A satellite of mass m revolves around the earth of radius R at a hight 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. `(gR^(2))/(R + x)`
B. `(gR)/(R - x)`
C. gx
D. `((gR^(2))/(R + x) )^(1//2)`

Answer

D

Explanation

`(mv^(2))/((R+x)) = (GmM)/((R + x)^(2)) `also g = `(GM)/(R^(2))`
`therefore (mv^(2))/((R + x)) = m ((GM)/(R^(2))) (R^(2))/((R + x)^(2))`
`therefore (mv^(2))/((R + x)) = mg (R^(2))/((R + x)^(2))`
`therefore v^(2) = (gR^(2))/(R + x) rArr v = ((gR^(2))/(R + x))^(1//2)`