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The radius of a planet is R. A Satellite revolves around it in a circle of radius r with angular velocity $ω_{0}$ . The acceleration due to the gravity on planet’s surface is

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$\frac{r^{3} ω_{0}^{3}}{R}$

$\frac{r^{3} ω_{0}^{2}}{R^{2}}$

$\frac{r^{3} ω_{0}}{R}$

$\frac{r^{3} ω_{0}^{2}}{R}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{GMm}{r^{2}} = {mω}_{0}^{2} r \\ \Rightarrow GM = ω_{0}^{2} r^{3} \\ \Rightarrow {gR}^{2} = ω_{0}^{2} r^{3} \\ \Rightarrow g = \frac{ω_{0}^{2} r^{3}}{R^{2}} \end{array}$

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