Question

Determine the height of a geo-stationary satellite from the earth surface.

Answer 1

Dear Student ,
here in this case let us take that the mass of the earth = M , radius of earth=R , mass of the geo stationary satellite = m and the distance of the satellite from the centre of the earth = r , speed of the satellite = v ,time period =T .
Now as the gravitational force supplies the centripetal force in the circular orbit for the motion of the satellite so ,
$$\begin{gathered} \frac{m{v}^2}{r}=\frac{GMm}{r^2} \\ \Rightarrow v^2=\frac{GM}{r} \\ Nowastheaccelerationduetogravityatthesurfaceoftheearthisgso, \\ g=\frac{GM}{R^2} \\ \Rightarrow GM=g{R}^2 \\ Therefore,v^2=\frac{g{R}^2}{r} \\ \Rightarrow v=\sqrt{\frac{g{R}^2}{r}} \\ Nowthetimeperiodofrevolutionis,T=\frac{2\pi r}{v}=2\pi r\sqrt{\frac{r}{g{R}^2}}=2\pi \sqrt{\frac{r^3}{g{R}^2}} \\ So,T^2=\frac{4{\pi }^2{R}^3}{g{R}^2} \\ \Rightarrow r^3=\frac{g{R}^2{T}^2}{4{\pi }^2} \\ \Rightarrow r={\left(\frac{g{R}^2{T}^2}{4{\pi }^2}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ Now,T=24h=24\times 60\times 60s,g=9·8m/s^2,R=6400Km=6·4\times {10}^{6}m \\ So,puttingthevaluesweget, \\ r=4·234\times {10}^{7}m=4·234\times {10}^{4}km=42340km \\ Thereforeattheequatortheheightofthegeostationarysatelliteis, \\ h=r-R=42340-6400=35940km\approx 36000km \end{gathered}$$
Regards