MHT CET · Physics · Gravitation
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)
- A \(\left[\frac{\mathrm{R}^2 \omega^2}{\mathrm{~g}}\right]^{1 / 3}\)
- B \(\left[\frac{\mathrm{Rg}}{\omega^2}\right]^{1 / 3}\)
- C \(\left[\frac{\mathrm{R}^2 \mathrm{~g}}{\omega}\right]^{1 / 3}\)
- D \(\left[\frac{\mathrm{R}^2 g}{\omega^2}\right]^{1 / 3}\)
Answer & Solution
Correct Answer
(D) \(\left[\frac{\mathrm{R}^2 g}{\omega^2}\right]^{1 / 3}\)
Step-by-step Solution
Detailed explanation
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, \(g=\frac{G M}{R^2}\)
\(\Rightarrow \mathrm{r}=\left(\frac{\mathrm{R}^2 \mathrm{~g}}{\omega^2}\right)^{1 / 3}\)
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, \(g=\frac{G M}{R^2}\)
\(\Rightarrow \mathrm{r}=\left(\frac{\mathrm{R}^2 \mathrm{~g}}{\omega^2}\right)^{1 / 3}\)
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