Question

An object is taken to height $2 R$ above the surface of earth, the increase in potential energy is $[R$ is radius of earth]

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

Answer 1

Answer: (C)

Potential energy at surface $=-\frac{G M m}{R}$
Potential energy at height, $2 R=-\frac{G M m}{3 R}$
Change in potential energy $=-\frac{G M m}{3 R}+\frac{G M m}{R}$
$=\frac{G M m}{R}\left(\frac{-1+3}{3}\right) $
$=\frac{2}{3} \frac{G M m}{R} $
$=\frac{2}{3}\left(\frac{G M}{R^{2}}\right) m R$
$=\frac{2}{3} m g R$