Q. A satellite moves in a circle around the earth. The radius of this circle is equal to one-half of the radius of the moon’s orbit. The satellite completes one revolution in
A) $\frac{1}{2}\ lunar\ month$
B) $\frac{2}{3}\ lunar\ month$
C) $2^{- 3/2}\ lunar\ month$
D) $2^{3/2}\ lunar\ month$
Explanation
(c) The period of revolution of a satellite at a height $h$ from the surface of earth is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = 2\pi\sqrt{\frac{\left( R_{e} + h \right)^{3}}{gR_{e}^{2}}}$
$Given,\ \ T_{m} = 1$ lunar month,
$\therefore\ \ \ \ \ \ \ T_{sat} = 2\pi\sqrt{\frac{\left( R + \frac{h}{2} \right)^{2}}{gR^{2}}}$
$\Longrightarrow \ \ \ \ T_{sat} = \frac{1}{2^{3/2}}$
$\ \ \ \ \ \ \ T_{moon} = 2^{- 3/2}$ lunar month
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.