Q. An earth satellite of mass $m$ revolves in a circular orbit of a height $h$ from the surface of the earth. $R$ is the radius of the earth and $g$ is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by
A) $\frac{gR^{2}}{R + h}$
B) $gR$
C) $\frac{gR}{R + h}$
D) $\sqrt{\frac{gR^{2}}{R + h}}$
Explanation
(d) $v_{0} = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^{2}}{R + h}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.