The mean radius of the earth is `R`, its angular speed on its own axis is `omega` and the acceleration due to gravity at earth's surface is `g`. The cube of the radius of the orbit of a geo-stationary satellite will be

Options

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

Answer

D

Explanation

`T=(2pir^(3//2))/(sqrt(GM))`
`omega=(2pi)/T=(sqrt(GM))/(r^(3//2))`
`omega^(2)=(gR^(2))/(r^(3)) implies r^(3)=(gR^(2))/(omega^(2))`