Q. Given radius of Earth $'R'$ and length of a day $'T'$ the height of a geostationary satellite is [G-Gravitational Constant, M-Mass of Earth]
A) $\left( \frac{4\pi^{2}GM}{T^{2}} \right)^{1/3}$
B) $\left( \frac{4\pi GM}{R^{2}} \right)^{1/3} - R$
C) $\left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} - R$
D) $\left( \frac{GMT^{2}}{4\pi^{2}} \right)^{1/3} + R$
Explanation
(c) $T = 2\pi\sqrt{\frac{r^{3}}{GM}} \Rightarrow T^{2} = \frac{4\pi^{2}}{GM}(R + h)^{3}$
$\Rightarrow R + h = \left\lbrack \frac{GMT^{2}}{4\pi^{2}} \right\rbrack^{1/3} \Rightarrow h = \left\lbrack \frac{GMT^{2}}{4\pi^{2}} \right\rbrack^{\frac{1}{3}} - R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.