2018年IAO理论低年组第4题-科伦坡地球静止轨道卫星

英文题目

Colombo. Geostationary satellite

4.1. At what minimun zenith distance can a geostationary satellite be observed from Colombo? Suppose that such a satellite is observed as a 2^m star in the night sky.

4.2. How long during a day(24^h) can we see this satellite with the naked eye (in a clear sky)?

4.3. Estimate teh size of the satellite, considering it a polished metallic sphere.

中文题目

科伦坡 地球静止卫星

4.1 从科伦坡可以观测到的地球静止卫星的天顶距最小为多少？ 假设这样的卫星与夜空中的2等星光度相等。

4.2 在一天（24小时）中我们可以用肉眼在晴朗的天空中看到这颗卫星多长时间？

4.3 估计卫星的大小，考虑它是抛光金属球（反射率为100%）。

英文解答

Geostationary satellite

4.1.Geostationary satellites are located above the same equator point. Terefore, each actual stellite is always visible at one point for a stationary observer on Earth.

It should be noted that the geostationary satellite should not be confused with a geosynchronous one, the orbit with a nonzero eccentricity or slight inclination with respect to the equator. Such a satellite in the sky describes loops and eights for observers on the Earth.

A geostationary satellite has a circular queatorial orbit with diameter R under the conditions:

$$\omega^{2}R_{st}\ =\ GM/R_{st}$$,

$$R_{st}^3\ =\ GM/\omega^2\ =\ GMT^2/4\pi^2$$,

where G is the gravitation constant, M is the mass of the Earth, T is the period of the Earth's revolution around its axis (23^h56^m04^s). The calculations give the result

$$R_{st}\ =\ 42\ 160\ km$$.

It is obvious that of all the possible geotaionary satellites, the smallest zenitn distance has a satellite located on the meridian of the observer. From the drawing we see that the zenith distance is z = φ + Ψ, where φ is the latitude,and Ψ is the angle at which the "equatorial pane - Colombo" segment is visble from the satellite. We can use the latitude os the plane in Colombo, which is indicated as"seafront in the center", 06°54′).

$$tan\psi\ =\ rsin\psi/(R-rcos\psi)$$,

The calculations give

$$\psi\ \approx\ 1°14′$$.

$$z\ \approx\ 08°08′$$.

4.2. This satellite can be observed all the dark time of day from the moment when 2^m stars become visible (apporximately the end of evening civil teilight) to the corresponding moment in the morning, except the possible period when the satellite falls into the shadow of the Earth. the indicated "dark time of day" is 11^h 12^m (12^h - 2×24 ^m) in average, and can continue in Colombo from 11^h00^m in summer to 11^h24^m in winter and in summer the satellite is never eclipsed by the Earth (and therefore visible all night), but in periods close to the equinoxes, such eclipses in the middle of the night are regular. In maximum they last approximately the time T = 24^h × 0.9·D_E / 2πR_st (we use here just 24^h, since the movement is aynodic, D_E is the diameter of the Earth, anf the coefficient 0.9 appears due to the narrowing of the earth shadow cone.

$$T_{max}\ =\ 62\ min$$.

This, in periods close to the equinoxes, when the "dark time of day" is 11^h12^m, the duration of visibility can be reduced by 62 minutes, that is, to 10^h20^m. Thus, the answer to the second question of the problem is: from 10^h20^m to 11^h24^m.

4.3. Let us compare the light fluxes coming to the observer (O) from the satellite (s) and from the full Moon (L).

If

W = light flux from the Sun near Earth,

a_i = albedo of the observed objects,

S_i = πr_i² = πd_i²/4, where r_i and d_i are the radius and the diameter of the observed objects respectively,

R_i = distances from the observed objects to the observer:

Flux from Moon to boserver: l_L = W·a_L·S_L/(2πR_LO²).

Flux from satellite to observer: l_S = W·a_S·S_S/(4πR_SO).

(We do not take into account effects related to diagrams of scattering and random/mirror scattering. In the student's solutions it's also not reallt necessary to understand the difference between the coefficients of 2πand the 4π in 2πR_LO and 4πR_SO².)

Since the satellite was made of polished metal, we may assume ·a_S = 1.

Thus,

$$l_{L}\ /\ l_{s}\ =\ a_{L}·(S_{L}\ /\ S_{S})·(4{\pi}{R_{SO}}^{2}\ /\ 2{\pi}{R_{LO}}^{2})\ =\ 2a_{L}·(d_{L}/d_{S})^{2}·(R_{SO}/R_{LO})^{2}$$.

Necessary data may be taken from the Table of planetary data. Calculations may then give us:

$$l_{S}/l_{L}\ =\ 10^{-0.4(m_{S}-m_{L})}\ =\ 10^{-0.4(2+12.7)}\ ≈\ 1.3·10^{-6}$$,

and

$$d_{S}\ =\ 3.475·10^{6}\ m\ ×\ (2·0.07·1.3·10^{-6})^{1/2}·3.6·10^{7}\ m\ ≈\ 140\ m$$.

中文解答

4.1

地球静止卫星的星下点轨迹为赤道上的固定一点。因此，对于地球上的静止观察者来说，每个地球静止卫星都总是在天空中固定的一个点可见。

应该指出的是，地球静止卫星不应与地球同步卫星相混淆，地球静止轨道（克拉克轨道）是地球同步轨道的一种特例（即轨道倾角为零的地球同步轨道）。当卫星运行在地球静止轨道上时，它的星下点轨迹为赤道上的固定一点。而当卫星运行在轨道倾角不为零的其他地球同步轨道上时，它的星下点轨迹为一个“8”字。

地球静止卫星在以下条件下具有直径为R的圆形赤道轨道：

$$\omega^{2}R_{st}\ =\ GM/R_{st}$$,

$$R_{st}^3\ =\ GM/\omega^2\ =\ GMT^2/4\pi^2$$，

其中G是引力常数，M是地球的质量，T是地球围绕其轴旋转的周期（23^h56^m04^s）。计算给出了结果

$$R_{st}\ =\ 42160\ km$$。

很明显，位于观察者的子午线上的卫星有着最小的天顶距。该天顶距Ζ=φ+ψ，其中φ是观测者的纬度，ψ则是科伦坡到赤道面距离对卫星的张角。则： $$tan\psi\ =\ rsin\psi/(R-rcos\psi)$$

计算给出

$$\psi\ \approx\ 1°14′$$.

$$z\ \approx\ 08°08′$$

4.2

假设该卫星可以在2^m 星可见的黑夜（大约是晚上的民用昏影之后，除去卫星可能进入地球阴影的情形外）。这一时长平均为每天11^h 12^m (即12^h - 2×24 ^m)，这一时长在夏天为11^h00^m到冬季则为11^h24^m。很容易理解，在夏季和冬季时，该卫星不会进入到地球的阴影里，因此整夜可见。

但在每年的春秋分前后，该卫星每天会进入地球阴影。它在阴影里的最大时长应为 T = 24^h × 0.9·D_E / 2πR_st

其中D_E为地球的直径，系数0.9源自地球阴影的锥形形状，计算可得

$$T_{max}\ =\ 62\ min$$

所以，在春秋分前后，卫星每晚的可见时间11^h12^m，还需减去62分钟，即 10^h20^m。因此，问题的第二个问题的答案是：从 10^h20^m到 11^h12^m。

4.3

让我们比较来自卫星和满月（L）的观察者（O）的光通量。

如果

W =来自地球附近太阳的光通量，

a_i =被观察物体的反照率，

S_i = πr_i² = πd_i²/4，其中 r_i和 d_i分别是被观察物体的半径和直径，

R_i =从观察对象到观察者的距离：

从月球到支架的通量： l_L = W·a_L·S_L/(2πR_LO²)

观测者收到的月球的光流量：

I_L=Wα_LS_L/(2πR_LO²)

观测者收到的卫星的光流量：

I_S=Wα_SS_S/(4πR_LO²)

对于光亮的金属表面：α_S=1

由上可得：

$$l_{L}\ /\ l_{s}\ =\ a_{L}·(S_{L}\ /\ S_{S})·(4{\pi}{R_{SO}}^{2}\ /\ 2{\pi}{R_{LO}}^{2})\ =\ 2a_{L}·(d_{L}/d_{S})^{2}·(R_{SO}/R_{LO})^{2}$$.

从所提供的行星数据表可得：

$$l_{S}/l_{L}\ =\ 10^{-0.4(m_{S}-m_{L})}\ =\ 10^{-0.4(2+12.7)}\ ≈\ 1.3·10^{-6}$$,

$$d_{S}\ =\ 3.475·10^{6}\ m\ ×\ (2·0.07·1.3·10^{-6})^{1/2}·3.6·10^{7}\ m\ ≈\ 140\ m$$.