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)

$\sqrt{v_{e}^{2} -2v_{0}^{2}}$

Answer 2

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 $- \frac{GMm}{R+h} = \frac{1}{2}mv^{2} - \frac{GMm}{R}$ $v^{2} = \frac{2GM}{R} - \frac{2GM}{R+h}$ $\because v_{0} = \sqrt{\frac{GM}{R+h}}, v_{e} = \sqrt{\frac{2GM}{R}}$ $\therefore v^{2} = \sqrt{v_{e}^{2} -2v_{0}^{2}}$