What height do geostationary satellites orbit above the Earths surface?

You are given the following values: M_e = 6x10²⁴ kgR_e = 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 (F_g = F_c). The formulae for these forces are:F_g = GMm/(r²)F_c = 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 R_e.h = r - R_e = cuberoot( GMT² / 4pi²) - R_eSubstituting 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.