Q. An artificial satellite is moving in a circular orbit around the Earth. The height of the satellite above the surface of Earth is R. Suppose the satellite is stopped suddenly in its orbit and allowed to fall freely. On reaching Earth, its speed will be
Explanation
(a) Using conservation of energy.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}{mv}^{2} - \frac{GMm}{R} = 0 - \frac{GMm}{2R}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2}{mv}^{2} = \frac{GMm}{R} - \frac{GMm}{2R}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = 2\frac{GM}{R}\left\lbrack 1 - \frac{1}{2} \right\rbrack$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v^{2} = \frac{GM}{R}\ \ \ $
$But\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {gR}^{2} = GM$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{\frac{{gR}^{2}}{R}}$
$or\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \sqrt{gR}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.