Question

A satellite revolving in a circular equatorial orbit of radius $R=2.00\times {10}^{4}km$ from west to east appears over a certain point at the equator every $\tau =11.6hours$. Using these data, calculate the mass of the Earth. The gravitational constant is supposed to be known.

Answer 1

We know that a satellite moving west to east at a distance $R=2.00\times {10}^{4}km$ from the centre of the earth will be revolving round the earth with an angular velocity faster than the earth's diurnal angular velocity. Let
$\omega =$ angular velocity of the satellite
${\omega }_0=\frac{2\pi }{T}=$ angular velocity of the earth. Then
$\omega -{\omega }_0=\frac{2\pi }{\tau }$
as the relative angular velocity with respect to earth. Now by Newton's law
$\frac{\gamma M}{R^2}={\omega }^{2}R$
So, $M=\frac{R^3}{\gamma }{(\frac{2\pi }{\tau }+\frac{2\pi }{T})}^2$
$=\frac{4{\pi }^2{R}^3}{\gamma T^2}{(1+\frac{T}{\tau })}^2$
Substitution gives $M=6.27\times {10}^{24}kg$