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 :

The formula to calculate the mass of the Earth considering it a sphere is given by

$M=\frac{4}{3}\pi R^{3}d ...\left(1\right)$

The formula to calculate the time period of revolution of the satellite can be written as

$T=2\pi \sqrt{\frac{R^3}{GM}} ...\left(2\right)$

Substitute the expression for the mass from equation (1) into equation (2) and simplify to obtain the expression for the required quantity.

$\begin{array}{rcl}T& =& 2\pi \sqrt{\frac{R^3}{G\frac{4}{3}\pi R^{3}d}}\\ & =& 2\pi \sqrt{\frac{3}{4\pi Gd}}\\ & =& \sqrt{\frac{12{\pi }^2}{4\pi Gd}}\\ & =& \sqrt{\frac{3\pi }{Gd}}\\ & \Rightarrow & \frac{3\pi }{Gd}=T^2\end{array}$