The three conditions for a geostationary orbit

Three conditions, three marks, and they are commonly asked as a straight recall. Period 24 h, direction west to east, plane of the equator. Learn them as a list and write all three even if the question seems to want fewer.

Condition 1 gives the radius through Kepler III. T = 86 400 s into r³ = GMT²/4π² gives r = 4.23 × 10⁷ m from the CENTRE, which is an altitude of about 35 800 km. Quote the radius or the altitude, but make clear which.

Condition 2 is often skipped. A 24-hour orbit going east to west would sweep past the ground observer twice a day rather than hovering. Matching the period is not enough; you have to match the sense of rotation too.

Condition 3 has the best explanation and it is worth learning properly. The gravitational force always points at the Earth's CENTRE, so the centre of any orbit must be the Earth's centre. That immediately rules out an orbit hovering above, say, Singapore's latitude, because such a circle is not centred on the Earth's centre. Even a correctly centred but tilted 24-hour orbit drifts north and south over a day, tracing a figure of eight on the ground, so only the equatorial orbit stays over one fixed point.

Everything follows from the radius: v = 3.07 km s⁻¹, ω = 7.27 × 10⁻⁵ rad s⁻¹, and g at that height is only about 0.22 N kg⁻¹. Those numbers turn up in the long chain question.

Uses: telecommunications, TV broadcast and weather monitoring, because a ground dish can be aimed once and never moved, giving permanent line of sight.