2018年IAO理论低年组第4题-科伦坡地球静止轨道卫星

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原文题目

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%）。

解答

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)