Question

At which height from the surface of the earth a satellite has to revolve around the earth if it is a geosynchronous satellite? It is given that R" = 6.4 × 10* m, g. = 9.8 m s2 ⁄ .

Answer 1

The Kepler's third law says

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

Hence, the radius of the satellite's orbit

$$r=\sqrt[3]{gR^2T^2/4\pi^2}$$ $$=\sqrt[3]{\frac{9.8\times (6.4\times10^6)^2\times(24\times 3600)^2}{4\pi^2}}=4.23\times 10^7\:\rm m.$$

Finally, the height from the surface of the earth to the satellite

$$h=r-R=4.23\times 10^7-6.4\times 10^6=3.6\times 10^7\:\rm m=36\:000\:\rm km.$$