The change in potential energy of the body when it is taken from the earth's surface to a height above its surface is :

Options

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

Answer

c

Explanation

At the surface of earth gravitational potential energy
`U_(1)=(-GMm)/(R)`
At height nR from the surface of earth,
`U_(2)=(-GMn)/((n+1)R)`
`:.` Change in G.P.E is `Delta U=U_(2)-U_(1)`
`DeltaU=(-GMn)/((n+1)R)+(GMn)/(R)`
`DeltaU=(GMn)/(R)[1-(1)/(n+1)]=(GMn)/(R)((n)/(n+1))`
Since `g=(GM)/(R^(2))implies GM=gR^(2)`
`DeltaU=(gR^(2)m)/(R)((n)/(n+1))=(n)/(n+1)mgR`