Question

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)

• A. $\left(\frac{n}{n+1}\right)mg R$
• B. $\left(\frac{n}{n-1}\right)mg R$
• C. $nmgR$
• D. $\frac{mgR}{n}$

Answer 1

The Correct Option is A

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