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}mv^{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}}$