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 is
[R = Radius of Earth, M – mass of earth, G = Gravitational constant]
Answer & explanation
Correct answer: option 2
Gravitational potential energy of a body of mass m at height h above the surface of earth is given as $\frac{GMm}{R+h}$
Here, initial gravitational potential energy
$U_1=\frac{GMm}{(R+2 R)}=-\frac{GMm}{3 R}$
Final gravitational potential energy,
$U_2=\frac{GMm}{(R+R)}=-\frac{GMm}{2 R}$
gain in K.E = loss in potential energy = $-\frac{GMm}{3 R}-\left(-\frac{GMm}{2 R}\right)$
$=\frac{GMm}{R}\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{GMm}{6 R}$