The change in the gravitational potential energy when a body of a mass `m` is raised to a height `nR` above the surface of the earth is (here `R` is the radius of the earth)

Options

A. `(mgR)/(n)`
B. `nmgR`
C. `mgR(n/(n-1))`
D. `mgR(n/(n+1))`

Explanation

(c ) Gravitational potential energy at any point at a distance r from the centre of the earth is
`U = -(GMm)/(r )`
where `M` and m be masses of earth and body respectively.
At the surface of the earth , `r=R`
`:. U_(1)=-(GMm)/(R)`
At a height h from the surface, `r=R+h=R+nR=(1+n)R`
`:. U_(2)=-(GMm)/((n+1)R)`
Change in potential energy is
`DeltaU=U_(2)-U_(1)=-(GMm)/((n+1)R)-(-(GMm)/(R))`
`=(GMm)/(R)(1-(1)/((n+1)))=(GMmn)/((n+1)R)`
`=mgR(n)/((n+1))( :.g=(GM)/(R)^(2)))`