# Time Period of Satellite

## Time Period of Satellite

A satellite is anything that orbits around a larger object. A natural satellite is any celestial body in space that orbits around a larger body. Moons are called natural satellites because they orbit planets. Time period of satellite is the time taken by a satellite to complete one revolution around the earth and is denoted by T. Thus,

$$T=\frac{Dis\tan ce\,\,travelled\,\,in\,\,one\,\,revolution}{Orbital\,\,Velocity}=\frac{2\pi r}{v}$$.

$$T=\frac{2\pi r}{R}\sqrt{\frac{r}{g}}=\frac{2\pi }{R}\sqrt{\frac{{{r}^{3}}}{g}}=\frac{2\pi r}{R}\sqrt{\frac{{{(R+h)}^{3}}}{g}}$$… (1)

If the earth is supposed to be a sphere of mean density ρ, then the mass of the earth is:

$$M=\frac{4}{3}\pi {{R}^{3}}\rho$$ and $$g=\frac{GM}{{{R}^{2}}}=\frac{G}{{{R}^{2}}}\left( \frac{4}{3}\pi {{R}^{3}}\rho \right)=\frac{4G\pi R\rho }{3}$$.

Substitute the value of g in the equation (1), we get:

$$T=\frac{2\pi }{R}\sqrt{\frac{3{{(R+h)}^{3}}}{4\pi GR\rho }}=\sqrt{\frac{4{{\pi }^{2}}}{{{R}^{2}}}\times \frac{3{{(R+h)}^{3}}}{4\pi GR\rho }}=\sqrt{\frac{3\pi {{(R+h)}^{3}}}{G\rho {{R}^{3}}}}$$ … (2)

For a satellite orbiting close to the surface of earth, h << R.

∴ h + R ≈ R

From equation (1), $$T=\sqrt{\frac{3\pi }{G\rho }}$$.

From equation (2),

$$T=\frac{2\pi }{R}\sqrt{\frac{{{R}^{3}}}{g}}=2\pi \sqrt{\frac{R}{g}}$$ … (3)

By substituting g = 9.8 m/sec² and R = 6.4 x 10⁶ m in equation (3), we get the value of T = 5.08 x 10³ sec = 84.6 Minutes. It means a satellite orbiting close to the surface of the earth has a time period of revolution about 84.6 Minutes.

It is clear from equation (2) that the period of revolution of a satellite depends only upon its height above the earth’s surface. The larger is the height of a satellite above the surface of earth, greater will be its period of revolution.