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

Answer

A

Explanation

At the surface of the earth, gravitational P.E. of the body,
`P.E.=-(GMm)/(R_(e))=(gR_(e)^(2)m)/(R_(e))=-mgR_(e)=U_(1)`
When it is lifted to a height `R_(e)`, the distance of the body from the centre of the earth `=R_(e)+R_(e)=2R_(e)` and
`therefore U_(2)=-(GMm)/(2R_(e))=-(mgR_(e))/(2)`