Question

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 $\frac{3\pi}{G}$ is equal to

• A. $T^2D$
• B. $3\pi\,T^2D$
• C. $3\pi\,D^2T$
• D. $D^2T$

Answer 1

Answer: (A)

Using $T = 2 \pi \sqrt{\frac{R^3}{GM}} = 2\pi \sqrt{\frac{R^3}{G \times \frac{4}{3} \times \pi \, R^3 \, D}}$
$T^2 = \frac{4 \pi^2 R^3}{G \frac{4}{3} \pi \, R \, D} = \frac{3 \pi}{DG}$
$\frac{3 \pi}{DG} = T^2 \, D$