A satellite of mass `m` revolves around the earth of radius `R` at a height `x` from its surface. If `g` is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options

A. `sqrt(gx)`
B. `sqrt((gR)/(R - x))`
C. `sqrt((gR^(2))/(R - x))`
D. `sqrt((gR^(2))/(R + x))`

Answer

D

Explanation

`g = (GM)/(R^(2)) rArr :. GM = gR^(2)`
`nu = sqrt((GM)/(r )) = sqrt((GM)/(R + x)) = sqrt((gR^(2))/(R + x))`