Answer to Question #116394 in Mechanics | Relativity for Foibe Kambala

Question

4. A geosynchronous satellite is one that stays above the same point on earth. Which is
possible only if it is above a point on the equator. Such Satellites are used for TV and
radio transmission, for weather forecasting, and as communication relays.
Determine:
a) the height above the earth’s surface such a satellite must orbit
b) Satellite’s speed
c) Compare to the speed of a satellite orbiting 200 km above earth surface.

Answer 1

a) according to the third Kepler's law, the radius of orbit r and the period of rotation T are related by the expression

$\dfrac{T^2}{r^3} = \dfrac{4\pi^2}{GM_{\oplus}}$ , where $M_{\oplus}$ ​is the Earth's mass. Therefore,

$r = \sqrt[3]{\dfrac{T^2GM_{\oplus}}{4\pi^2}} \approx 42.2\cdot10^3\,\mathrm{km}.$ The height is$h=r-R_{\oplus} = 42.2\cdot10^3\,\mathrm{km} - 6.4\cdot10^3\,\mathrm{km} = 35.8\cdot10^3\,\mathrm{km}$ .

b) The velocity can be calculated as the ratio of the length of an orbit and the period

$v = \dfrac{2\pi r}{T} \approx 3.07\,\mathrm{km/s}$ .

с) Let us calculate the period of the second satellite:

$\dfrac{T_2^2}{T^2} = \dfrac{(200+R_{\oplus})^3}{r^3}.$

So $T_2 = 5.33\cdot10^3\,\mathrm{s}.$ The velocity $v_2$ ​ is

$v_2 = \dfrac{2\pi(200+R_{\oplus})}{T_2} \approx 7.78\,\mathrm{km/s}.$ We can see that

$\dfrac{v}{v_2} = \dfrac{3.07}{7.78} \approx 0.39.$