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The mean radius of earth is $R$, and its angular speed on its axis is $\omega$. What will be the radius of orbit of a geostationary satellite?
$\left(\frac{R g}{\omega^2}\right)^{1 / 3}$
$\left(\frac{R^2 g}{\omega^2}\right)^{1 / 3}$
$\left(\frac{R^2 g}{\omega}\right)^{1 / 3}$
$\left(\frac{R^2 \omega^2}{g}\right)^{1 / 3}$
Solution
(b)
Time period of rotation of earth $=\frac{2 \pi}{\omega}$
(Duration of one day)
Geostationary satellite has same time period, $T=\frac{2 \pi}{\omega}$. Let $r$ be the radius of orbit of satellite $\Rightarrow$ Time period of satellite $=\frac{2 \pi r^{3 / 2}}{\sqrt{G M_\theta}}$
Also, $g=\frac{G M_e}{R_e^2}$
$\Rightarrow T=\frac{2 \pi r^{3 / 2}}{\sqrt{g}\left(R_e\right)}=\frac{2 \pi r^{3 / 2}}{R_e \sqrt{g}}=\frac{2 \pi}{\omega}$
$\Rightarrow r^{3 / 2}=\frac{R_e}{\omega} \sqrt{g}$
$\Rightarrow r=\left(\frac{R_e^2}{\omega^2} g\right)^{1 / 2}$
