# A body is released from height equal to the radius of earth the velocity of the object when it strikes the surface is? 1) ✓gr 2) ✓2gr 3) 2✓gr

Jun 12, 2018

$\sqrt{g R}$

#### Explanation:

The potential energy of a mass $m$ at a distance $r$ from the center of the earth (assumed to be a spherically symmetric object of mass $M$ and radius $R$ ) is

$V \left(r\right) = - \frac{G M m}{r}$

Initially the mass is at rest at $r = 2 R$. So, its total energy is

$V \left(2 R\right) = - \frac{G M m}{2 R}$

Thus, from the law of conservation of energy, its speed $v$ when it hits the earth (at $r = R$) is given by

$\frac{1}{2} m {v}^{2} - \frac{G M m}{R} = - \frac{G M m}{2 R} \implies$

$\frac{1}{2} m {v}^{2} = G M m \left(\frac{1}{R} - \frac{1}{2 R}\right) = \frac{G M m}{2 R} \implies$

${v}^{2} = G \frac{M}{R}$

Now, the force on the mass at the surface of the earth is given by

$m g = \frac{G M m}{R} ^ 2 \implies g = \frac{G M}{R} ^ 2$

Thus

${v}^{2} = g R \implies v = \sqrt{g R}$