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 3 π G d represents :

Question

Aicharya · Accepted Answer

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

$M = \frac{4}{3} π R^{3} d . . . (1)$

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

$T = 2 π \sqrt{\frac{R^{3}}{G M}} . . . (2)$

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

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

A satellite is orbiting just above the surface of the earth with period T . If d is the density of the earth…

Solution