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 eart’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

The correct option is B

$\frac{1}{6}\frac{GMm}{R}$

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 = $(-\frac{GMm}{r_1})-(-\frac{GMm}{r_2})$

$\therefore$ Final kinetic energy $=\left(\frac{-GMm}{R+h_1}\right)-(-\frac{GMm}{R+h_2})=(-\frac{GMm}{R+2R})-(-\frac{GMm}{R+R})=-\frac{GMm}{3R}+\frac{GMm}{2R}=\frac{1}{6}\frac{GMm}{R}$