Question

A body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h=3R$ when $R$ is the radius of the earth. The change in gravitational potential energy of the body is

• A. $\frac{3}{2}mgR$
• B. $\frac{3}{4}mgR$
• C. $\frac{1}{2}mgR$
• D. $\frac{1}{4}mgR$

Answer 1

Answer: (B)

PE at the surface of earth
$PE=\frac{- G M m}{R}$
PE at a height $3R_{e}$
$PE=\frac{- G M m}{R + 3 \, R}=\frac{- G M_{m}}{4 R}$
$\Delta PE=PE_{f}-PE_{i}=\left(- \frac{G M m}{4 R} \, \right)-\left(\frac{- G M_{m}}{R} \, \right)=\frac{3}{4} \, \frac{G M_{m}}{R}$
$\Rightarrow \Delta PE=\frac{3}{4} \, mgR$