Periodic time of a satellite revolving above Earth's surface at a height equal to R, radius of Earth, is : [g is acceleration due to gravity at Earth's surface]

Options

A. `2 pi sqrt((2R)/(g))`
B. `4sqrt2 pi sqrt(R/g)`
C. `2 pi sqrt(R/g)`
D. `8pisqrt(R/g)`

Answer

B

Explanation

`T=2pisqrt(((R+h)^3)/(gR^2))=2pisqrt(((2R)^3)/(gR^2))=4sqrt2pisqrt(R/g)`