Question

A body of mass $m \,kg$ starts falling from a point $2R$ above the earth’s surface. Its kinetic energy when it has fallen to a point ‘$R$’ above the earth’s surface [$R$-Radius of earth, $M$-Mass of earth, $G$-Gravitational constant]

• A. $\frac{1}{2} \frac{GMm}{R}$
• B. $\frac{1}{6} \frac{GMm}{R}$
• C. $\frac{2}{3} \frac{GMm}{R}$
• D. $\frac{1}{3} \frac{GMm}{R}$

Answer 1

Answer: (B)

When body starts falling toward earth’s surface its potential energy decreases so kinetic energy increases.
Increase in kinetic energy = Decrease in potential energy
Final kinetic energy - Initial kinetic energy = Initial potential energy - Final potential energy
Final kinetic energy - $0$
$ = \left(-\frac{GMm}{r_{1}}\right) - \left(-\frac{GMm}{r_{2}}\right)$
$\therefore $ Final kinetic energy $ = \left(\frac{-GMm}{R +h_{1}}\right) - \left(-\frac{GMm}{R +h_{2}}\right)$
$ =\left(-\frac{GMm}{R+2R}\right)-\left(-\frac{GMm}{R+R}\right) $
$ = - \frac{GMm}{3R} + \frac{GMm}{2R} $
$ = \frac{1}{6} \frac{GMm}{R}$