{{由ai生成}}
==英文题目==

'''T10. Greatest Eclipse (75 points)'''

The greatest eclipse is defined as the instant when the axis of the Moon's shadow cone gets closest to the centre of the Earth in a solar eclipse. This problem explores the geometry of this phenomenon, using the solar eclipse of 29$^\text{th}$ May 1919 as an example, as it has great historical significance for being the first time astronomers were able to observationally verify general relativity. One of the scientific expeditions to observe this eclipse took place in the Brazilian city of Sobral.

The two following tables show the Cartesian and spherical coordinates of the Sun and the Moon at the time of the greatest eclipse. The system used for these coordinates is right-handed and has the origin at the centre of the Earth, the positive $x$-axis pointing towards the Greenwich meridian, and the positive $z$-axis pointing towards the North Pole. For the rest of this problem, this will be referred to as System I.

Spherical Coordinates:

| | Centre of the Sun | Centre of the Moon |
|---|---|---|
| Radial Distance ($r$) | $1.516 \times 10^{11}$ m | $3.589 \times 10^8$ m |
| Polar Angle ($\theta$) | $68^{\circ}29'44.1''$ | $68^{\circ}47'41.6''$ |
| Azimuthal Angle ($\varphi$) | $-1^h11^{m}28.2^s$ | $-1^h11^{m}22.9^s$ |

Cartesian Coordinates:

| | Centre of the Sun | Centre of the Moon |
|---|---|---|
| $x$ | $1.342 \times 10^{11}$ m | $3.185 \times 10^8$ m |
| $y$ | $-4.327 \times 10^{10}$ m | $-1.025 \times 10^8$ m |
| $z$ | $5.557 \times 10^{10}$ m | $1.298 \times 10^8$ m |

For this problem, assume that the Earth is a perfect sphere.

Note: The spherical coordinates of a point $P$ are defined as follows:
- Radial distance ($r$): distance between the origin ($O$) and $P$ (range: $r \geq 0$).
- Polar angle ($\theta$): angle between the positive $z$-axis and the line segment $OP$ (range: $0^{\circ} \leq \theta \leq 180^{\circ}$).
- Azimuthal angle ($\varphi$): angle between the positive $x$-axis and the projection of the line segment $OP$ onto the $xy$-plane (range: $-12^h \leq \varphi < 12^h$).

===== Part I: Geographic Coordinates (25 points) =====

(a) (3 points) Determine the declination of the Sun and the Moon during the greatest eclipse for a geocentric observer.

(b) (3 points) Determine the right ascension of the Sun and the Moon at the time of the greatest eclipse for a geocentric observer. The local sidereal time at Greenwich at that same moment was $5^h32^m35.5^s$.

(c) (4 points) Find a unit vector that indicates the direction of the axis of the Moon’s shadow cone. This vector should point from the Moon to the vicinity of the centre of the Earth.

(d) (15 points) Determine the latitude and the longitude of the point where the axis of the Moon's shadow cone crosses the surface of the Earth during the greatest eclipse.

===== Part II: Duration of the Totality (50 points) =====

Precisely determining the duration of totality of a solar eclipse involves complex calculations that would be beyond the scope of this problem. However, it is possible to obtain a reasonable approximation for this value using the two following assumptions:
- The size of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location.
- The velocity of the umbra on the surface of the Earth remains roughly constant throughout totality at a given location.

(e) (10 points) Estimate the radius of the umbra during the greatest eclipse. In order to simplify the calculations, assume that the umbra is small enough that it can be considered approximately flat and that the axis of the Moon's shadow cone is extremely close to the centre of the Earth during the greatest eclipse.

(f) (3 points) Calculate the velocity of the Earth's rotation at the latitude of the centre of the umbra.

(g) (4 points) Determine the orbital velocity of the Moon at the instant of the greatest eclipse. Neglect the changes in the semi-major axis of the Moon’s orbit.

For the remaining items of this problem, assume that the tangential velocity of the Moon is roughly the same as the orbital velocity and neglect its radial component.

(h) (14 points) Using System II, determine the velocity vector of the Moon during the greatest eclipse. Note that the intersection between the Celestial Equator and the lunar orbit that is closer to the position of the eclipse has a right ascension of $23^{h}07^{m}59.2^s$.

(i) (10 points) Write the velocity vector of the Moon in System III. Note that in System I, the azimuthal angle difference between the positions of the origins $O_{\text{II}}$ and $O_{\text{III}}$ is negligible, so you should only take into account the difference in the polar angles.

(j) (6 points) Calculate the speed of the centre of the umbra along the surface of the Earth at the instant of the greatest eclipse.

(k) (3 points) Estimate the duration of the totality of the eclipse at the location with the coordinates found in part (d).

[[分类:日月食]]
[[分类:天体力学]]
[[分类:由ai生成]]

==中文翻译==

'''T10. 最大食分（75分）'''

最大食分定义为月球本影锥轴最接近地球中心的时刻。本问题以1919年5月29日日食为例探讨该现象的几何学，此次日食因首次通过观测验证广义相对论而具有重大历史意义。科学考察队之一在巴西索布拉尔市观测了此次日食。

下表展示了最大食分时刻太阳和月球的球面坐标与笛卡尔坐标。坐标系为右手系，原点在地球中心，$x$轴正向指向格林尼治子午线，$z$轴正向指向北极点。下文中称此坐标系为**系统I**。

'''球面坐标系：'''
{| class="wikitable"
|-
! !! 太阳中心 !! 月球中心
|-
! 径向距离（$r$）
| $1.516 \times 10^{11}$ 米 || $3.589 \times 10^{8}$ 米
|-
! 极角（$\theta$）
| $68^\circ29'44.1''$ || $68^\circ47'41.6''$
|-
! 方位角（$\varphi$）
| $-1^h11^m28.2^s$ || $-1^h11^m22.9^s$
|}

'''笛卡尔坐标系：'''
{| class="wikitable"
|-
! !! 太阳中心 !! 月球中心
|-
! $x$
| $1.342 \times 10^{11}$ 米 || $3.185 \times 10^{8}$ 米
|-
! $y$
| $-4.327 \times 10^{10}$ 米 || $-1.025 \times 10^{8}$ 米
|-
! $z$
| $5.557 \times 10^{10}$ 米 || $1.298 \times 10^{8}$ 米
|}

==官方解答==

=== Part I: 地理坐标 ===
(a) 太阳和月球的赤纬：
$$$\delta = 90^\circ - \theta$$$
$$$\delta_\odot = 21^\circ30'15.9''$$$
$$$\delta_{Moon} = 21^\circ12'18.4''$$$

(b) 太阳和月球的赤经（格林尼治恒星时为$5^h32^m35.5^s$）：
$$$\alpha = \varphi + LST_{Greenwich}$$$
$$$\alpha_\odot = 4^h21^m7.3^s$$$
$$$\alpha_{Moon} = 4^h21^m12.6^s$$$

(c) 月球本影轴方向单位向量：
$$$\vec{u} = \left\langle -0.8855, 0.2855, -0.3666 \right\rangle$$$

(d) 本影轴与地球表面交点坐标：
通过方程$$|\vec{M} + k\vec{u}| = R_\oplus$$解得：
$$$\phi = 4^\circ22'N$$$
$$$\lambda = 16^\circ42'W$$$

=== Part II: 全食持续时间 ===
(e) 本影半径估算：
$$$r_{umbra} = 1.196 \times 10^5 \, \text{m}$$$

(f) 地球自转线速度：
$$$v_{rot} = 463.7 \, \text{m/s}$$$

(g) 月球轨道速度（忽略半长轴变化）：
$$$v_{Moon} = 1088 \, \text{m/s}$$$

(h) 系统II中月球速度向量：
$$$\mathbf{v_{II}} = \begin{bmatrix} 1085 \, \text{m/s} \\ 81.25 \, \text{m/s} \\ 0 \end{bmatrix}$$$

(i) 系统III中月球速度向量：
$$$\mathbf{v_{III}} = \begin{bmatrix} 1085 \, \text{m/s} \\ 77.76 \, \text{m/s} \\ -23.54 \, \text{m/s} \end{bmatrix}$$$

(j) 本影中心地表移动速度：
$$$v_{umbra} = \sqrt{621.4^2 + 77.76^2} = 626.3 \, \text{m/s}$$$

(k) 全食持续时间：
$$$\Delta t = \frac{2r_{umbra}}{v_{umbra}} = 6\,\text{分钟}\,21.4\,\text{秒}$$$

[[分类:日月食]]
[[分类:球面三角]]
[[分类:由ai生成]]

2025-03-06T14:54:57Z

Quan787：创建页面，内容为“{{由ai生成}} ==英文题目== '''T3. Asteroid (10 points)''' A peculiar asteroid of mass, $m$, was spotted at a distance, $d$, from a star with mass, $M$.…”

{{由ai生成}}
==英文题目==

'''T3. Asteroid (10 points)'''

A peculiar asteroid of mass, $m$, was spotted at a distance, $d$, from a star with mass, $M$. The magnitude of the asteroid’s velocity at the time of the observation was $v = \sqrt{\frac{GM}{d}}$, where $G$ is the universal gravitational constant. The distance $d$ is much larger than the radius of the star.

For both of the following items, express your answers in terms of $M$, $d$, and physical or mathematical constants.

(a) (8 points) If the asteroid is initially moving exactly towards the star, how long will it take for it to collide with the star?

(b) (2 points) If the asteroid is instead initially moving exactly away from the star, how long will it now take for it to collide with the star?

==中文翻译==

'''T3. 小行星（10分）'''

一颗质量为$m$的特殊小行星在距离质量为$M$的恒星$d$处被发现。观测时小行星的速度大小为$v = \sqrt{\frac{GM}{d}}$，其中$G$为万有引力常数。距离$d$远大于恒星半径。

对以下两问，答案需用$M$、$d$和物理/数学常数表示。

(a) (8分) 若小行星初始直接朝向恒星运动，它需要多长时间会与恒星碰撞？

(b) (2分) 若小行星初始直接背离恒星运动，它需要多长时间会与恒星碰撞？

== 官方解答 ==

(a) 该情况下小行星将沿退化椭圆轨道运动。根据开普勒第三定律：

$$T = 2\pi d \sqrt{\frac{d}{GM}}$$

利用开普勒第二定律计算碰撞时间：

$$\Delta t = \left( \frac{\pi}{2} - 1 \right) d \sqrt{\frac{d}{GM}}$$

(b) 当小行星初始背离恒星运动时，需遍历轨道面积II和III后返回：

$$\Delta t = \frac{\pi}{2} d \sqrt{\frac{d}{GM}}$$

[[分类:天体力学]]
[[分类:由ai生成]]

2024年IOAA理论第2题-星系团

2025-03-05T16:58:21Z

Quan787：

==英文题目==
{{由ai生成}}

'''T2. Galaxy Cluster (10 points)'''

An astrophysical survey mapped all the galaxies in a small region of the sky, of angular diameter $\Delta \theta = 0.01$ rad, where many galaxies seemed to be concentrated around the central area of the image. When the positions and redshifts of all the galaxies in this cluster were measured, an interesting distribution emerged, which is shown in the plot below.

Using these observations, estimate the total mass of the galaxy cluster and express your answer in solar masses. Assume that this galaxy cluster is in dynamical equilibrium, with a root-mean-square redshift dispersion $\sigma_z = \sqrt{\langle (z - 0.7)^2 \rangle} = 0.0005$. Feel free to make reasonable approximations when considering the average velocities, masses, and spatial distribution of the galaxies.

Consider that the distance to $\bar{z} = 0.7$ in the standard cosmological model is $D_A = 1500$ Mpc. Ignore cosmological effects on the distance.

[[文件:IOAA2024T2-1.jpg|缩略图|居中|图一星系红移分布示意图]]

==中文翻译==
{{由ai生成}}

'''T2. 星系团（10分）'''

一项天体物理巡天观测了天空中一个小区域（角直径$\Delta \theta = 0.01$弧度）内的所有星系，发现许多星系集中在图像的中心区域。测量该星系团中所有星系的位置和红移后，呈现出有趣的分布（如下图所示）。

利用这些观测数据，估算该星系团的总质量，并以太阳质量为单位表示答案。假设该星系团处于动力学平衡状态，红移弥散的均方根值为$\sigma_z = \sqrt{\langle (z - 0.7)^2 \rangle} = 0.0005$。在考虑星系的平均速度、质量和空间分布时，可作合理近似。

已知标准宇宙学模型中$\bar{z} = 0.7$对应的共动距离为$D_A = 1500$ Mpc，忽略距离计算中的宇宙学效应。

[[文件:IOAA2024T2-1.jpg|缩略图|居中|图一星系红移分布示意图]]

==官方解答==
{{由ai生成}}

根据维里定理，总引力势能$U$和总动能$K$满足：
$$U + 2K = 0$$

假设星系团中有$N$个质量为$m_i$的星系，其相对于星系团中心的运动速度为$\vec{u}_i$，则动能为：
$$K = \frac{1}{2}M_T \sigma_v^2$$
其中$\sigma_v = \sqrt{\langle u_i^2 \rangle}$为速度弥散。红移弥散对应的径向速度弥散为：
$$\sigma_{v_r} = c \cdot \sigma_z = 2.998 \times 10^8 \, \text{m/s} \times 0.0005 = 1.499 \times 10^5 \, \text{m/s}$$

假设三维速度各向同性，则$\sigma_v^2 = 3\sigma_{v_r}^2$。引力势能近似为均匀球体的结合能：
$$U = -\frac{3}{5} \frac{GM_T^2}{R}$$
其中星系团半径：
$$R = \frac{D_A \cdot \Delta \theta}{2} = \frac{1500 \times 10^6 \times 206265 \times 1.496 \times 10^{11} \times 10^{-2}}{2} = 2.3143 \times 10^{23} \, \text{m}$$

代入维里定理：
$$\frac{5R\sigma_{v_r}^2}{G} = M_T$$
计算得：
$$M_T = \frac{5 \times 2.314 \times 10^{23} \times (1.499 \times 10^5)^2}{6.67 \times 10^{-11}} \, \text{kg} \approx 3.9 \times 10^{44} \, \text{kg} = 2.0 \times 10^{14} \, M_\odot$$

[[分类:宇宙学]]
[[分类:天体力学]]
[[分类:由ai生成]]

2024年IOAA理论第2题-星系团

2025-03-05T16:50:00Z

Quan787：

==英文题目==
{{由ai生成}}

'''T2. Galaxy Cluster (10 points)'''

An astrophysical survey mapped all the galaxies in a small region of the sky, of angular diameter $\Delta \theta = 0.01$ rad, where many galaxies seemed to be concentrated around the central area of the image. When the positions and redshifts of all the galaxies in this cluster were measured, an interesting distribution emerged, which is shown in the plot below.

Using these observations, estimate the total mass of the galaxy cluster and express your answer in solar masses. Assume that this galaxy cluster is in dynamical equilibrium, with a root-mean-square redshift dispersion $\sigma_z = \sqrt{\langle (z - 0.7)^2 \rangle} = 0.0005$. Feel free to make reasonable approximations when considering the average velocities, masses, and spatial distribution of the galaxies.

Consider that the distance to $\bar{z} = 0.7$ in the standard cosmological model is $D_A = 1500$ Mpc. Ignore cosmological effects on the distance.

[[文件:IOAA2024T2-1.jpg|缩略图|居中|图一星系红移分布示意图]]

==中文翻译==
{{由ai生成}}

'''T2. 星系团（10分）'''

一项天体物理巡天观测了天空中一个小区域（角直径$\Delta \theta = 0.01$弧度）内的所有星系，发现许多星系集中在图像的中心区域。测量该星系团中所有星系的位置和红移后，呈现出有趣的分布（如下图所示）。

利用这些观测数据，估算该星系团的总质量，并以太阳质量为单位表示答案。假设该星系团处于动力学平衡状态，红移弥散的均方根值为$\sigma_z = \sqrt{\langle (z - 0.7)^2 \rangle} = 0.0005$。在考虑星系的平均速度、质量和空间分布时，可作合理近似。

已知标准宇宙学模型中$\bar{z} = 0.7$对应的共动距离为$D_A = 1500$ Mpc，忽略距离计算中的宇宙学效应。

[[文件:IOAA2024T2-1.jpg|缩略图|居中|图一星系红移分布示意图]]

==官方解答==
{{由ai生成}}

根据维里定理，总引力势能$U$和动能$K$满足：
$$ U + 2K = 0 $$

假设星系团中所有星系平均质量为$\langle m \rangle = M_T/N$，动能可近似为：
$$ K = \frac{1}{2} M_T \sigma_v^2 $$

其中径向速度弥散$\sigma_{v_r} = c \cdot \sigma_z = 1.499 \times 10^5 \, \text{m/s}$，三维速度弥散满足$\sigma_v^2 = 3\sigma_{v_r}^2$。

引力势能采用均匀球体近似：
$$ U = -\frac{3}{5} \frac{G M_T^2}{R} $$
其中星系团半径：
$$ R = \frac{D_A \cdot \Delta \theta}{2} = 2.314 \times 10^{23} \, \text{m} $$

联立维里定理得质量公式：
$$ M_T = \frac{5 R \sigma_{v_r}^2}{G} $$
代入数据计算得：
$$ M_T \approx 2.0 \times 10^{14} \, M_\odot $$

[[分类:宇宙学]]
[[分类:天体力学]]
[[分类:由ai生成]]

2024年IOAA理论第2题-星系团

2025-03-05T16:17:23Z

Quan787：

2024年IOAA理论第2题-星系团质量估计

==英文题目==

{{由ai生成}}

'''T2. Galaxy Cluster (10 points)'''

An astrophysical survey mapped all the galaxies in a small region of the sky, of angular diameter $\Delta\theta=0.01$ rad, where many galaxies seemed to be concentrated around the central area of the image. When the positions and redshifts of all the galaxies in this cluster were measured, an interesting distribution emerged, which is shown in the plot below.

Using these observations, estimate the total mass of the galaxy cluster and express your answer in solar masses. Assume that this galaxy cluster is in dynamical equilibrium, with a root-mean-square redshift dispersion $\sigma_{z}=\sqrt{\langle(z-0.7)^{2}\rangle}=0.0005$. Feel free to make reasonable approximations when considering the average velocities, masses, and spatial distribution of the galaxies.

Consider that the distance to $\bar{z}=0.7$ in the standard cosmological model is $D_{A}=1500$ Mpc. Ignore cosmological effects on the distance.

==中文翻译==

{{由ai生成}}

'''第2题星系团质量估计（10分）'''

某天体物理巡天项目对天空小区域（角直径$\Delta\theta=0.01$弧度）内所有星系进行了测绘，发现许多星系集中在图像中心区域。通过测量该星团中所有星系的位置和红移，得到下图所示的分布。

假设该星系团处于动力学平衡状态，其红移分布的均方根弥散为$\sigma_{z}=\sqrt{\langle(z-0.7)^{2}\rangle}=0.0005$。请估算该星系团的总质量，并以太阳质量为单位表示。在考虑星系平均速度、质量和空间分布时可采用合理近似。

已知标准宇宙学模型中红移$\bar{z}=0.7$对应的角直径距离$D_{A}=1500$ Mpc。忽略宇宙学效应对距离的影响。

== 官方解答 ==

{{由ai生成}}

根据维里定理，总引力势能$U$和动能$K$满足：
$$U + 2K = 0$$

考虑星系平均质量$\langle m\rangle = M_T/N$，动能可近似为：
$$K = \frac{1}{2}M_T\sigma_v^2$$

红移弥散对应的径向速度弥散：
$$\sigma_{v_r} = c\cdot\sigma_z = 2.998 \times 10^8 \, \text{m/s} \times 0.0005 = 1.499 \times 10^5 \, \text{m/s}$$

假设速度各向同性，三维速度弥散：
$$\sigma_v^2 = 3\sigma_{v_r}^2$$

星系团半径估算：
$$R = \frac{D_A \cdot \Delta\theta}{2} = \frac{1500 \times 10^6 \times 206265 \times 1.496 \times 10^{11} \times 10^{-2}}{2} = 2.3143 \times 10^{23} \, \text{m}$$

均匀球体引力势能：
$$U = -\frac{3}{5}\frac{GM_T^2}{R}$$

代入维里定理得质量公式：
$$M_T = \frac{5R\sigma_{v_r}^2}{G} = \frac{5 \times 2.314 \times 10^{23} \times (1.499 \times 10^5)^2}{6.67 \times 10^{-11}} \, \text{kg}$$

计算结果：
$$M_T \approx 2.0 \times 10^{14} \, M_\odot$$

[[分类:宇宙学]]
[[分类:天体力学]]
[[分类:由ai生成]]

分类:由ai生成

2025-03-05T16:01:02Z

Quan787：创建页面，内容为“以下是由ai从pdf中提取题目和答案生成的页面。目前ai的工作只限于翻译题目和官方答案，并不会给出自己的答案。如果你认…”

以下是由ai从pdf中提取题目和答案生成的页面。目前ai的工作只限于翻译题目和官方答案，并不会给出自己的答案。如果你认为ai的翻译或格式有问题，请在评论区提出或联系管理员。

经过用户校对后，可以去掉页面上的这个分类标记。

2022-05-06T20:27:03Z

Quan787：

<table bgcolor='Olive' border=1 width=100%>
<tr >
<td width=5% >
<center><big>❤</big></center>
</td>
<td width=75%>
'''用户答案'''
贡献者：{{{1|请查询编辑记录}}}
</td>
<tr>
</table>

模板:用户答案

2022-05-06T20:26:16Z

Quan787：