Q. The mean radius of the earth is $R$, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth’s surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be
A) $R^{2}g/\omega$
B) $R^{2}\omega^{2}/g$
C) $Rg/\omega^{2}$
D) $R^{2}g/\omega^{2}$
Explanation
(d) Orbital velocity $v_{0} = \sqrt{\frac{GM}{r}} = \sqrt{\frac{gR^{2}}{r}}$ and $v_{0} = r\omega$
This gives $r^{3} = \frac{R^{2}g}{\omega^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.