Question

A body which is initially at rest at a height $h=6400 \,km$ above the surface of the earth of radius $R(6400 \,km )$, 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 $=10 \, m / s ^{2}$ )

Answer 1

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