Q.Derive the time period of a satellite orbiting the Earth.

Equating gravitational force to the required centripetal force for a satellite in circular orbit gives v = √(GM/r), and dividing the orbit's circumference by v gives T = 2π√(r^3/GM).

Consider a satellite of mass m moving in a stable circular orbit of radius r (measured from the centre of the Earth, so r = R + h for a satellite at height h above the Earth's surface) around the Earth of mass M.

Derivation of orbital velocity:

For the satellite to stay in a circular orbit, the gravitational force of attraction between the Earth and the satellite must exactly supply the centripetal force needed for circular motion:

Gravitational force = Centripetal force

GMm/r^2 = mv^2/r

Cancelling m and one factor of r from both sides:

GM/r = v^2

v = √(GM/r)

This is the orbital velocity. (It can also be written using g = GM/R^2, i.e. GM = gR^2, as v = R√(g/r).)

Derivation of time period:

The time period T is the time taken to complete one full orbit, i.e. to travel the circumference 2πr at speed v:

T = (circumference)/(speed) = 2πr/v

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