The orbit of a geostationary satellite is concentric and coplanar with the equator of Earth and rotates along the direction of rotation of Earth. Calculate the height and speed. Take mass of Earth `= 6 xx 10^(-11) Nm^(2) kg^(-2)`. Given `pi^(2) = 10`.

Explanation

Here, `T = 24 h = 24 xx 3600 s, M = 6 xx 10^(24) kg`,
`G = 6.67 xx 10^(-11) Nm^(2) kg^(-2), R = 6.4 xx 10^(6) m`
`h = [(T^(2) GM)/(4pi^(2))]^(1//3) - R`
`= [((24 xx 3600)^(2) xx (6.67 xx 10^(-11)) xx 6 xx 10^(24))/(4 xx 10)]^(1//2)`
`- 6.4 xx 10^(6)`
`= 42.11 xx 10^(6) - 6.4 xx 10^(6)`
`= 35.71 xx 10^(6) m`.
Orbital velocity, `upsilon_(0) = (2pi (R + h))/(T)`
`= (2 xx 3.14 xx 42.11 xx 10^(6))/(24 xx 3600) = 3.062 xx 10^(3) ms^(-1)`