Question

A body which is initially at rest at a height $h=6400km$ above the surface of the earth of radius $R(6400km)$, falls freely towards the earth. If its velocity on reaching the surface of earth is $2^N\times 10^{3}m/s$, then calculate $N$.
(Take $g=$ acceleration due to gravity on the surface of the earth $=10m/s^2$ )

Answer 1

Answer: 3

Potential energy at the surface of ground $=\frac{-GMm}{R}$
Potential energy at a height $h=\frac{-GMm}{R+h}$
When a body comes to ground,
Loss in potential energy $=$ Gain in kinetic energy
$\Rightarrow \frac{-GMm}{R+h}-\left(\frac{-GMm}{R}\right)=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{GMm}{2R}=\frac{1}{2}m{v}^2$
$(\because h=R)$
$\Rightarrow gR=v^2$
$\left(\because \frac{GM}{R^2}=g\right)$
$\Rightarrow v=\sqrt{gR}$
$v=\sqrt{10\times 64\times 10^5}$
$=8\times 10^{3}m/s$
$=2^3\times 10^{3}m/s;N=3$