A geostationary satellite is revolving around the earth. If radius of the earth is ' $R$ ' and the angular…

A geostationary satellite is revolving around the earth. If radius of the earth is ' $R$ ' and the angular speed about its own axis is ' $\omega$ ' then the radius of the orbit of the geostationary satellite is ( $\mathrm{g}=$ acceleration due to gravity)
  1. $\left[\frac{\mathrm{R}^2 \omega^2}{\mathrm{~g}}\right]^{1 / 3}$
  2. $\left[\frac{\mathrm{Rg}}{\omega^2}\right]^{1 / 3}$
  3. $\left[\frac{\mathrm{R}^2 \mathrm{~g}}{\omega}\right]^{1 / 3}$
  4. $\left[\frac{\mathrm{R}^2 g}{\omega^2}\right]^{1 / 3}$

Solution

Considering force balance: Centrifugal force $=$ Gravitation force, $\begin{aligned} & \therefore m \omega^2 r=\frac{G M m}{r^2} \\ & \Rightarrow r=\left\{\frac{G M}{\omega^2}\right\}^{1 / 3}\end{aligned}$ We know, $g=\frac{G M}{R^2}$ $\Rightarrow \mathrm{r}=\left(\frac{\mathrm{R}^2 \mathrm{~g}}{\omega^2}\right)^{1 / 3}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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