A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity $\frac{3\pi}{Gd}$ represents
Answer & explanation
Correct answer: option 3
Time Period of revolution of satellite is $ T = 2\pi \sqrt {\frac{R^3}{GM}} $
$ \text{If d is the density then } M = \frac{4\pi}{3} R^3 d$
$\Rightarrow T = 2\pi \sqrt {\frac{R^3}{Gd\frac{4\pi}{3} R^3}}= 2\pi \sqrt{\frac{3}{4\pi Gd}}$
$\Rightarrow T^2 = \frac{3\pi}{Gd}$