A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of `g_(0)`, the value of acceleration due to gravity at the earth's surface is

Options

A. `(mg_(0)R^(2))/(2(R+h))`
B. `-(mg_(0)R^(2))/(2(R+h))`
C. `(2mg_(0)R^(2))/(R+h)`
D. `-(2mg_(0)R^(2))/(R+h)`

Answer

B

Explanation

Total enery of a satellite at height `h` is
`=KE+PE=(GMm)/(2(R+h))-(GMm)/((R+h))`
`=(-GMm)/(2(R+h))=(-GMmR^(2))/(2R^(2)(R+h))=(-mg_(0)R^(2))/(2(R+g))" "(becauseg_(0)=(GM)/(R^(2)))`.