Question

The change in Gravitational potential energy when a body of mass $m$ is raised to a height $nR$ from the Earth’s surface is ($R$= radius of the earth, $g$= acceleration due to gravity on the surface of Earth)

• A. $mgR\frac{n}{(n-1)}$
• B. $mgR$
• C. $mgR\frac{n}{(n+1)}$
• D. $mgR\frac{n^2}{(n^2+1)}$

Answer 1

The correct option is C $mgR\frac{n}{(n+1)}$

At surface gravitational potential energy, $U_1=\frac{-GMm}{\left(R\right)}$
At height $nR,U_2=-\frac{GMm}{(R+nR)}$

Change in PE
$\Delta U=U_2-U_1=-\frac{GMm}{(R+nR)}+\frac{GMm}{R}=\frac{GMm}{R}(1-\frac{1}{n+1})=\frac{\left(R^{2}g\right)m}{R}\times \frac{n}{(n+1)}=mgR\left[\frac{n}{n+1}\right]$