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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?

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Solution

## $\text{T=(}\frac{g{R}^{2}{T}^{2}}{4{\mathrm{\pi }}^{2}}{\right)}^{\frac{1}{3}}-R\phantom{\rule{0ex}{0ex}}T=\frac{4{\mathrm{\pi }}^{2}\left(\mathrm{h}+\mathrm{R}{\right)}^{3}}{g{R}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4×3.{14}^{2}×\left(36000+6400{\right)}^{3}×{10}^{9}}{9.8×\left(6400×{10}^{3}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\text{24.097 Hr}\phantom{\rule{0ex}{0ex}}\text{Which implies that it is a geostationary sattelite with time period=24 Hrs.}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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