An asteroid is moving directly towards the centre of the earth. When at a distance of $10R$ ($R$ is the radius of the earth) from the centre of the earth, it has a speed of $12$ km s$^{-1}$. Neglecting the effect of earth's atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is $11.2$ km s$^{-1}$)? Give your answer to the nearest integer in km s$^{-1}$ ______.
Numerical value type. Enter your answer.
Answer: 16
Energy conservation per unit mass:
$$\frac{1}{2}v^2 - \frac{GM}{R} = \frac{1}{2}u^2 - \frac{GM}{10R}$$
$$v^2 = u^2 + \frac{2GM}{R}\left(1 - \frac{1}{10}\right) = u^2 + 0.9\,v_e^2$$
since $v_e^2 = 2GM/R$.
$$v^2 = 144 + 0.9\times125.44 = 256.9 \Rightarrow v \approx 16\ \text{km s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics