Gravitation - Result Question 33
34. 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
[2002]
(a) $3 m g R$
(b) $\frac{3}{4} m g R$
(c) $1 mg R$
(d) $\frac{3}{2} m g R$
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Answer:
Correct Answer: 34. (b)
Solution:
- (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
$=E_2-E_1=-\frac{G M m}{4 R}+\frac{G M m}{R}=\frac{3 G M m}{4 R}$
$=\frac{3 g R^{2} m}{4 R} \quad(\because g=\frac{G M}{R^{2}})$
$=\frac{3}{4} m g R$
