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))`

Answer

`6xx10^(24)kg`

Explanation

Angular speed of the satellite relative to the earth
`omega_(SE)=omega_(S)=omega_(E)`
(because they are moving in same direction west to east)
`omega_(S)=omega_(SE)=omega_(E)=(2pi)/T_(SE)+(2pi)/(T_(E))`
Given `T_(SE)=11.6h` and `T_(E)=24h`
`omega_(s)=2pi[1/11.6+1/24]h^(-1)=(2pixx(35.6))/((24xx10.6xx3600) rads^(-1)`
`=2.23xx10^(-4)rads^(-1)`
The condition of the circular orbit around the earth is
`(GM_(e)m)/(R^(2))=momega_(s)^(2)R` or `GM_(e)=omega_(s)^(2)R^(3)`
`R=2.0xx10^(4)km=2.0xx10^(7)m`
`M_(e)=(2.23xx10^(-4))^(2)xx(2.0xx10^(7)).67xx10^(-11)~~6xx10^(24)kg`