The mean radius of Earth is R,and its angular | vedclass

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(B) For a geostationary satellite,the time period $T$ must be equal to the time period of Earth's rotation,which is $T = \frac{2\pi}{\omega}$.The orbi

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$\left(\frac{R^2 g}{\omega^2}\right)^{1/3}$

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(B) For a geostationary satellite,the time period $T$ must be equal to the time period of Earth's rotation,which is $T = \frac{2\pi}{\omega}$.The orbital period of a satellite at a distance $r$ from the center of the Earth is given by Kepler's Third Law: $T = 2\pi \sqrt{\frac{r^3}{GM}}$.We know that the acceleration due to gravity at the Earth's surface is $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.Substituting $GM$ into the time period formula: $T = 2\pi \sqrt{\frac{r^3}{gR^2}} = \frac{2\pi r^{3/2}}{R\sqrt{g}}$.Equating the two expressions for $T$: $\frac{2\pi}{\omega} = \frac{2\pi r^{3/2}}{R\sqrt{g}}$.Simplifying the equation: $\frac{1}{\omega} = \frac{r^{3/2}}{R\sqrt{g}} \Rightarrow r^{3/2} = \frac{R\sqrt{g}}{\omega}$.Squaring both sides: $r^3 = \frac{R^2 g}{\omega^2}$.Therefore,the radius of the orbit is $r = \left(\frac{R^2 g}{\omega^2}\right)^{1/3}$.

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?

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