Question

The time period of a satellite very close to the earth is $ T $ . The time period of a geo-synchronous satellite will be
(A) $ 2\sqrt 2 \left( T \right) $
(B) $ 6\sqrt 6 \left( T \right) $
(C) $ 7\sqrt 7 \left( T \right) $
(D) $ \dfrac{1}{{7\sqrt 7 }}\left( T \right) $

Answer 1

Hint : A geosynchronous orbit is an orbit which has an orbital period exactly to that of the earth. A geosynchronous orbit has a height of about 36000km above the earth's surface (usually the equator). We need to assume that the radius of a satellite close to the earth has the same orbital radius as the radius of the earth. The radius of the earth is about 6400 km

Formula used: In this solution we will be using the following formula;
 $ \dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}} $ where $ {T_1} $ is the period of a period of one satellite and $ {T_2} $ is period of another.
The $ {R_1} $ and $ {R_2} $ are their corresponding orbital radius.

Complete step by step answer
In general, a geosynchronous satellite is a satellite whose period of revolution around the earth is exactly equal to the period of rotation of the earth. The orbit of rotation is called a geosynchronous orbit. The height above the earth for a geosynchronous orbit is given as about 36000 km. Now, since the radius of the earth is about 6400 km, the height of the satellite hence is about six times the radius of the earth as in
 $ \dfrac{{36000}}{{6400}} = 5.625 \approx 6 $
Hence, by Kepler’s third law of planetary motion, we have that
 $ \dfrac{{T_1^2}}{{T_2^2}} = \dfrac{{R_1^3}}{{R_2^3}} $
 $ \Rightarrow \dfrac{{T_1^2}}{{T_2^2}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^3} $
Hence, by letting 2 be the satellite close to the earth’s orbit and 1 the geosynchronous satellite, we have that
 $ T_1^2 = {\left( {\dfrac{{6{R_E}}}{{{R_E}}}} \right)^3}T_2^2 $ ( by making the radius close to earth equal to the radius of the earth)
Hence, $ T_1^2 = T_2^2{6^3} $
By square-rooting both sides, we have
 $ {T_1} = \sqrt {T_2^2{6^3}} = 6\sqrt 6 {T_2} $
Since, from question, the period $ {T_2} = T $
Then, period of geosynchronous orbit is
 $ {T_1} = 6\sqrt 6 T $
Hence, the correct option is B.

Note
For clarity, the height of a geosynchronous orbit can be gotten from the general formula of the height of a satellite above the earth surface, which is given by
 $ h = {\left( {\dfrac{{G{M_E}}}{{4\pi }}{T^2}} \right)^{\dfrac{1}{3}}} - {R_E} $
Hence, in the case of a geosynchronous satellite $ T $ will be equal to the period of the earth rotation which is about 24 hours.