Question

If $g$ is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass $m$ raised from the surface of the Earth to the height equal to the radius $R$ of the Earth is

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

Answer 1

Answer: (B)

Potential energy of an object at the surface of the Earth is
$U_{1}=-\frac{G M m}{R}$ ..... (i)
Potential energy of the object at a height, $h=R$ from the surface of the Earth.
$U_{2}=-\frac{G M m}{R + h}=-\frac{G M m}{R + R}$
Gain in potential energy
$\Delta U=U_{2}-U_{1}=\left(\right. - \frac{G M m}{2 R} \left.\right) - \left(\right. - \frac{G M m}{R} \left.\right)=\frac{1}{2}\frac{G M m}{R}$
Also, $GM=gR^{2}$
$\Delta U=\frac{1}{2}\frac{g R^{2} m}{R}=\frac{m g R}{2}$