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. gx
B. `(gR)/(R-x)`
C. `(gR^(2))/(R+x)`
D. `((gR^(2))/(R+x))^(1//2)`

Answer

D

Explanation

`F=(mv^(2))/((R+x))`
`(GMm)/((R+x)^(2))=(mv^(2))/((R+x))`
`v=sqrt((GM)/((R+x)))=sqrt((GM)/(R^(2))(R^(2))/((R+x)))`
`=sqrt((gR^(3))/((R+x)))=((gR^(2))/(R+x))^(1//2)`