Question

An asteroid is moving directly towards tire centre of the earth. When at a distance of $10R$ ($R$ 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

Answer: 16.00

$U_1+K_1=U_2+K_2$
$-\frac{G{M}_{e}m}{10R}+\frac{1}{2}m{v}_0^2=-\frac{G{M}_{e}m}{R}+\frac{1}{2}m{v}^2$
$+\frac{9}{10}\times \frac{G{M}_{e}m}{R}+\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2$
$\frac{9}{10}\times \frac{1}{2}M\times v_e^2+\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2$
$v^2=\frac{9}{10}{v}_e^2+v_0^2<br/>=\frac{9}{10}\times {\left(11.2\right)}^2+{\left(12\right)}^2$
$v^2=112.896+144$
$v=16.027$
$v=16km/s$