The change in the gravitational potential energy when a body of 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. `((n)/(n+1))mgR`
B. `((n)/(n-1))mgR`
C. nmgR
D. `(mgR)/(n)`

Answer

A

Explanation

Gravitational potential energy of mass m at any point at a distance r from the centre of earth is
`U=-(GM m)/(r )`
At the surface of earth r=R,
`:. U_(S)=-(GM m)/(R )=-mgR" "(`:.`g=(GM)/(R^(2))`
At the height h=nR from the surface of earth
`r=R+h=R+nR-R(1+n)`
`U_(h)=-(GM m)/(R(1+n))=-(mgR)/((1+n))`
`:.` Change in gravitational potential energy is
`Delta U=U_(h)-U_(S)=-(mgR)/((1+n))-(-mgR)`
`=-(mgR)/(1+n)+mgR=mgR(1-(1)/(1+n))=mgR((n)/(1+n))`