Q. An artificial satellite of the earth moves at an altitude to $h = 670$ km along a circular orbit. The velocity of the satellite is
A) $7.5\ kms^{- 1}$
B) $8.5\ kms^{- 1}$
C) $11.2\ kms^{- 1}$
D) $4.5\ kms^{- 1}$
Explanation
(a) For the satellite to move along closed orbit (a circle with a radius $R + h$) it should be acted upon by a force directed towards the centre. In this case, this is the force of earth’s attraction. According to Newton’s Second law
$\frac{mv^{2}}{R + h} = \frac{GMm}{(R + h)^{2}}$
At the earth’s surface, $\frac{GMm}{R^{2}} = mg$
Therefore, $v = \sqrt{\frac{gR^{2}}{R + h}} = 7.5\ kms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.