The gravitational potential energy of body of mass 'm' at the earth's surface -`mgR_(e)`. Its gravitational potential energy at a height `R_(e)` from the earth's surface will be (here `R_(e)` is the radius of the earth)

Options

A. `-2mgR_(e)`
B. `2mgR_(e)`
C. `1/2 mgR_(e)`
D. `-1/2 mgR_(e)`

Answer

D

Explanation

`DeltaU=U_(2)-U_(1)=(mgR_(e))/(1+h/(R_(e)))=(mgR_(e))/(1+(R_(e))/(R_(e)))=(mgR_(e))/2`
`implies U_(2)-(-mgR_(e))=(mgR_(e))/2implies U_(2)=1/2mgR_(e)`