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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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T ²D

$3 π T^{2} D$

$3 π D^{2} T$

D²T

answer is A.

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Detailed Solution

Using $T = 2 π \sqrt{\frac{R^{3}}{GM}} = 2 π \sqrt{\frac{R^{3}}{G \times \frac{4}{3} {πR}^{3} D}}$
$T^{2} = \frac{4 π^{2} R^{3}}{G \frac{4}{3} π R^{3} D} = \frac{3 π}{D G} \Rightarrow \frac{3 π}{G} = T^{2} D$

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