Question

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

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

Answer 1

Answer: (C)

Potential energy at surface $=-\frac{GMm}{R}$
Potential energy at height, $2R=-\frac{GMm}{3R}$
Change in potential energy $=-\frac{GMm}{3R}+\frac{GMm}{R}$
$=\frac{GMm}{R}\left(\frac{-1+3}{3}\right)$
$=\frac{2}{3}\frac{GMm}{R}$
$=\frac{2}{3}\left(\frac{GM}{R^2}\right)mR$
$=\frac{2}{3}mgR$