Question

A body of mass $mkg$ starts falling from a distance $2R$ above the earth's surface. What is its kinetic energy when it reaches a distance ${}^{\prime }{R}^{\prime }$ above the surface of the earth of radius $R$ and mass $M$?

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

Answer 1

The correct option is D $\frac{1}{6}\frac{GMm}{R}$

When body starts to fall towards the earth's surface its potential energy decreases and kinetic energy increases. Increase in kinetic energy $=$ Decrease in potential energy

Final kinetic energy $-$ Initial kinetic energy $=$ Initial potential energy $-$ Final potential energy

Final $O=\frac{-GMm}{R+2R}-\left(\frac{-GMm}{R+R}\right)$

Final $=\frac{-GMm}{3R}+\frac{GMm}{2R}=\frac{GMm}{6R}=\frac{1}{6}\frac{GMm}{R}$

Option (D) correct.