The mean radius of earth is R and its angular speed about its own axis is `omega` and acceleration due to gravity is 'g'. The cube of the radius of the orbit of geostationary satellite is :

Options

A. `(R^(2)g)/(omega)`
B. `(R^(2)omega^(2))/(g)`
C. `(Rg)/(omega^(2))`
D. `(R^(2)g)/(omega^(2))`

Answer

D

Explanation

Here time period of a satellite is
`T=2pi sqrt((r^(3))/(GM))`
`therefore (T)/(2pi)=sqrt((r^(3))/(GM)) or (1)/(omega)=sqrt((r^(3))/(gR^(2)))`
`(1)/(omega^(2)) = (r^(3))/(gR^(2)) or r^(3)=(gR^(2))/(omega^(2))`.
Correct choice is (d).