Earth's standard gravitational parameter is its mass M x G (the gravitational constant), which is roughly μ = 398,600.4418. The standard gravitational parameters for planets are better known than either M or G individually (the best way we have at our disposal to weigh extremely massive objects is by observing their gravitational effect on other bodies like orbiting satellites). Acceleration due to gravity is μr2. The equatorial radius of Earth is
6,378.14 km
. If we plug in that figure, we'd see acceleration on the surface is the familiar 0.009798285 km/s
. 422 km above the surface (in the orbital sphere of the ISS), it's just 0.008619904 km/s
(87.97%). The further out we go, the lower the acceleration and the longer the orbital period.
To find out how far away we have to place a geostationary satellite, consider that it will need to be in a circular orbit with an orbital period of one sidereal day. In one second, it will move 7.29212×10−5radians; therefore r×(1−cos(7.29212×10−5)) will be the distance that it falls towards the Earth in that time. We know from the SUVAT equation s=ut+12at2 that in a second, with no initial velocity, that the distance travelled is half the acceleration. We now have an identity r(1−cos(7.29212×10−5))=μr2. Rearrange it and solve: r=3√μ2−2cos(7.29212×10−5) It turns out that the altitude of 42164.15974 km above the center of the Earth (35786.02274 km above the equatorial surface) is where you'll find geostationary satellites - and they can only exist in that one orbital plane.