Question

Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth's surface. Will any satellite moving at this height be a geostationary satellite? Will any satellite moving at this height have a time period of 24 hours?

Answer 1

$$\begin{gathered} \text{T=(}\frac{g{R}^2{T}^2}{4{\mathrm{\pi }}^2}{)}^{\frac{1}{3}}-R \\ T=\frac{4{\mathrm{\pi }}^2(\mathrm{h}+\mathrm{R}{)}^3}{g{R}^2} \\ =\frac{4\times 3.{14}^2\times (36000+6400{)}^3\times {10}^9}{9.8\times (6400\times {10}^3{)}^2} \\ =\text{24.097 Hr} \\ \text{Which implies that it is a geostationary sattelite with time period=24 Hrs.} \end{gathered}$$