A satellite is revolving very close to a planet of density D. The time period of revolution of that planet is

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$\sqrt{\frac{3 π}{2 D G}}$

$\sqrt{\frac{3 π}{D G}}$

$\sqrt{\frac{3 πG}{D}}$

${(\frac{3 π}{D G})}^{3 / 2}$

answer is A.

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Detailed Solution

$F = \frac{G m M}{R^{2}} ∴ m ω^{2} R = \frac{G M m}{R^{2}} \Rightarrow \frac{2 π}{T} = \frac{g}{R} \Rightarrow T = 2 π \sqrt{\frac{R}{g}} = 2 π \sqrt{\frac{R}{\frac{4}{3} πGDR}}$

$\Rightarrow T = 2 π \sqrt{\frac{3 π}{DG}}$

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