Question

The change in potential energy, when a body of mass m is raised to a height nR from the earth's surface is
(R = Radius of earth)

• A.
• B.
• C.
• D.

Answer 1

The correct option is

D

$△U=\frac{mgh}{1+\frac{h}{R}}=\frac{mgnR}{1+\frac{nR}{R}}=\frac{nmgR}{n+1}$

alt method:

$\Delta U=U_f-U_i$

$U_f=-\frac{GMm}{nR+R}$

$U_i=-\frac{GMm}{R}$

$\Delta U=(-\frac{GMm}{nR+R})-(-\frac{GMm}{R})$

$\Delta U=\frac{GMm}{R}(1-\frac{1}{n+1})$

$\Delta U=mgR\left(\frac{n}{n+1}\right)$