MHT CET · Physics · Gravitation
The radius of the orbit of a geostationary satellite is (mean radius of earth is ' \(R\) ', angular velocity about own axis is ' \(\omega\) ' and acceleration due to gravity on earth's surface is ' \(\mathrm{g}\) ')
- A \(\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}\)
- B \(\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}\)
- C \(\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}\)
- D \(\frac{\mathrm{gR}^2}{\omega^2}\)
Answer & Solution
Correct Answer
(A) \(\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{mr} \omega^2=\frac{\mathrm{GMm}}{\mathrm{r}^2}\)
\(\omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots\left(\because g=\frac{G M}{R^2}\right)\)
\(\therefore \quad\) Radius of the orbit of the satellite is:
\(\mathrm{r}=\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}\)
\(\omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots\left(\because g=\frac{G M}{R^2}\right)\)
\(\therefore \quad\) Radius of the orbit of the satellite is:
\(\mathrm{r}=\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}\)
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