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^(2)GM)/(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=2pi sqrt((r^(3))/(GM))impliesT^(2)=(4pi^(2))/(GM)(R+h)^(3)`
`h=[(GMT^(2))/(4pi^(2))]^(1//3)-R`