Solveeit Logo
Login

Question

Physics Question on Gravitational Potential Energy

The change in the gravitational potential energy when a body of mass mm is raised to a height nRnR above the surface of the earth is (here RR is the radius of the earth)

A

(nn+1)mgR\left(\frac{n}{n+1}\right)mg R

B

(nn−1)mgR\left(\frac{n}{n-1}\right)mg R

C

nmgRnmgR

D

mgRn\frac{mgR}{n}

Answer

(nn+1)mgR\left(\frac{n}{n+1}\right)mg R

Explanation

Solution

Gravitational potential energy of mass m at any point at a distance r from the centre of earth is U=−GMmrU=-\frac{GMm}{r} At the surface of earth r = R, ∴Us=−GMmR=−mgR(∵  g=GMR2)\therefore\quad U_{s}=-\frac{GMm}{R}=-mgR\quad\quad\left(\because \,\,g=\frac{GM}{R^{2}}\right) At the height h = nR from the surface of earth r = R + h = R + nR = R(1 + n) ∴Uh=−GMmR(1+n)=−mgR(1+n)\therefore\quad U_{h}=-\frac{GMm}{R\left(1+n\right)}=-\frac{mgR}{\left(1+n\right)} Change in gravitational potential energy is ΔU=Uh−Us=−mgR(1+n)−(−mgR)\Delta U=U_{h}-U_{s}=-\frac{mgR}{\left(1+n\right)}-\left(-mgR\right) =−mgR1+n+mgR=-\frac{mgR}{1+n}+mgR =mgR(1−11+n)=mgR(n1+n)=mgR\left(1 -\frac{1}{1+n}\right)=mgR\left(\frac{n}{1+n}\right)