NEET-XII-Physics
11: Gravitation
- #15Consider 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?Ans : `` \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.}``
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