A satellite revolving in a circular equatorial orbit of radius `R = 2.0 xx 10^(4)` km from west to east appears over a certain point at the equator every `11.6 h`. From these data, calculate the mass of the earth. `(G = 6.67 xx 10^(-11) N m^(2))`

Explanation

We know from the previous problem that a satellite moving west to east at a difference `R=2*00xx10^4km` 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=(2pi)/(T)`=angular velocity of the earth. Then
`omega-omega_0=(2pi)/(tau)`
as the relative angular velocity with respect to earth. Now by Newton's law
`(gammaM)/(R^2)=omega^2R`
So, `M=(R^3)/(gamma)((2pi)/(tau)+(2pi)/(T))^2`
`=(4pi^2R^3)/(gammaT^2)(1+T/tau)^2`
Substitution gives
`M=6*27xx10^(24)kg`