Question

An asteroid is falling towards the centre of earth under the influence of earth's gravitational field. At $10 R_e\left(R_e=\right.$ radius of earth) it has speed of $12 \,km / s$. What will be the speed of the asteroids when it hits the earth's surface?

• A. $12.4\, km / s$
• B. $14.6 \,km / s$
• C. $16 \,km / s$
• D. $18 \,km / s$

Answer 1

Answer: (C)

According to conservation of energy,
Total energy of asteroid at $10 R_{\varepsilon}=$ Total energy of asteroid at surface of earth
$\Rightarrow U_1+K_1=U_2+K_2$
$ \Rightarrow \frac{-G M_{\varepsilon} m}{10 R_{\varepsilon}}+\frac{1}{2} m v_0^2=\frac{-G m_e m}{R_{\varepsilon}}+\frac{1}{2} m v^2$
$ \Rightarrow \frac{9}{10} \frac{G M_e m}{R_e}+\frac{1}{2} m v_0^2=\frac{1}{2} m v^2 $
$ \Rightarrow \frac{9}{10} \times \frac{2 G M}{R_e}+v_0^2=v^2 $
$ \Rightarrow v^2=\frac{9}{10}\left(v_e\right)^2+v_0^2$
where $v_e=$ escape velocity $=11.2 \,km / s$
$ \Rightarrow v^2=\frac{9}{10} \times(112)^2+(12)^2$
$ =\frac{9}{10} \times(11.2)^2+144=16\, kms ^{-1}$