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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?
Solution
\[T=(\frac{g R^2 T^2}{4 \pi^2} )^\frac{1}{3} - R\]
\[T = \frac{4 \pi^2 (h + R )^3}{g R^2}\]
\[ = \frac{4 \times 3 . {14}^2 \times (36000 + 6400 )^3 \times {10}^9}{9 . 8 \times (6400 \times {10}^3 )^2}\]
= 24.097 Hr
Which implies that it is a geostationary satellite with a time period = 24 Hrs.
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