What height do geostationary satellites orbit above the Earths surface?
You are given the following values: Me = 6x1024 kgRe = 6.37x106 mA geostationary satellite appears to an observer on the ground to always be at the same position on the sky. This means its orbit must be circular with a period of T = 24hr. [Strictly geostationary orbits are also equatorial, otherwise the position would oscillate north/south].For a circular orbit the centripetal force is provided by the gravitational force from the Earth (Fg = Fc). The formulae for these forces are:Fg = GMm/(r2)Fc = mw2r (or mv2/r then use v = wr)Equating leads to:GMm/(r2) = mw2r Rearrangement gives:r3 = GM/(w2)Substituting w = 2pi*f = 2pi/T and taking the cube root gives:r = cuberoot( GMT2/ 4pi2)BUT this is the distance of the orbit from the centre of the Earth, for its height above the surface we have to subtract Re.h = r - Re = cuberoot( GMT2 / 4pi2) - ReSubstituting the values given at the start of the question, the value of G, and converting T = 24hr = 86400s gives:3.59x106 m or ~36,000 km.
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