A satellite is orbiting just above the surface of a planet of average density D with period T. If G is the universal gravitational constant, the quantity $\frac{3 π}{G}$ is equal to

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$3 π T^{2} D$

$T^{2} D$

$D^{2} T$

$3 π D^{2} T$

answer is D.

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Detailed Solution

$T = 2 π \sqrt{\frac{R^{3}}{G m}} = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4}{3} π R^{3} D}} T^{2} = \frac{4 π R \times 3}{4 π G D} = \frac{3 π}{G D} ∴ \frac{3 π}{9} = T^{2} D$

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