# Q 3.5 What is the relationship between orbital velocity, time period and height of satellite?

A first-order behaviour of satellite motion can be explained by considering only the gravitational force of the Earth as this is dominant; however a number of external forces act on a satellite causing its motion to deviate from the ideal. Simplified relationships, assuming a perfectly spherical Earth and only the Earth’s gravitational force, is as follows:

For an elliptical orbit, the relationship between the period T of a satellite and the semi-major axis, a, can be obtained by Kepler’s Third Law:

T^{2} = 4π^{2}a^{3}/μ

For a circular orbit the equation reduces to,

T^{2} = 4π^{2} (R+h)^{3/μ}where R = Earth radius, h = satellite altitude and μ= gravitational parameter = 398600.5 km3s-2,

The velocity, V, of a satellite in an elliptical orbit can be obtained as follows:V2 = μ[(2/r) – (1/a)]

Where r = distance of satellite from the Earth’s centre

For a circular orbit, setting a = rV2 =μ/r