2015IOAA理论短问题第5题-卫星轨道

英文题目

An observer is trying to determine an approximate value of the orbital eccentricity of a man-made satellite. When the satellite was at apogee, it was observed to have moved by $$∆𝜃_{1}$$ = 2′44" in a short time. When the radius vector connecting Earth and the satellite is perpendicular to the major axis (true anomaly is equal to 90o), within the same duration of time, it was observed to have moved by $$∆𝜃_{2}$$ = 21′17". Assume that the observer is located at the center of the Earth. Find an approximate value of the eccentricity of the satellite’s orbit.

中文翻译

一个观察者正在尝试确定一颗人造卫星轨道的偏心率。当卫星位于远地点时，观察到它在短时间内移动了 $$∆𝜃_{1}$$ = 2′44"。当连接地球和卫星的半径向量与长轴垂直（真近点角为90°）时，在相同时间内，观察到它移动了 $$∆𝜃_{2}$$ = 21′17"。假设观察者位于地球中心。求卫星轨道离心率的近似值。

解答

官方答案

✔	官方答案

Identify expression for $$r(\theta)$$. We know that $$r(\theta) = \frac{a(1 - e^2)}{1 - e \cos \theta} $$.Hence

\[r(0) = \frac{a(1 - e^2)}{1 - e} = a(1 + e)\]

\[r(\frac{\pi}{2}) = \frac{a(1 - e^2)}{1 - e \times 0} = a(1 - e^2)\]

Estimate area of sectors using area of triangles. Estimate the area of swept out sectors as the area of sectors of circles \[A(S_1) \approx 0.5 \times \Delta \theta_1 \times (r(0))^2 = \Delta \theta_1 (a(1 + e))^2\] \[A(S_2) \approx 0.5 \times \Delta \theta_2 \times (r(\frac{\pi}{2}))^2 = \Delta \theta_2 (a(1 - e^2))^2\]

Use Kepler's second law to find a relation between the ratio $$\frac{\Delta \theta_1}{\Delta \theta_2}$$ and the eccentricity $$e$$. Use Kepler's second law to obtain that $$A(S_1) = A(S_2)$$ \[\Delta \theta_1 \times (a(1 + e))^2 = \Delta \theta_2 \times (a(1 - e^2))^2\]

Thus, \[\frac{\Delta \theta_1}{\Delta \theta_2} = (\frac{1 - e^2}{1 + e})^2 = (1 - e)^2 = \frac{2' 44''}{21' 17''} \approx 0.12843\]

4 Obtain an estimate value of the eccentricity. Thus, the eccentricity is $$e \approx 1 - \sqrt{0.12843} = 0.64$$

用户答案

❤	用户答案 贡献者：RT.x

设卫星的远地点为A点,对应物理量角标为$$A$$，卫星真近点角为90°时B点，对应物理量角标为$$B$$。椭圆轨道的半长轴为$$a$$，半短轴为$$b$$，半焦距为$$c$$，离心率为$$e$$。

卫星位于A点时，与地球（中心天体，视作质点）的距离:

$$r_A = a + c$$

卫星位于B点时，与地球的距离:

$$r_B = \frac{b^2}{a}$$（椭圆通径$$d = \frac{2b^2}{a}$$的一半）

由卫星的角动量守恒可得：

$$L_A = L_B$$

根据角动量的表达式：

$$L = m\omega r^2$$

可得:

$$m\omega_A r_A = m\omega_B r_B$$

约去质量$$m$$，将上式变形，得：

$$\frac{\omega_A}{\omega_B} = \frac{r_B^2}{r_A^2}$$

则有：

$$\frac{r_B}{r_A} = \sqrt{\frac{\omega_A}{\omega_B}}$$

在观测的短时间t内，有如下关系：

$$\frac{\omega_A}{\omega_B} = \frac{∆𝜃_{1}}{∆𝜃_{2}} = \frac{2′44"}{21′17"}$$

所以

$$\frac{r_B}{r_A} = \sqrt{\frac{2′44"}{21′17"}} \approx 0.358$$

接下来，我们运用椭圆的相关知识，用含$$a+c$$和$$\frac{b^2}{a}$$的式子表示$$e$$（椭圆离心率，$$e = \frac{c}{a}$$）

$$r_B = \frac{b^2}{a} = \frac{a^2-c^2}{a} = \frac{(a+c)(a-c)}{a} = (a+c)(1-\frac{c}{a})$$

移相，得：

$$e = \frac{c}{a} = 1-\frac{\frac{b^2}{a}}{a+c}$$

把$$r_A$$和$$r_B$$代入，得：

$$e = 1-\frac{r_A}{r_B} = 0.642$$

故卫星轨道离心率的近似值为$$0.642$$