KEAM 2015 Physics and Chemistry Paper

since, the initial velocity of body is zero, its total energy is $E\;=\frac{-G m M}{I}$....(i) where, $m$ is the mass of body, $M$ is the mass of the Earth and $I$ its distance from the centre of the Earth. When the body reaches the Earth, let its, velocity be $v$ and its distance from the centre of the Earth equals the Earths radius $R$. Therefore, the energy is $E\;=\frac{1}{2} m v^{2} - \frac{G M m}{R}$....(ii) Equating Eqs. (i) and (ii), we get $v^{2}=2 G M\left(\; \frac{1}{R} - \frac{1}{I} \right)$ Also, $g\;=\frac{G M}{R^{2}}$. Therefore, $G M = g R^{2}$. Using this in the above equation, we get $v = R \left[ 2 g \left( \; \frac{1}{R} - \frac{1}{r} \right) \right]^{\frac{1}{2}}$ Now, $r = 2 R$ (given). Therefore, $v = R \left[ 2 g \left( \; \frac{1}{R} - \frac{1}{2 R} \right) \right]^{\frac{1}{2}}$ $\Rightarrow$ Velocity of a body which strike the Earth's surface, $v = \sqrt{g R}$