The radius of the orbit of a geostationary satellite is (mean radius of the earth is $\mathrm{R}$, angular…
The radius of the orbit of a geostationary satellite is (mean radius of the earth is $\mathrm{R}$,
angular velocity about an axis in $\omega$ and accleration due to gravity on earth's surface
is g)
$\left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 3}$
$\frac{g R^{2}}{\omega^{2}}$
$\left(\frac{g R^{2}}{\omega^{2}}\right)^{2 / 3}$
$\left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 2}$
Solution
$\begin{aligned} m r \omega^{2} &=\frac{G M m}{r^{2}} \\ r \omega^{2} &=\frac{G M}{r^{2}} \\ r^{3} &=\frac{G M}{\omega^{2}}=\frac{G M}{R^{2}} \times \frac{R^{2}}{\omega^{2}}=g \frac{R^{2}}{\omega^{2}} \\ r &=\left(\frac{R^{2} g}{\omega^{2}}\right)^{1 / 3} \end{aligned}$