br />
<br />
[Allow calculating the solar absolute magnitude, $$M_{\odot}=4.29$$, as an alternative first mark] <br />
<br />
'''a. (ii)''' For thermal balance, power absorbed must equal power emitted, so<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15252019年BAAO第1题-帕克太阳探测器2019-10-07T08:22:36Z<p>更方更正的物理：/* 英文解答 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
'''Q1 - Parker Solar Probe''' '''[25 marks]'''<br />
<br />
'''a. (i)''' ratio of brightness, $$\frac{b_1}{b_0} = \frac{(1\rm\ au)^2}{9(.86\rm\ R_{\odot})^2}=\frac{(1.50\times10^{11})^2}{(9.86\times6.96\times10^8)^2}=478$$ [''1 mks'']<br />
$$\therefore m_{new}=m_{\odot}-2.5\rm log(\frac{b_1}{b_0})=-26.74-6.698=-33.44$$ [''1 mks''] ['''total 2 mks''']<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15242019年BAAO第1题-帕克太阳探测器2019-10-07T07:33:44Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
'''d.''' Derive a formula for the distance from the focus for an elliptical orbit, ''r'' (''SP'' in the figure) in terms of the semi-major axis ''a'', the eccentricity ''e'', and the eccentric anomaly ''E''.<br />
<br />
The eccentric anomaly can be related to the mean anomaly through Kepler’s Equation,<br />
<br />
''M = E − e sin E'' .<br />
<br />
'''e.''' Calculate how long PSP spends doing primary science in its final orbit. Give your answer in days.<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15232019年BAAO第1题-帕克太阳探测器2019-10-07T06:56:24Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
<br />
'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle ''θ'' is called the true anomaly and is not needed for this question. Credit: Wikipedia.<br />
<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15222019年BAAO第1题-帕克太阳探测器2019-10-07T06:46:23Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
[[文件:BAAO2019 fig3.png|替代= Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.|左|无框|400x400像素|'''Figure 3:''' The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled ''S'' and the probe ''P'' . ''M'' and ''E'' are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia. ]]<br />
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<br />
<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=%E6%96%87%E4%BB%B6:BAAO2019_fig3.png&diff=1521文件:BAAO2019 fig3.png2019-10-07T06:45:00Z<p>更方更正的物理：</p>
<hr />
<div>Figure 3: The definitions of the anomalies needed to get the position of an object in an ellipse as a function of time. The Sun (located at the focus) is labeled S and the probe P . M and E are the mean and eccentric anomalies respectively. The angle θ is called the true anomaly and is not needed for this question. Credit: Wikipedia.</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15112019年BAAO第1题-帕克太阳探测器2019-10-06T02:08:25Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
Close to the Sun the communications equipment is very sensitive to the extreme environment, so the mission is planned for the probe to take all of its primary science measurements whilst within 0.25 au of the Sun, and then to spend the rest of the orbit beaming that data back to Earth, as shown in Figure 2.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
When considering the position of an object in an elliptical orbit as a function of time, there are two important angles (called ‘anomalies’) necessary to do the calculation, and they are defined in Figure 3. By constructing a circular orbit centred on the same point as the ellipse and with the same orbital period, the eccentric anomaly, ''E'', is then the angle between the major axis and the perpendicular projection of the object (some time ''t'' after perihelion) onto the circle as measured from the centre of the ellipse (∠''xcz'' in the figure). The mean anomaly, ''M'', is the angle between the major axis and where the object would have been at time ''t'' if it was indeed on the circular orbit (∠''ycz'' in the figure, such that the shaded areas are the same).<br />
<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15102019年BAAO第1题-帕克太阳探测器2019-10-06T02:03:01Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
'''c.''' After the first flyby of Venus on 3<sup>rd</sup> October 2018 it was moved into an orbit with a 150 day period, and the subsequent first perihelion on 6<sup>th</sup> November 2018 was at a distance of 35.7 $$R_⊙$$. Given its mass at launch was 685 kg, calculate the total amount of energy that had to be lost by the probe to get from this first orbit (ignoring the orbital properties prior to the Venus flyby) to the final orbit. Ignore any change in the mass of the probe due to burning fuel.<br />
<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br />
'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.<br />
<br />
<br />
<br />
<br />
<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15092019年BAAO第1题-帕克太阳探测器2019-10-06T01:58:53Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
[[文件:BAAO2019 fig2.png|替代= Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.|左|无框|600x600像素|'''Figure 2:''' The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL. ]]<br />
<br /><br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=%E6%96%87%E4%BB%B6:BAAO2019_fig2.png&diff=1508文件:BAAO2019 fig2.png2019-10-06T01:58:00Z<p>更方更正的物理：</p>
<hr />
<div>Figure 2: The way PSP is planned to split each orbit into taking measurements and sending data back. Credit: NASA / Johns Hopkins APL.</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15072019年BAAO第1题-帕克太阳探测器2019-10-06T01:50:53Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
'''b.''' The speed, $$v$$, of an object in an elliptical orbit of semi-major axis $$a$$ around an object of mass $$M$$ when a distance $$r$$ away can be calculated as<br />
<br />
$$v^2=GM(\frac{2}{r}-\frac{1}{a})$$.<br />
<br />
Given that in its final orbit PSP has a orbital period of 88 days, calculate the speed of the probe as it passes through the minimum perihelion. Give your answer in km s<sup>−1</sup>.<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15062019年BAAO第1题-帕克太阳探测器2019-10-06T01:43:48Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15052019年BAAO第1题-帕克太阳探测器2019-10-06T01:43:26Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15042019年BAAO第1题-帕克太阳探测器2019-10-06T01:42:45Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
'''a.''' When the probe is at its closest perihelion:<br />
'''(i)''' Calculate the apparent magnitude of the Sun, given that from Earth $$m_{⊙} = −26.74$$.<br />
'''(ii)''' Calculate the temperature the heat shield must be able to survive. Assume that the heat shield of the probe absorbs all of the incident radiation, radiates as a perfect black body, and that only one side of the probe ever faces the Sun (to protect the instruments) such that the emitting (surface) area is double the absorbing (cross-sectional) area.<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15032019年BAAO第1题-帕克太阳探测器2019-10-06T01:39:18Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
'''Figure 1:''' ''Left:'' The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right:'' The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15022019年BAAO第1题-帕克太阳探测器2019-10-05T15:30:11Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=15012019年BAAO第1题-帕克太阳探测器2019-10-05T15:25:42Z<p>更方更正的物理：/* 英文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
[[文件:BAAO2019 fig1.png|替代= Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.|左|无框|600x600像素|'''Figure 1''': ''Left'': The journey PSP will take to get from the Earth to the final orbit around the Sun. ''Right'': The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman. ]]<br />
<br />
<br />
(To be continued)<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=%E6%96%87%E4%BB%B6:BAAO2019_fig1.png&diff=1500文件:BAAO2019 fig1.png2019-10-05T15:23:18Z<p>更方更正的物理：</p>
<hr />
<div>Figure 1: Left: The journey PSP will take to get from the Earth to the final orbit around the Sun. Right: The probe just after assembly in the John Hopkins University Applied Physics Laboratory. Credit: NASA / John Hopkins APL / Ed Whitman.</div>更方更正的物理https://www.astro-init.top/index.php?title=2019%E5%B9%B4BAAO%E7%AC%AC1%E9%A2%98-%E5%B8%95%E5%85%8B%E5%A4%AA%E9%98%B3%E6%8E%A2%E6%B5%8B%E5%99%A8&diff=14992019年BAAO第1题-帕克太阳探测器2019-10-05T15:16:26Z<p>更方更正的物理：创建页面，内容为“==英文题目== '''Qu 1. Parker Solar Probe''' The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first…”</p>
<hr />
<div>==英文题目==<br />
'''Qu 1. Parker Solar Probe'''<br />
<br />
The Parker Solar Probe (PSP) is part of a mission to learn more about the Sun, named after the scientist that first proposed the existence of the solar wind, and was launched on 12th August 2018. Over the course of the 7 year mission it will orbit the Sun 24 times, and through 7 flybys of Venus it will lose some energy in order to get into an ever tighter orbit (see Figure 1). In its final 3 orbits it will have a perihelion (closest approach to the Sun) of only $$r_{peri} = 9.86 R_{⊙}$$, about 7 times closer than any previous probe, the first of which is due on 24th December 2024. In this extreme environment the probe will not only face extreme brightness and temperatures but also will break the record for the fastest ever spacecraft.<br />
<br />
(To be continued)<br />
<br />
==英文解答==<br />
<br />
==中文题目==<br />
<br />
==中文解答==</div>更方更正的物理https://www.astro-init.top/index.php?title=BAAO2019&diff=1487BAAO20192019-10-05T03:50:44Z<p>更方更正的物理：/* 题目列表 */</p>
<hr />
<div>==赛前须知==<br />
<br />
==重要常数==<br />
<br />
==重要公式==<br />
<br />
==题目列表==<br />
<br />
[[2019年BAAO第1题-帕克太阳探测器]]<br />
<br />
==出题者==<br />
<br />
==合作机构==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=BAAO2019&diff=1486BAAO20192019-10-05T03:46:04Z<p>更方更正的物理：创建页面，内容为“==赛前须知== ==重要常数== ==重要公式== ==题目列表== ==出题者== ==合作机构== ==相关链接==”</p>
<hr />
<div>==赛前须知==<br />
<br />
==重要常数==<br />
<br />
==重要公式==<br />
<br />
==题目列表==<br />
<br />
==出题者==<br />
<br />
==合作机构==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO1996&diff=1485IAO19962019-10-05T03:14:48Z<p>更方更正的物理：/* 理论考试 */</p>
<hr />
<div>{{内容需要完善}}<br />
==理论考试==<br />
低年组：<br />
<br />
[[1996年IAO理论低年组第1题]]<br />
<br />
高年组：<br />
<br />
[[1996年IAO理论高年组第1题]]<br />
<br />
低年组和高年组共同：<br />
<br />
==实测考试==<br />
<br />
==观测考试==<br />
[[1996年IAO观测题]]<br />
<br />
==参与国家==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=1996%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E9%AB%98%E5%B9%B4%E7%BB%84%E7%AC%AC1%E9%A2%98&diff=14841996年IAO理论高年组第1题2019-10-05T03:14:27Z<p>更方更正的物理：创建页面，内容为“==中文题目== '''1.''' 为什么有的星在蓝光里表现为双星而红光里却无法分辨而是单星？ ==中文解答== '''1.''' 这是由于望远镜的…”</p>
<hr />
<div>==中文题目==<br />
'''1.''' 为什么有的星在蓝光里表现为双星而红光里却无法分辨而是单星？<br />
<br />
==中文解答==<br />
'''1.''' 这是由于望远镜的分辨角与口径成正比，而与观测的波长成反比，其表达式为：分辨角$$\delta=1.22\frac{\lambda}{D}$$，其中$$\lambda$$为波长，D为望远镜的口径。从这一公式很容易看出望远镜对于波长较短的蓝光分辨率较高。</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO1996&diff=1474IAO19962019-10-04T15:42:06Z<p>更方更正的物理：/* 理论考试 */</p>
<hr />
<div>{{内容需要完善}}<br />
==理论考试==<br />
低年组：<br />
<br />
[[1996年IAO理论低年组第1题]]<br />
<br />
高年组：<br />
<br />
低年组和高年组共同：<br />
<br />
==实测考试==<br />
<br />
==观测考试==<br />
[[1996年IAO观测题]]<br />
<br />
==参与国家==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO1996&diff=1473IAO19962019-10-04T15:41:37Z<p>更方更正的物理：/* 理论考试 */</p>
<hr />
<div>{{内容需要完善}}<br />
==理论考试==<br />
低年组：<br />
[[1996年IAO理论低年组第1题]]<br />
<br />
高年组：<br />
<br />
低年组和高年组共同：<br />
<br />
==实测考试==<br />
<br />
==观测考试==<br />
[[1996年IAO观测题]]<br />
<br />
==参与国家==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=1996%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC1%E9%A2%98&diff=14721996年IAO理论低年组第1题2019-10-04T15:41:19Z<p>更方更正的物理：创建页面，内容为“==中文题目== '''1.''' 为什么有时候使用在地球轨道上的小望远镜比使用在山顶上的大望远镜更好？ ==中文解答== 这道题考查的…”</p>
<hr />
<div>==中文题目==<br />
'''1.''' 为什么有时候使用在地球轨道上的小望远镜比使用在山顶上的大望远镜更好？<br />
<br />
==中文解答==<br />
这道题考查的是参赛者的天文常识。<br />
<br />
大家知道，一般来说天文望远镜的口径越大越好，口径增大回答来两个直接的好处，一是增加了集光本领，而是提高了空间分辨率。但是由于地球大气扰动的影响，对于安放在地面上口径较大的天文望远镜，其实际分辨率无法达到理论值，而对于在太空中的望远镜（如哈勃空间望远镜）由于在大气层之上，其分辨率可以达到理论值。所以对于需要高空分辨率的观测项目，使用在环绕地球的轨道上的小望远镜观测的效果要优于使用山顶上的大望远镜。</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO1996&diff=1466IAO19962019-10-04T13:26:05Z<p>更方更正的物理：</p>
<hr />
<div>{{内容需要完善}}<br />
==理论考试==<br />
低年组：<br />
<br />
高年组：<br />
<br />
低年组和高年组共同：<br />
<br />
==实测考试==<br />
<br />
==观测考试==<br />
[[1996年IAO观测题]]<br />
<br />
==参与国家==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=1996%E5%B9%B4IAO%E8%A7%82%E6%B5%8B%E9%A2%98&diff=14651996年IAO观测题2019-10-04T13:25:40Z<p>更方更正的物理：创建页面，内容为“{{需要解答}} ==中文题目== （不计入总分） 1. 在1996年11月2日星期六的什么时间，你能看到金星？ 2. 当时它离太阳的角距离为…”</p>
<hr />
<div>{{需要解答}}<br />
==中文题目==<br />
（不计入总分）<br />
<br />
1. 在1996年11月2日星期六的什么时间，你能看到金星？<br />
<br />
2. 当时它离太阳的角距离为多少？</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO&diff=1464IAO2019-10-04T13:16:55Z<p>更方更正的物理：/* 往届赛事 */</p>
<hr />
<div>IAO（国际天文奥林匹克竞赛）是天文方面历史悠久的国际赛事，参赛年龄为14-18岁的中学学生。自1996年成立以来已经连续举办过23届。<br />
<br />
==历史==<br />
IAO的起源具有浓厚的苏联背景，学科“奥赛”的概念最初就发源于苏联。在1990-1991年，苏联科学院组织了两次次多语言多学科的奥赛活动，有俄罗斯、莫斯科和爱沙尼亚派队参加。这次比赛的成功举办促使欧亚天文学会决心设立一个年度比赛，并且成立了专门的协调委员会。在1996年正式举办了第一次国际天文奥赛。<br />
<br />
此后IAO多次在俄罗斯和乌克兰举办。参赛国家也多局限于东欧国家和东亚国家。<br />
<br />
自成立以来，IAO委员会主席一直由Dr. Michael G. Gavrilov担任。<br />
<br />
==规则==<br />
IAO的比赛分为两个组：α组和β组。其中α组相当于低年组，为15岁以下的选手；β组年龄上限是18岁。详细的年龄限制如下：<br />
<br />
*年龄均以赛事举办当年1月1日计算；<br />
*参赛的年龄下限为14岁；<br />
*第一次参加且未满15岁的选手是α组，不满足的为β组；<br />
*第一次参赛的年龄上限是18岁，第二次参赛的年龄上限是17岁，第三次参赛的年龄上限是16岁。<br />
*在之前的IAO赛事得过一等奖和二等奖的选手可以在年龄限制允许范围内不受名额限制地参赛。<br />
<br />
每个参赛队由3名α组选手和2名β组选手和两名领队组成。得过一二等奖的“保送选手”不在此限制之内，但对于中国选手，想继续参加IAO依然需要通过国家队的选拔。<br />
<br />
每届IAO的组委会需要负责选手在比赛国的所有开销，两次比赛间IAO委员会的开销，以及比赛期间IAO委员会工作人员的劳务。但组委会允许向参赛选手收取参赛费。参赛费的相当一部分用于支付IAO委员会的实际开销。<br />
<br />
IAO允许区域和城市组队参加。因此莫斯科队与俄罗斯队经常一同出现在IAO赛场上。<br />
<br />
==流程==<br />
<br />
===赛程===<br />
IAO同样包括理论、实测和观测三部分。理论部分的试题由委员会提供，实测和观测试题由组委会提供。试题均被分为α组和β组。理论部分的试题为两个组各5道，其中可能有公用的题目或是略有差异的题目。实测部分为1-2道题目，观测部分为1-3道题目。<br />
<br />
理论部分的考试时间是4.5小时，实测部分为3-4小时，由组委会决定。<br />
<br />
===试题翻译与判分===<br />
原始的试题同时提供英语和俄语两个版本，共领队选择使用。在比赛开始几个小时前，领队才开始试题的翻译工作。由于参赛选手的通信设备不受限制，因此翻译过程中领队的通信设备会被封存，而只能使用纸质词典或是专门的电子词典。同时领队提供的翻译件也只能是手写版。<br />
<br />
比赛结束之后，领队需要将选手的答卷翻译回英语。选手只允许在答题本的左侧页面答题，右侧留给翻译和判分。<br />
<br />
试卷的判分工作同样由领队承担。领队在判分时可以直接与对应的领队讨论，因此省掉了改分步骤。<br />
<br />
===奖项设置===<br />
IAO设置三个主要奖项，分别为一等奖、二等奖和三等奖（对应金牌、银牌和铜牌）。三个奖项之下的选手可获得参赛证书。出于弱化功利性、提倡思维和创新性的目的，IAO不会颁发奖牌，而是纸质的奖项证书，证书的设计也是固定的，只有背景图和logo会随举办地进行一些定制。<br />
<br />
三个奖项的划定没有固定的标准，而是由当届领队开会投票决定的。开会时领队们会看到一个只有分数排名的表格，并从中选取较大的分数空当划定分数线。当然大部分领队会尽可能将分数线降低，但存在某种人数限制阻止他们这么做。<br />
<br />
两个年龄组各设置一个最佳成绩奖和三个单项奖。第一次参加IAO的选手中成绩最佳的会获得一个单项奖；第一次参加IAO的国家的最佳成绩也会获得一个单项奖。<br />
<br />
另外，作为IAO的文化传统，每届IAO理论部分都包含一道题目要求参赛选手绘制一副带有熊和其他动物的示意图。绘制最美观的选手将获得“最佳画熊奖”，两个年龄组只设置一个。<br />
<br />
==往届赛事==<br />
{| class="wikitable"<br />
| rowspan="2" |'''IAO Nr.'''<br />
| rowspan="2" |'''Year'''<br />
| rowspan="2" |'''Dates'''<br />
| rowspan="2" |'''Country'''<br />
| rowspan="2" |'''Observatory or Institute'''<br />
| rowspan="2" |'''Town'''<br />
| colspan="2" |'''Number of states'''<br />
| rowspan="2" |'''Lo-calte-am**'''<br />
|-<br />
|'''partici-pating'''<br />
|'''obser-ving'''<br />
|-<br />
|Preli-<br />
minary<br />
|1990<br />
|October<br />
20-22<br />
|'''Moscow Land *'''<br />
|ISSP RAS<br />
|Chernogolovka<br />
|3<br />
|1<br />
| +<br />
|-<br />
|Preli-<br />
minary<br />
|1991<br />
|December<br />
13-16<br />
|'''Moscow Land *'''<br />
|ISSP RAS<br />
|Chernogolovka<br />
|4<br />
|1<br />
| +<br />
|-<br />
|I<br />
|[[IAO1996|1996]]<br />
|November<br />
1-9<br />
|'''Russia'''<br />
|SAO RAS<br />
|Nizhnij Arkhyz<br />
|4<br />
|1<br />
| -<br />
|-<br />
|II<br />
|1997<br />
|October<br />
21-28<br />
|'''Russia'''<br />
|SAO RAS<br />
|Nizhnij Arkhyz<br />
|4<br />
| -<br />
| -<br />
|-<br />
|III<br />
|1998<br />
|October<br />
20-27<br />
|'''Russia'''<br />
|SAO RAS<br />
|Nizhnij Arkhyz<br />
|5<br />
|1<br />
| -<br />
|-<br />
|IV<br />
|1999<br />
|September 25<br />
- October 2<br />
|'''Crimea *'''<br />
|CrLab SAI,<br />
CrAO<br />
|Nauchnyj<br />
|7<br />
|2<br />
| -<br />
|-<br />
|V<br />
|2000<br />
|October<br />
20-27<br />
|'''Russia'''<br />
|SAO RAS<br />
|Nizhnij Arkhyz<br />
|8<br />
| -<br />
| -<br />
|-<br />
|VI<br />
|2001<br />
|September 26<br />
- October 3<br />
|'''Crimea *'''<br />
|CrLab SAI,<br />
CrAO<br />
|Nauchnyj<br />
|7<br />
|1<br />
| -<br />
|-<br />
|VII<br />
|2002<br />
|October<br />
22-29<br />
|'''Russia'''<br />
|SAO RAS<br />
|Nizhnij Arkhyz<br />
|11<br />
| -<br />
| +<br />
|-<br />
|VIII<br />
|2003<br />
|October<br />
3-8<br />
|'''Sweden'''<br />
|Stockholm Observatory<br />
|Stockholm<br />
|13<br />
|2<br />
| +<br />
|-<br />
|IX<br />
|2004<br />
|October<br />
1-9<br />
|'''Crimea *'''<br />
|CrLab SAI,<br />
CrAO<br />
|Simeiz<br />
|18<br />
| -<br />
| -<br />
|-<br />
|X<br />
|2005<br />
|October 25 -<br />
November 2<br />
|'''China'''<br />
|Beijing Planetaruim<br />
|Beijing<br />
|15<br />
|2<br />
| +<br />
|-<br />
|XI<br />
|2006<br />
|November<br />
10-19<br />
|'''India'''<br />
|Tata Institute for Fundamental Research<br />
|Bombay<br />
( Mumbai )<br />
|16<br />
|3<br />
| rowspan="11" |n<br />o<br />r<br />u<br />l<br />e<br />
|-<br />
|XII<br />
|2007<br />
|September 29<br />
- October 7<br />
|'''Crimea *'''<br />
|CrLab SAI,<br />
CrAO<br />
|Simeiz<br />
|23<br />
|1<br />
|-<br />
|XIII<br />
|2008<br />
|October<br />
13-21<br />
|'''Italy'''<br />
|Trieste Astronomical Observatory<br />
|Trieste<br />
|19<br />
| -<br />
|-<br />
|XIV<br />
|2009<br />
|November<br />
8-16<br />
|'''China'''<br />
|Beijing Planetaruim<br />
|Hangzhou<br />
|17<br />
|1<br />
|-<br />
|XV<br />
|2010<br />
|October<br />
16-24<br />
|'''Crimea *'''<br />
|CrLab SAI,<br />
CrAO<br />
|Sudak<br />
|19<br />
| -<br />
|-<br />
|XVI<br />
|2011<br />
|September<br />
22-30<br />
|'''Kazakhstan'''<br />
|AFI<br />
|Alma-Ata<br />
|19<br />
|2<br />
|-<br />
|XVII<br />
|2012<br />
|October<br />
16-24<br />
|'''Korea'''<br />
|SNU<br />
|Gwangju<br />
|20<br />
|2<br />
|-<br />
|XVIII<br />
|2013<br />
|September<br />
6-14<br />
|'''Lithuania'''<br />
|<br />
|Vilnius<br />
|20<br />
| -<br />
|-<br />
|XIX<br />
|2014<br />
|October<br />
12-21<br />
|'''Kyrgyzstan'''<br />
|AFI<br />
|Bishkek -<br />
Cholpon-Ata<br />
|17<br />
| -<br />
|-<br />
|XX<br />
|2015<br />
|October<br />
15-23<br />
|'''Russia'''<br />
|Kazan University Observatory<br />
|Kazan<br />
|13<br />
| -<br />
|-<br />
|XXI<br />
|[[IAO2016|2016]]<br />
|October<br />
5-13<br />
|'''Bulgaria'''<br />
|Rozhen Observatory<br />
|Pamporovo -<br />
Smolyan<br />
|<br />
|<br />
|-<br />
|XXII<br />
|[[IAO2017|2017]]<br />
|Oct 27-Nov 4<br />
|China<br />
|<br />
|Weihai<br />
|<br />
|<br />
|<br />
|-<br />
|XXIII<br />
|[[IAO2018|2018]]<br />
|October 6-14<br />
|Sri Lanka<br />
|<br />
|Colombo<br />
|<br />
|<br />
|<br />
|-<br />
|XXIV<br />
|[[IAO2019|2019]]<br />
|October 19-27<br />
|Romania<br />
|<br />
|<br />
|<br />
|<br />
|<br />
|}<br />
<br />
==相关链接==<br />
IAO章程 http://www.issp.ac.ru/iao/statutes/list_en.html<br />
<br />
过往IAO列表 http://www.issp.ac.ru/iao/list_e.html</div>更方更正的物理https://www.astro-init.top/index.php?title=IAO1996&diff=1463IAO19962019-10-04T13:16:11Z<p>更方更正的物理：创建页面，内容为“{{内容需要完善}} ==理论考试== 低年组： 高年组： 低年组和高年组共同： ==实测考试== ==观测考试== ==参与国家== ==相关…”</p>
<hr />
<div>{{内容需要完善}}<br />
==理论考试==<br />
低年组：<br />
<br />
高年组：<br />
<br />
低年组和高年组共同：<br />
<br />
==实测考试==<br />
<br />
==观测考试==<br />
<br />
<br />
==参与国家==<br />
<br />
==相关链接==</div>更方更正的物理https://www.astro-init.top/index.php?title=2018%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC2%E9%A2%98-%E7%81%AB%E6%98%9F%E5%A4%A7%E5%86%B2&diff=11682018年IAO理论低年组第2题-火星大冲2019-09-17T12:04:15Z<p>更方更正的物理：修改链接到的知识点网页</p>
<hr />
<div>==英文题目==<br />
'''Great oppositions of Mars.''' On the morning of July 27, 2018, Mars was in a great oposition (see attached ephemeris), The great oppositons of Mars (the situations when this planet is seen most brightly in comparison with the visibility during other oppositions) are repeated every 15 or 17 years. So the previous opposition (which is ales called the "greatest", since Marswas so bright only once in many centuries) was on August 28, 2003, and the next one will occur in 2035.<br />
<br />
'''2.1.''' Determine the constellation, in which Mars was during this year opposition.<br />
<br />
'''2.2.''' Considering the orbits of the Earth and Mars as circular, and based on the date of opposition 2003, calculate the dates the oppositions in 2018 and 2035 would occur.<br />
<br />
'''2.3''' Explain, why the difference appeared between the calculated and real dates of the opposition 2018. (Draw a figure that clearly demonstrates your explanation.)<br />
<br />
'''2.4.''' In which of the oppositions, 2018 or 2035 is Mars brightest? (The answer must be justified by drawing a figure).<br />
<br />
==中文题目==<br />
'''火星大冲''' 2018年7月27日上午，火星到达大冲位置（见附录星历表【注：暂无】）。火星大冲（大冲时行星的视亮度在所有冲日事件中是最亮的）每15或17年发生一次。所以上一次大冲（也被称为“最”大冲，因为这一次的火星是许多世纪以来最亮的）发生在2003年8月28日，下一次大冲将发生在2035年。<br />
<br />
'''2.1.''' 写出今年火星大冲时，火星位于哪个星座。<br />
<br />
'''2.2.''' 假设地球和火星的轨道为圆轨道，基于题目中给出的2003年火星大冲的发生日期，计算2018年和2035年发生火星大冲的日期。<br />
<br />
'''2.3.''' 试解释计算得到的2018年火星大冲日期与实际日期间存在差别的原因。（请画草图说明）<br />
<br />
'''2.4.''' 2018年与2035年的火星大冲，哪一次更亮？（请画图证明你的答案）<br />
<br />
==解答==<br />
<br />
'''英文解答'''<br />
<br />
'''2.1.''' During the opposition, the planet is at a point approximately (up to an angular distance from the ecliptic at this moment) opposite to the Sun, that is, located where the Sun was six monthe ago. The Sun is in the constellation Capricorn on January 27.<br />
<br />
'''2.2.''' The oppositions are repeatde every synodic period. The synodic of Mars is:<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
Approximately 15 years passed from 2003 to 2018, and there are 15·T<sub>E</sub> / T<sub>S</sub> ≈ 7.02 synodic periods of Mars during this period. So, the opposition 2018 was to occur after 7 synodic periods earlier than August 28.<br />
<br />
The easiest way to calculate the date is as follows. For 7 synodic periods of Mars 7×779.93 ≈ 5459 days pass. From August 28 2003, to August 28, 2018, 365×15+4 = 5479 days pass. Thus, it turns out that in 2018 the opposition of Mars should be 20 days earlier than August 28, that is, August 8.<br />
<br />
【注：此段英文解答不全，详细过程在中文解答中补全】<br />
Thus, it turns out that in 2035 the opposition of Mars should be 11 days later than August 28, that is, Septmeber 8.<br />
<br />
'''2.3.''' The difference appears from face that in our calculations we take the average angular velocity of Mars in orbit, and in previous close to great oppositions, Mars is near the perihelion of its orbit, and its speed is larger. As a result, the difference in the angular velocities of the Earth and Mars is smaller. Therefore, to change the position of Mars relative to the Earth takes more time. That is why, the farther the opposition from the perihelion point of Mars orbit (which is at the ecliptic longitude that the Earth passes around August 28), the greater the error. Calculations in the approximation of circular orbits give about 1.5 times the underestimated interval between the day of the calculated opposition and the date on which the "greatest oppositions" take place (that is, August 28).<br />
<br />
'''2.4.''' The position of Mars during the opposition of 2035 is closer to the positiom of the perihelion of the orbit than in 2018, therefore in the opposition of 2035 Mars will be brighter.<br />
<br />
'''中文解答'''<br />
<br />
'''2.1.''' 摩羯座。冲日时，行星大约正好在太阳对侧（此时的行星位置距离黄道只有一些角距离）。所以，火星在太阳半年前所在的位置。1月27日太阳位于摩羯座。<br />
<br />
'''2.2.''' 行星与地球的每个[[:分类:会合周期|会合周期]]会发生一次冲日。火星的会合为：<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
其中T<sub>S</sub>代表火星的会合周期，T<sub>E</sub>代表地球的公转周期，T<sub>M</sub>代表火星的公转周期。<br />
<br />
从2003年到2018年经过了大概15年，这大约相当于15×T<sub>E</sub> / T<sub>S</sub> ≈ 7.02个火星与地球的会合周期。所以2018年的火星冲日应发生在2003年火星冲日的7个会合周期之后，比8月28日更早的日子。<br />
<br />
以下是最简单的计算方法。火星的7个会合周期大约经过了7×779.93 ≈ 5459 天。从2003年8月28日到2018年8月28日，经过了365×15+4 = 5479 天。所以2018年的火星冲日发生的日期应该比8月28日早20天，也就是在8月8日。<br />
<br />
同理，从2003年到2035年经过了大概32年，这大约相当于32×T<sub>E</sub> / T<sub>S</sub> ≈ 14.99个火星会合周期。火星的15个会合周期大约经过了15×779.93 ≈ 11699 天。从2003年8月28日到2035年8月28日经过了365×32+8 = 11688 天。所以2035年的火星冲日发生的日期应该比8月28日晚11天，也就是在9月8日。<br />
<br />
'''2.3. '''造成冲日日期计算与实际上有差异的原因是在我们的计算中，假设了火星的公转角速度为一个恒定的平均角速度，但实际上在上一次2003年火星冲日的时候，火星更接近它的近日点，所以轨道角速度更大。因此，那时候地球和火星的轨道角速度差更小。火星相对于地球位置的改变需要花更多时间。这就是当大冲发生时火星（此时火星的黄经大约即地球在8月28日左右所在的黄经位置）离其轨道近日点越远，计算与实际大冲日期间的误差就越大。如果取近似圆轨道来计算2018年的大冲日期，所得的日期与2003年“最”大冲的8月28日间的日期差会被低估1.5倍。（译注：圆轨道计算该日期差为20天，而实际日期差为约31天）<br />
<br />
'''2.4.''' 2035年火星大冲时火星的在轨道上位置比2018年火星大冲时更靠近近日点。所以2035年的火星大冲更亮。</div>更方更正的物理https://www.astro-init.top/index.php?title=2018%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC2%E9%A2%98-%E7%81%AB%E6%98%9F%E5%A4%A7%E5%86%B2&diff=11672018年IAO理论低年组第2题-火星大冲2019-09-17T12:03:01Z<p>更方更正的物理：链接入知识点</p>
<hr />
<div>==英文题目==<br />
'''Great oppositions of Mars.''' On the morning of July 27, 2018, Mars was in a great oposition (see attached ephemeris), The great oppositons of Mars (the situations when this planet is seen most brightly in comparison with the visibility during other oppositions) are repeated every 15 or 17 years. So the previous opposition (which is ales called the "greatest", since Marswas so bright only once in many centuries) was on August 28, 2003, and the next one will occur in 2035.<br />
<br />
'''2.1.''' Determine the constellation, in which Mars was during this year opposition.<br />
<br />
'''2.2.''' Considering the orbits of the Earth and Mars as circular, and based on the date of opposition 2003, calculate the dates the oppositions in 2018 and 2035 would occur.<br />
<br />
'''2.3''' Explain, why the difference appeared between the calculated and real dates of the opposition 2018. (Draw a figure that clearly demonstrates your explanation.)<br />
<br />
'''2.4.''' In which of the oppositions, 2018 or 2035 is Mars brightest? (The answer must be justified by drawing a figure).<br />
<br />
==中文题目==<br />
'''火星大冲''' 2018年7月27日上午，火星到达大冲位置（见附录星历表【注：暂无】）。火星大冲（大冲时行星的视亮度在所有冲日事件中是最亮的）每15或17年发生一次。所以上一次大冲（也被称为“最”大冲，因为这一次的火星是许多世纪以来最亮的）发生在2003年8月28日，下一次大冲将发生在2035年。<br />
<br />
'''2.1.''' 写出今年火星大冲时，火星位于哪个星座。<br />
<br />
'''2.2.''' 假设地球和火星的轨道为圆轨道，基于题目中给出的2003年火星大冲的发生日期，计算2018年和2035年发生火星大冲的日期。<br />
<br />
'''2.3.''' 试解释计算得到的2018年火星大冲日期与实际日期间存在差别的原因。（请画草图说明）<br />
<br />
'''2.4.''' 2018年与2035年的火星大冲，哪一次更亮？（请画图证明你的答案）<br />
<br />
==解答==<br />
<br />
'''英文解答'''<br />
<br />
'''2.1.''' During the opposition, the planet is at a point approximately (up to an angular distance from the ecliptic at this moment) opposite to the Sun, that is, located where the Sun was six monthe ago. The Sun is in the constellation Capricorn on January 27.<br />
<br />
'''2.2.''' The oppositions are repeatde every synodic period. The synodic of Mars is:<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
Approximately 15 years passed from 2003 to 2018, and there are 15·T<sub>E</sub> / T<sub>S</sub> ≈ 7.02 synodic periods of Mars during this period. So, the opposition 2018 was to occur after 7 synodic periods earlier than August 28.<br />
<br />
The easiest way to calculate the date is as follows. For 7 synodic periods of Mars 7×779.93 ≈ 5459 days pass. From August 28 2003, to August 28, 2018, 365×15+4 = 5479 days pass. Thus, it turns out that in 2018 the opposition of Mars should be 20 days earlier than August 28, that is, August 8.<br />
<br />
【注：此段英文解答不全，详细过程在中文解答中补全】<br />
Thus, it turns out that in 2035 the opposition of Mars should be 11 days later than August 28, that is, Septmeber 8.<br />
<br />
'''2.3.''' The difference appears from face that in our calculations we take the average angular velocity of Mars in orbit, and in previous close to great oppositions, Mars is near the perihelion of its orbit, and its speed is larger. As a result, the difference in the angular velocities of the Earth and Mars is smaller. Therefore, to change the position of Mars relative to the Earth takes more time. That is why, the farther the opposition from the perihelion point of Mars orbit (which is at the ecliptic longitude that the Earth passes around August 28), the greater the error. Calculations in the approximation of circular orbits give about 1.5 times the underestimated interval between the day of the calculated opposition and the date on which the "greatest oppositions" take place (that is, August 28).<br />
<br />
'''2.4.''' The position of Mars during the opposition of 2035 is closer to the positiom of the perihelion of the orbit than in 2018, therefore in the opposition of 2035 Mars will be brighter.<br />
<br />
'''中文解答'''<br />
<br />
'''2.1.''' 摩羯座。冲日时，行星大约正好在太阳对侧（此时的行星位置距离黄道只有一些角距离）。所以，火星在太阳半年前所在的位置。1月27日太阳位于摩羯座。<br />
<br />
'''2.2.''' 行星与地球的每个[[会合周期]]会发生一次冲日。火星的会合为：<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
其中T<sub>S</sub>代表火星的会合周期，T<sub>E</sub>代表地球的公转周期，T<sub>M</sub>代表火星的公转周期。<br />
<br />
从2003年到2018年经过了大概15年，这大约相当于15×T<sub>E</sub> / T<sub>S</sub> ≈ 7.02个火星与地球的会合周期。所以2018年的火星冲日应发生在2003年火星冲日的7个会合周期之后，比8月28日更早的日子。<br />
<br />
以下是最简单的计算方法。火星的7个会合周期大约经过了7×779.93 ≈ 5459 天。从2003年8月28日到2018年8月28日，经过了365×15+4 = 5479 天。所以2018年的火星冲日发生的日期应该比8月28日早20天，也就是在8月8日。<br />
<br />
同理，从2003年到2035年经过了大概32年，这大约相当于32×T<sub>E</sub> / T<sub>S</sub> ≈ 14.99个火星会合周期。火星的15个会合周期大约经过了15×779.93 ≈ 11699 天。从2003年8月28日到2035年8月28日经过了365×32+8 = 11688 天。所以2035年的火星冲日发生的日期应该比8月28日晚11天，也就是在9月8日。<br />
<br />
'''2.3. '''造成冲日日期计算与实际上有差异的原因是在我们的计算中，假设了火星的公转角速度为一个恒定的平均角速度，但实际上在上一次2003年火星冲日的时候，火星更接近它的近日点，所以轨道角速度更大。因此，那时候地球和火星的轨道角速度差更小。火星相对于地球位置的改变需要花更多时间。这就是当大冲发生时火星（此时火星的黄经大约即地球在8月28日左右所在的黄经位置）离其轨道近日点越远，计算与实际大冲日期间的误差就越大。如果取近似圆轨道来计算2018年的大冲日期，所得的日期与2003年“最”大冲的8月28日间的日期差会被低估1.5倍。（译注：圆轨道计算该日期差为20天，而实际日期差为约31天）<br />
<br />
'''2.4.''' 2035年火星大冲时火星的在轨道上位置比2018年火星大冲时更靠近近日点。所以2035年的火星大冲更亮。</div>更方更正的物理https://www.astro-init.top/index.php?title=2018%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC2%E9%A2%98-%E7%81%AB%E6%98%9F%E5%A4%A7%E5%86%B2&diff=11662018年IAO理论低年组第2题-火星大冲2019-09-17T11:59:17Z<p>更方更正的物理：/* 中文题目 */</p>
<hr />
<div>==英文题目==<br />
'''Great oppositions of Mars.''' On the morning of July 27, 2018, Mars was in a great oposition (see attached ephemeris), The great oppositons of Mars (the situations when this planet is seen most brightly in comparison with the visibility during other oppositions) are repeated every 15 or 17 years. So the previous opposition (which is ales called the "greatest", since Marswas so bright only once in many centuries) was on August 28, 2003, and the next one will occur in 2035.<br />
<br />
'''2.1.''' Determine the constellation, in which Mars was during this year opposition.<br />
<br />
'''2.2.''' Considering the orbits of the Earth and Mars as circular, and based on the date of opposition 2003, calculate the dates the oppositions in 2018 and 2035 would occur.<br />
<br />
'''2.3''' Explain, why the difference appeared between the calculated and real dates of the opposition 2018. (Draw a figure that clearly demonstrates your explanation.)<br />
<br />
'''2.4.''' In which of the oppositions, 2018 or 2035 is Mars brightest? (The answer must be justified by drawing a figure).<br />
<br />
==中文题目==<br />
'''火星大冲''' 2018年7月27日上午，火星到达大冲位置（见附录星历表【注：暂无】）。火星大冲（大冲时行星的视亮度在所有冲日事件中是最亮的）每15或17年发生一次。所以上一次大冲（也被称为“最”大冲，因为这一次的火星是许多世纪以来最亮的）发生在2003年8月28日，下一次大冲将发生在2035年。<br />
<br />
'''2.1.''' 写出今年火星大冲时，火星位于哪个星座。<br />
<br />
'''2.2.''' 假设地球和火星的轨道为圆轨道，基于题目中给出的2003年火星大冲的发生日期，计算2018年和2035年发生火星大冲的日期。<br />
<br />
'''2.3.''' 试解释计算得到的2018年火星大冲日期与实际日期间存在差别的原因。（请画草图说明）<br />
<br />
'''2.4.''' 2018年与2035年的火星大冲，哪一次更亮？（请画图证明你的答案）<br />
<br />
==解答==<br />
<br />
'''英文解答'''<br />
<br />
'''2.1.''' During the opposition, the planet is at a point approximately (up to an angular distance from the ecliptic at this moment) opposite to the Sun, that is, located where the Sun was six monthe ago. The Sun is in the constellation Capricorn on January 27.<br />
<br />
'''2.2.''' The oppositions are repeatde every synodic period. The synodic of Mars is:<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
Approximately 15 years passed from 2003 to 2018, and there are 15·T<sub>E</sub> / T<sub>S</sub> ≈ 7.02 synodic periods of Mars during this period. So, the opposition 2018 was to occur after 7 synodic periods earlier than August 28.<br />
<br />
The easiest way to calculate the date is as follows. For 7 synodic periods of Mars 7×779.93 ≈ 5459 days pass. From August 28 2003, to August 28, 2018, 365×15+4 = 5479 days pass. Thus, it turns out that in 2018 the opposition of Mars should be 20 days earlier than August 28, that is, August 8.<br />
<br />
【注：此段英文解答不全，详细过程在中文解答中补全】<br />
Thus, it turns out that in 2035 the opposition of Mars should be 11 days later than August 28, that is, Septmeber 8.<br />
<br />
'''2.3.''' The difference appears from face that in our calculations we take the average angular velocity of Mars in orbit, and in previous close to great oppositions, Mars is near the perihelion of its orbit, and its speed is larger. As a result, the difference in the angular velocities of the Earth and Mars is smaller. Therefore, to change the position of Mars relative to the Earth takes more time. That is why, the farther the opposition from the perihelion point of Mars orbit (which is at the ecliptic longitude that the Earth passes around August 28), the greater the error. Calculations in the approximation of circular orbits give about 1.5 times the underestimated interval between the day of the calculated opposition and the date on which the "greatest oppositions" take place (that is, August 28).<br />
<br />
'''2.4.''' The position of Mars during the opposition of 2035 is closer to the positiom of the perihelion of the orbit than in 2018, therefore in the opposition of 2035 Mars will be brighter.<br />
<br />
'''中文解答'''<br />
<br />
'''2.1.''' 摩羯座。冲日时，行星大约正好在太阳对侧（此时的行星位置距离黄道只有一些角距离）。所以，火星在太阳半年前所在的位置。1月27日太阳位于摩羯座。<br />
<br />
'''2.2.''' 行星与地球的每个会合周期会发生一次冲日。火星的会合为：<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
其中T<sub>S</sub>代表火星的会合周期，T<sub>E</sub>代表地球的公转周期，T<sub>M</sub>代表火星的公转周期。<br />
<br />
从2003年到2018年经过了大概15年，这大约相当于15×T<sub>E</sub> / T<sub>S</sub> ≈ 7.02个火星与地球的会合周期。所以2018年的火星冲日应发生在2003年火星冲日的7个会合周期之后，比8月28日更早的日子。<br />
<br />
以下是最简单的计算方法。火星的7个会合周期大约经过了7×779.93 ≈ 5459 天。从2003年8月28日到2018年8月28日，经过了365×15+4 = 5479 天。所以2018年的火星冲日发生的日期应该比8月28日早20天，也就是在8月8日。<br />
<br />
同理，从2003年到2035年经过了大概32年，这大约相当于32×T<sub>E</sub> / T<sub>S</sub> ≈ 14.99个火星会合周期。火星的15个会合周期大约经过了15×779.93 ≈ 11699 天。从2003年8月28日到2035年8月28日经过了365×32+8 = 11688 天。所以2035年的火星冲日发生的日期应该比8月28日晚11天，也就是在9月8日。<br />
<br />
'''2.3. '''造成冲日日期计算与实际上有差异的原因是在我们的计算中，假设了火星的公转角速度为一个恒定的平均角速度，但实际上在上一次2003年火星冲日的时候，火星更接近它的近日点，所以轨道角速度更大。因此，那时候地球和火星的轨道角速度差更小。火星相对于地球位置的改变需要花更多时间。这就是当大冲发生时火星（此时火星的黄经大约即地球在8月28日左右所在的黄经位置）离其轨道近日点越远，计算与实际大冲日期间的误差就越大。如果取近似圆轨道来计算2018年的大冲日期，所得的日期与2003年“最”大冲的8月28日间的日期差会被低估1.5倍。（译注：圆轨道计算该日期差为20天，而实际日期差为约31天）<br />
<br />
'''2.4.''' 2035年火星大冲时火星的在轨道上位置比2018年火星大冲时更靠近近日点。所以2035年的火星大冲更亮。</div>更方更正的物理https://www.astro-init.top/index.php?title=Topic:V7ftow3jgql3co7n&topic_postId=v7ftow3jguj5ks5v&topic_revId=v7ftow3jguj5ks5v&action=single-viewTopic:V7ftow3jgql3co7n2019-09-16T14:57:56Z<span class="plainlinks"><a href="/index.php?title=%E7%94%A8%E6%88%B7:%E6%9B%B4%E6%96%B9%E6%9B%B4%E6%AD%A3%E7%9A%84%E7%89%A9%E7%90%86&action=edit&redlink=1" class="new mw-userlink" title="用户:更方更正的物理（页面不存在）"><bdi>更方更正的物理</bdi></a><span class="mw-usertoollinks">（<a href="/index.php?title=%E7%94%A8%E6%88%B7%E8%AE%A8%E8%AE%BA:%E6%9B%B4%E6%96%B9%E6%9B%B4%E6%AD%A3%E7%9A%84%E7%89%A9%E7%90%86&action=edit&redlink=1" class="new mw-usertoollinks-talk" title="用户讨论:更方更正的物理（页面不存在）">讨论</a> | <a href="/index.php?title=%E7%89%B9%E6%AE%8A:%E7%94%A8%E6%88%B7%E8%B4%A1%E7%8C%AE/%E6%9B%B4%E6%96%B9%E6%9B%B4%E6%AD%A3%E7%9A%84%E7%89%A9%E7%90%86" class="mw-usertoollinks-contribs" title="特殊:用户贡献/更方更正的物理">贡献</a>）</span><a rel="nofollow" class="external text" href="https://www.astro-init.top/index.php?title=Topic:V7ftow3jgql3co7n&topic_showPostId=v7ftow3jguj5ks5v#flow-post-v7ftow3jguj5ks5v">已评论</a>"关于2.2.的英文解答"的话题(<em>本题貌似2.2.英文解答有疏漏之处，翻译时凭个人理解予以计算和补全，希望能有大佬来检查一下是否正确_(:зゝ∠)_</em>)</span>更方更正的物理https://www.astro-init.top/index.php?title=2018%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC2%E9%A2%98-%E7%81%AB%E6%98%9F%E5%A4%A7%E5%86%B2&diff=10922018年IAO理论低年组第2题-火星大冲2019-09-16T14:56:17Z<p>更方更正的物理：/* 解答 */</p>
<hr />
<div>==英文题目==<br />
'''Great oppositions of Mars.''' On the morning of July 27, 2018, Mars was in a great oposition (see attached ephemeris), The great oppositons of Mars (the situations when this planet is seen most brightly in comparison with the visibility during other oppositions) are repeated every 15 or 17 years. So the previous opposition (which is ales called the "greatest", since Marswas so bright only once in many centuries) was on August 28, 2003, and the next one will occur in 2035.<br />
<br />
'''2.1.''' Determine the constellation, in which Mars was during this year opposition.<br />
<br />
'''2.2.''' Considering the orbits of the Earth and Mars as circular, and based on the date of opposition 2003, calculate the dates the oppositions in 2018 and 2035 would occur.<br />
<br />
'''2.3''' Explain, why the difference appeared between the calculated and real dates of the opposition 2018. (Draw a figure that clearly demonstrates your explanation.)<br />
<br />
'''2.4.''' In which of the oppositions, 2018 or 2035 is Mars brightest? (The answer must be justified by drawing a figure).<br />
<br />
== 中文题目 ==<br />
'''火星大冲''' 2018年7月27日上午，火星到达大冲位置（见附录星历表【注：暂无】）。火星大冲（大冲时行星的视亮度在所有冲日事件中是最亮的）每15或17年发生一次。所以上一次大冲（也被称为“最”大冲，因为这一次的火星是许多世纪以来最亮的）发生在2003年8月28日，下一次大冲将发生在2035年。<br />
<br />
'''2.1.''' 写出今年火星大冲时，火星位于哪个星座。<br />
<br />
'''2.2.''' 假设地球和火星的轨道为圆轨道，基于题目中给出的2003年火星大冲的发生日期，计算2018年和2035年发生火星大冲的日期。<br />
<br />
'''2.3.''' 试解释计算得到的2018年火星大冲日期与实际日期间存在差别的原因。（请画草图说明）<br />
<br />
'''2.4.''' 2018年与2035年的火星大冲，哪一次更亮？（请画图证明你的答案）<br />
<br />
==解答==<br />
<br />
'''英文解答'''<br />
<br />
'''2.1.''' During the opposition, the planet is at a point approximately (up to an angular distance from the ecliptic at this moment) opposite to the Sun, that is, located where the Sun was six monthe ago. The Sun is in the constellation Capricorn on January 27.<br />
<br />
'''2.2.''' The oppositions are repeatde every synodic period. The synodic of Mars is:<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
Approximately 15 years passed from 2003 to 2018, and there are 15·T<sub>E</sub> / T<sub>S</sub> ≈ 7.02 synodic periods of Mars during this period. So, the opposition 2018 was to occur after 7 synodic periods earlier than August 28.<br />
<br />
The easiest way to calculate the date is as follows. For 7 synodic periods of Mars 7×779.93 ≈ 5459 days pass. From August 28 2003, to August 28, 2018, 365×15+4 = 5479 days pass. Thus, it turns out that in 2018 the opposition of Mars should be 20 days earlier than August 28, that is, August 8.<br />
<br />
【注：此段英文解答不全，详细过程在中文解答中补全】<br />
Thus, it turns out that in 2035 the opposition of Mars should be 11 days later than August 28, that is, Septmeber 8.<br />
<br />
'''2.3.''' The difference appears from face that in our calculations we take the average angular velocity of Mars in orbit, and in previous close to great oppositions, Mars is near the perihelion of its orbit, and its speed is larger. As a result, the difference in the angular velocities of the Earth and Mars is smaller. Therefore, to change the positoin of Mars relative to the Earth takes more time. That is why, the farther the opposition from the perihelion point of Mars orbit (which is at the ecliptic longitude that the Earth passes around August 28), the greater the error. Calculations in the approximation of circular orbits give about 1.5 times the underestimated interval between the day of the calculated opposition and the date on which the "greatest oppositions" take place (that is, August 28).<br />
<br />
'''2.4.''' The position of Mars during the opposition of 2035 is closer to the positiom of the perihelion of the orbit than in 2018, therefore in the opposition of 2035 Mars will be brighter.<br />
<br />
'''中文解答'''<br />
<br />
'''2.1.''' 摩羯座。冲日时，行星大约正好在太阳对侧（此时的行星位置距离黄道只有一些角距离）。所以，火星在太阳半年前所在的位置。1月27日太阳位于摩羯座。<br />
<br />
'''2.2.''' 行星与地球的每个会合周期会发生一次冲日。火星的会合为：<br />
<br />
$$T_S = 1/(1/T_E - 1/T_M),$$<br />
<br />
$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
<br />
其中T<sub>S</sub>代表火星的会合周期，T<sub>E</sub>代表地球的公转周期，T<sub>M</sub>代表火星的公转周期。<br />
<br />
从2003年到2018年经过了大概15年，这大约相当于15×T<sub>E</sub> / T<sub>S</sub> ≈ 7.02个火星与地球的会合周期。所以2018年的火星冲日应发生在2003年火星冲日的7个会合周期之后，比8月28日更早的日子。<br />
<br />
以下是最简单的计算方法。火星的7个会合周期大约经过了7×779.93 ≈ 5459 天。从2003年8月28日到2018年8月28日，经过了365×15+4 = 5479 天。所以2018年的火星冲日发生的日期应该比8月28日早20天，也就是在8月8日。<br />
<br />
同理，从2003年到2035年经过了大概32年，这大约相当于32×T<sub>E</sub> / T<sub>S</sub> ≈ 14.99个火星会合周期。火星的15个会合周期大约经过了15×779.93 ≈ 11699 天。从2003年8月28日到2035年8月28日经过了365×32+8 = 11688。所以2035年的火星冲日发生的日期应该比8月28日晚11天，也就是在9月8日。<br />
<br />
'''2.3. '''</div>更方更正的物理https://www.astro-init.top/index.php?title=2018%E5%B9%B4IAO%E7%90%86%E8%AE%BA%E4%BD%8E%E5%B9%B4%E7%BB%84%E7%AC%AC2%E9%A2%98-%E7%81%AB%E6%98%9F%E5%A4%A7%E5%86%B2&diff=10512018年IAO理论低年组第2题-火星大冲2019-09-16T12:12:44Z<p>更方更正的物理：补充题目的中文翻译。</p>
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<div>==英文题目==<br />
'''Great oppositions of Mars.''' On the morning of July 27, 2018, Mars was in a great oposition (see attached ephemeris), The great oppositons of Mars (the situations when this planet is seen most brightly in comparison with the visibility during other oppositions) are repeated every 15 or 17 years. So the previous opposition (which is ales called the "greatest", since Marswas so bright only once in many centuries) was on August 28, 2003, and the next one will occur in 2035.<br />
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'''2.1.''' Determine the constellation, in which Mars was during this year opposition.<br />
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'''2.2.''' Considering the orbits of the Earth and Mars as circular, and based on the date of opposition 2003, calculate the dates the oppositions in 2018 and 2035 would occur.<br />
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'''2.3''' Explain, why the difference appeared between the calculated and real dates of the opposition 2018. (Draw a figure that clearly demonstrates your explanation.)<br />
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'''2.4.''' In which of the oppositions, 2018 or 2035 is Mars brightest? (The answer must be justified by drawing a figure).<br />
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== 中文题目 ==<br />
'''火星大冲''' 2018年7月27日上午，火星到达大冲位置（见附录星历表【注：暂无】）。火星大冲（大冲时行星的视亮度在所有冲日事件中是最亮的）每15或17年发生一次。所以上一次大冲（也被称为“最”大冲，因为这一次的火星是许多世纪以来最亮的）发生在2003年8月28日，下一次大冲将发生在2035年。<br />
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'''2.1.''' 写出今年火星大冲时，火星位于哪个星座。<br />
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'''2.2.''' 假设地球和火星的轨道为圆轨道，基于题目中给出的2003年火星大冲的发生日期，计算2018年和2035年发生火星大冲的日期。<br />
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'''2.3.''' 试解释计算得到的2018年火星大冲日期与实际日期间存在差别的原因。（请画草图说明）<br />
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'''2.4.''' 2018年与2035年的火星大冲，哪一次更亮？（请画图证明你的答案）<br />
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==解答==<br />
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'''英文解答'''<br />
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'''2.1.''' During the opposition, the planet is at a point approximately (up to an angular distance form the ecliptic at this moment) opposite to the Sun, that is, located where the Sun was six monthe ago. The Sun is in the constellation Campricorn on January 27.<br />
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'''2.2.''' The oppositions are repeatde every synodic period. The eyniduc of Mars is:<br />
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$$T_S = 1/(1/T_E - 1/T_M),$$<br />
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$$T_S = 1/(1/365.256 - 1/686.98)\quad 779.93\ days,$$<br />
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Approximately 15 years passed from 2003 to 2018, and there are 15·T<sub>E</sub> / T<sub>S</sub> ≈ 7.02 synodic periods of Mars during this period. So, the opposition 2018 was to occur after 7 synodic periods earlier on August 28.<br />
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The easiest way to calculate the date is as follows. For 7 synodic periods of Mars 7×779.93 ≈ 5459 days pass. From August 28 2003, to Augusr 28, 2018, 365×15+4 = 5479 days pass. Thus, it turns out that in 2035 the opposition of Mars should be 11 days later than ugust 28, that is, Septmeber 8.<br />
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'''2.3.''' The difference appears from face that in our calculations we take the avrage angular velocitu of Mars in orbut, and iin perious close to great oppositions, Mars is near the perihelion of its orbit, and its speed is larger. As a result, the difference in the angular velocities of the Earth and Mars is smaller. Therefore, to change the positoin of Mars relative to the Earth takes more time. That is why, the farther the opposition from the perihelion point of Mars orbit (which is at the ecliptic longitude that the Earth passes aroune August 28), the greater the error. Calculations inthe approximation of circular orbits give about 1.5 times the underestimated interval between the dae of the calculated opposition and the date on which the "greatest oppositions" take place (that is, August 28).<br />
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'''2.4.''' The position of Mars during the opposition of 2035 is closer to the positiom of the perihelion of the orbit than in 2018, therefore in the opposition of 2035 Mars will be brighter.</div>更方更正的物理