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

Here the absolute angular velocity of satellite is given by
`omega = omega_(s) + omega_(E)`
Where `omega_(E)` is the angular velocity of earth, which is from west to east.
or `omega = (2pi)/(t) + (2pi)/(T)`
[Where t = 11.6 hr. and T = 24 hr.]
From Kepler's III law have `omega = (sqrt(GM))/(r^(3//2))`
Thus we have `(sqrt(GM))/(r^(3//2)) = (2pi)/(t) + (2pi)/(T)`
or `M = (4pi^(2) r^(3))/(G) [(1)/(t) + (1)/(T)]^(2)`
`= (4pi^(2)(2 xx 10^(7))^(3))/((6.67 xx 10^(-11)))[(1)/(11.6 xx 3600)+(1)/(24 xx 3600)]^(2)`
`= 6.0 xx 10^(24) kg`