### Time period of a satellite

The time period of a satellite is the time taken by it to go once around the earth. Therefore

T= Circumference of the orbit/Orbital velocity=2πr/v_{0}=[2π(R_{e} +h )]/v_{0}

Putting the value of v_{0} from the equation (1) at orbital speed of satellite, we get :

T = 2π(R_{e} +h )/[GM_{e}/(R_{e} +h )]^{1/2} = 2π√[(R_{e} +h )^{3}/gM_{e}]…….(4)

But GM_{e}=gR_{e}^{2},

hence we can also write

**T=2π√[(R _{e} +h )^{3}/gR_{e}^{2}]**

**Note:**a) It is evident from eq.(1) and (2) that the period of revolution of a satellite depends only upon its height above the earth’s surface. Greater is the height of a satellite above the earth’s surface.

b) Moon(the natural satellite of the earth), which is at a height of 380,000 km. above the earth’s surface, completes one revolution of earth in nearly 27 days, while an artificial satellite revolving near the earth’s surface completes 10 to 20 revolutions in a day.

c) For a satellite revolving very near to earth’s surface the time period of revolution is given by :

T = 2π√(R_{e}/g) = 2 ×3.14√[(6.37× 10^{6})/9.8]=5063 seconds

≈ 84 minutes ≈ 1 hour 24 minutes