Question

An asteroid is moving directly towards tire centre of the earth. When at a distance of $10\, R$ ($R$ is the radius of the earth) from the earths centre, it has a speed of $12$ km/ s. Neglecting the effect of earths atmosphere, what w ill be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2$ km / s) ? Give your answer to the nearest integer in kilometer/s _______ .

Answer 1

$U_{1}+K_{1}=U_{2}+K_{2}$
$-\frac{GM_{e}m}{10\,R}+\frac{1}{2}mv^{2}_{0}=-\frac{GM_{e}m}{R}+\frac{1}{2}mv^{2}$
$+\frac{9}{10}\times\frac{GM_{e}m}{R}+\frac{1}{2}mv^{2}_{0}=\frac{1}{2}mv^{2}$
$\frac{9}{10}\times\frac{1}{2}M\times v^{2}_{e}+\frac{1}{2}mv^{2}_{0}=\frac{1}{2}mv^{2}$
$v^{2}=\frac{9}{10}v^{2}_{e}+v^{2}_{0}
=\frac{9}{10}\times\left(11.2\right)^{2}+\left(12\right)^{2}$
$v^{2}=112.896+144$
$v=16.027$
$v=16\,km/s$