**Velocity of the artificial satellite**

Suppose the height of a satellite is h from the surface of the earth and if R is the radius of the earth, then the radius of the orbit of the satellite will be r = R + h. The gravitational attraction of the earth provides necessary centripetal force on the satellite for its rotation. So, if the velocity of the satellite is v, GMm/r^{2} = mv^{2}/r. Here M, m are respectively the masses of the earth and the satellite,

v^{2} = GM/r

or, v = √(GM/r)

= √ [(GM)/(R + h)] … … … (1)

**Time period of the satellite:** If the time period of the satellite is T, i.e., if the time to revolve around the earth once at a height h from the surface of the earth is T, then its linear velocity will be,

v = [2π (R + h)] / T

so, [2π (R + h)] / T = √ [(GM)/(R + h)]

or, T = [2π (R + h)] [√ (R + h) / GM] … … … (2)

**Height of the satellite:** Suppose the height of the artificial satellite from earth’s surface = h, then by squaring both sides of equation (1) we get,

T^{2} = [4π^{2} (R + h)^{3}] / GM

or, (R + h)^{3} = GMT^{2} / 4π^{2}

so, h = [GMT^{2} / 4π^{2}]^{1/3} – R.