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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 ' $\mathrm{g}$ ')
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The correct answer is:
$\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$
$\mathrm{mr} \omega^2=\frac{\mathrm{GMm}}{\mathrm{r}^2}$
$\omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots\left(\because g=\frac{G M}{R^2}\right)$
$\therefore \quad$ 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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