Practicing Success
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] |
$\frac{1}{2} \frac{GMm}{R}$ $\frac{1}{6} \frac{GMm}{R}$ $\frac{2}{3} \frac{GMm}{R}$ $\frac{1}{3} \frac{GMm}{R}$ |
$\frac{1}{6} \frac{GMm}{R}$ |
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 = $-\frac{GMm}{3 R}-\left(-\frac{GMm}{2 R}\right)$ $=\frac{GMm}{R}\left(\frac{1}{2}-\frac{1}{3}\right)=\frac{GMm}{6 R}$ |