Given raduis of earth `'R'` and length of a day `'T'` the height of a geostationary satellite is [`G`-Gravitational constant `M`-mass of earth]

Options

A. `((4pi^2GM)/(T^2))^(1//3)`
B. `((4piGM)/(R^2))^(1//3)-R`
C. `((GMT^2)/(4pi^2))^(1//3)-R`
D. `((GMT^2)/(4pi^2))^(1//3)+R`

Answer

C

Explanation

`T=2pisqrt(((R+h)^3)/(GM))rArrT^2=(4pi^2)/(GM)(R+h)^3`
`rArrR+h=((GMT^2)/(4pi^2))^(1/3)rArrh=((GMT^2)/(4pi^2))^(1/3)-R`