Question

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

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

Answer 1

Answer: (C)

According to conservation of energy,
Total energy of asteroid at $10R_{\epsilon }=$ Total energy of asteroid at surface of earth
$\Rightarrow U_1+K_1=U_2+K_2$
$\Rightarrow \frac{-G{M}_{\epsilon }m}{10R_{\epsilon }}+\frac{1}{2}m{v}_0^2=\frac{-G{m}_{e}m}{R_{\epsilon }}+\frac{1}{2}m{v}^2$
$\Rightarrow \frac{9}{10}\frac{G{M}_{e}m}{R_e}+\frac{1}{2}m{v}_0^2=\frac{1}{2}m{v}^2$
$\Rightarrow \frac{9}{10}\times \frac{2GM}{R_e}+v_0^2=v^2$
$\Rightarrow v^2=\frac{9}{10}{\left(v_e\right)}^2+v_0^2$
where $v_e=$ escape velocity $=11.2km/s$
$\Rightarrow v^2=\frac{9}{10}\times (112{)}^2+(12{)}^2$
$=\frac{9}{10}\times (11.2{)}^2+144=16km{s}^{-1}$