IAO 2018, Теория, Група α — Задача 4. αβ-4. Коломбо. Геостационарен спътник

4.1.

Geostationary satellites are located above the same equator point. Therefore, each actual satellite 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 orbital period of which is exactly equal to the orbital period of the Earth, but which has an 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 equatorial orbit with diameter $R$ under the conditions:

where $G$ is the gravitational constant, $M$ is the mass of the Earth, $T$ is the period of the Earth’s revolution around its axis ($23^{\mathrm h}56^{\mathrm m}04^{\mathrm s}$). The calculations give the result

It is obvious that of all the possible geostationary satellites, the smallest zenith distance has a satellite located on the meridian of the observer. From the drawing we see that the zenith distance is $z=\varphi+\psi$, where $\varphi$ is the latitude, and $\psi$ is the angle at which the “equatorial plane – Colombo” segment is visible from the satellite. We can use the latitude of the place in Colombo, which is indicated as “seafront in the center”, $06^\circ54'$).

The calculations give

4.2.

This satellite can be observed all the dark time of day from the moment when $2^{\mathrm m}$ stars become visible (approximately the end of evening civil twilight) 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 $11^{\mathrm h}12^{\mathrm m}$ ($12^{\mathrm h}-2\times24^{\mathrm m}$) in average, and can continue in Colombo from $11^{\mathrm h}00^{\mathrm m}$ in summer to $11^{\mathrm h}24^{\mathrm m}$ in winter (seasons are indicated according to the European calendar). It is easy to understand that 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 $\tau=24^{\mathrm h}\times0.9\cdot D_E/2\pi R_{st}$ (we use here just $24^{\mathrm h}$, since the movement is synodic, $D_E$ is the diameter of the Earth, and the coefficient 0.9 appears due to the narrowing of the earth shadow cone.

Thus, in periods close to the equinoxes, when the “dark time of day” is $11^{\mathrm h}12^{\mathrm m}$, the duration of visibility can be reduced by 62 minutes, that is, to $10^{\mathrm h}20^{\mathrm m}$. Thus, the answer to the second question of the problem is: from $10^{\mathrm h}20^{\mathrm m}$ to $11^{\mathrm h}24^{\mathrm m}$.

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,

$\alpha_i$ = albedo of the observed objects,

$S_i=\pi r_i^2=\pi d_i^2/4$, 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 observer: $I_L=W\cdot\alpha_L\cdot S_L/(2\pi R_{LO}^2)$.

Flux from satellite to observer: $I_S=W\cdot\alpha_S\cdot S_S/(4\pi R_{SO}^2)$.

(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 really necessary to understand the difference between the coefficients of $2\pi$ and the $4\pi$ in $2\pi R_{LO}^2$ and $4\pi R_{SO}^2$.)

Since the satellite was made of polished metal, we may assume $\alpha_S=1$.

Thus,

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

and

Basic criteria