Question

A satellite is revolving round the earth with orbital speed $v_0$. If it stops suddenly, the speed with which it will strike the surface of earth would be
($v_e=$ escape velocity of a body on earth's surface)

• A. $\frac{v_e^2}{v_0}$
• B. $v_0$
• C. $\sqrt{v_e^2-2v_0^2}$
• D. $\sqrt{v_e^2-v_0^2}$

Answer 1

Answer: (C)

Let $v$ be the velocity with which the satellite strikes the surface of the earth. According to law of conservation of total mechanical energy, we get
$-\frac{GMm}{R+h}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$
$v^2=\frac{2GM}{R}-\frac{2GM}{R+h}$
$\because v_0=\sqrt{\frac{GM}{R+h}},v_e=\sqrt{\frac{2GM}{R}}$
$\therefore v^2=\sqrt{v_e^2-2v_0^2}$