If mean radius of earth is R, its angular velocity is `omega` and the acceleration due to gravity at the surface of the earth is `g`. Then the cube of the radius of the orbit of geostationary satellite will be

Options

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

Answer

D

Explanation

Time period of satellite, `T=(2pir^(3//2))/(sqrt(GM))or sqrt(GM)=(2pi)/(T).r^(3//2)`
`(2pi)/(T)=omega`
`:. omegar^(3//2)=sqrt(GM)or r^(3) =(GM)/(omega^(2))=(gR^(2))/(omega^(2))`