Question

what is the minimum velocity of geostationary satellite

Answer 1

A geostationary orbit can be achieved only at an altitude very close to 35,786 km (22,236 mi) and directly above the equator. This equates to an orbital velocity of 3.07 km/s (1.91 mi/s) and an orbital period of 1,436 minutes, which equates to almost exactly one sidereal day (23.934461223 hours)......

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Answer 2

A geostationary orbit is a circular orbit directly above the Earth's equator approximately 35,786 km above ground. Any point on the equator plane revolves about the Earth in the same direction and with the same period as the Earth's rotation.
The period of the satellite is one day or approximately 24 hours. To find the speed of the satellite in orbit we use Newton’s law of gravity and his second law of motion along with that we know about centripetal acceleration. The inward and outward forces on the satellite must equal each another (by Newton's first law of motion).
Fcentripetal = Fcentrifugal
By Newton's second law of motion:
F = ma
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