Answer to Question #88585 in Mechanics | Relativity for deepak

Question

Artificial satellite revolves around the earth in 2.5 hours in a circular orbit . Find the height of satellite above the earth assuming earth as a sphere of radius 6370 kilometre.

Answer 1

The Kepler's third law states

$$T^2=\frac{4\pi^2 a^3}{GM}$$

So, the radius of satellite orbit

$$a=\sqrt[3]{\frac{GMT^2}{4\pi^2}}$$

$$=\sqrt[3]{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}\times(2.5\times 3600)^2}{4\pi^2}}$$

$$=9.352\times 10^6\:\rm{m}$$

The height of satellite above the earth surface

$$H=a-R$$

$$=9.352\times 10^6-6.370\times 10^6=2.982\times 10^6\:\rm{m}$$

$$=2982\:\rm{km}$$