Question

The mean radius of earth is $R$, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth's surface is $g$. What will be the radius of the orbit of a geostationary satellite ?

• A. $(R^2 g / {\omega}^2 ) ^{1/3}$
• B. $(R g / {\omega}^2 ) ^{1/3}$
• C. $(R^2 {\omega}^2 /g ) ^{1/3}$
• D. $(R^2 g / \omega) ^{1/3}$

Answer 1

Answer: (A)

$\frac{G M m}{r^{2}}=m \omega^{2} r $
$\Rightarrow r^{3}=\frac{G M}{\omega^{2}}=\frac{g R^{2}}{\omega^{2}}$
$\therefore r=\left(g R^{2} / \omega^{2}\right)^{1 / 3}$