Question

Assuming the radius of the earth as $R$, the change in gravitational potential energy of a body of mass $m$, when it is taken from the earth's surface to a height $3 R$ above its surface, is

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

Answer 1

Answer: (B)

Gravitational potential energy (GPE) on the surface of earth,
$E_1=-\frac{G M m}{R}$
GPE at $3 R, E_2=-\frac{G M m}{(R+3 R)}=-\frac{G M m}{4 R}$
$\therefore$ Change in GPE
$=E_2-E_1=-\frac{G M m}{4 R}+\frac{G M m}{R}=\frac{3 G M m}{4 R}$
$=\frac{3 g R^2 m}{4 R}\left(\because g=\frac{G M}{R^2}\right)=\frac{3}{4} m g R$