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

\[T=(\frac{g R^2 T^2}{4 \pi^2} )^\frac{1}{3} - R\]

\[T = \frac{4 \pi^2 (h + R )^3}{g R^2}\]

\[ = \frac{4 \times 3 . {14}^2 \times (36000 + 6400 )^3 \times {10}^9}{9 . 8 \times (6400 \times {10}^3 )^2}\]

 = 24.097 Hr

Which implies that it is a geostationary satellite with a time period = 24 Hrs.