Question

The radius of the orbit of geosynchronous satellite is 36000 km. Then, find the period of revolution of a satellite with its orbital radius 9000 km.
A. 24 hours
B. 12 hours
C. 6 hours
D. 3 hours

Answer 1

Geosynchronous satellite is placed in the geosynchronous orbit with an orbital period matching the Earth's rotation period. These satellites take 24 hours to complete one rotation around the earth. These satellites appear to be stationary above a particular point which is due to the synchronization. In other words, we can simply say that these remain fixed with respect to the earth and thus their time period is 24 hours.

Complete step by step solution:
We can deduce the following facts according to the question:
The radius of the orbit of geosynchronous satellite $(Rg)$= 36000 km
Time period of geosynchronous satellite $(Tg)$= 24 hour
Radius of satellite whose time period is to be calculated $(Rs)$= 9000 km
Let the time period of satellite which is to be calculated be $(Ts)$
Now,
We know that from Kepler’s third law: ${T^2}\alpha {R^3}$ (where T is the time period and R is the radius of orbiting satellite
Thus we can write:

{\left( {\dfrac{{Ts}}{{Tg}}} \right)^2} = {\left( {\dfrac{{Rs}}{{Rg}}} \right)^3} \\\ \Rightarrow \left( {\dfrac{{Ts}}{{24}}} \right) = {\left( {\dfrac{{9000}}{{36000}}} \right)^{3/2}} \\\ \Rightarrow Ts = 24 \times {\left( {\dfrac{1}{4}} \right)^{3/2}} \\\ \Rightarrow Ts = 24 \times \left( {\dfrac{1}{8}} \right) \\\ \Rightarrow Ts = 3hrs. \\\ \end{gathered} $$ **Thus the required time period of the satellite is 3 hours.** **Note:** One should be clear with the approach of Kepler’s law. However, sometimes the questions might be tricked by creating confusion between the distance of the satellite from the earth and its radius. To avoid this, be clear that in Kepler’s third law in which radius of orbiting satellite is considered not its distance.