Question

A satellite orbiting the earth in a circular orbit of radius R completes one revolution in 3 h. If orbital radius of geostationary satellite is 36,000 km, orbital radius of earth is :

• A. 6000 km
• B. 9000 km
• C. 12000 km
• D. 15000 km

Answer 1

Answer: (B)

Time period, $ {{T}^{2}}=\frac{4{{\pi }^{2}}}{GM}{{r}^{3}} $ or $ {{T}^{2}}=\frac{4{{\pi }^{2}}}{g{{R}^{2}}}{{r}^{3}} $ or $ T=\frac{2\pi }{R}\sqrt{\frac{{{r}^{3}}}{g}} $ or $ T\propto {{r}^{3/2}} $ $ \therefore $ $ \frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{3/2}} $ Given, $ {{T}_{1}}=3h,\,{{T}_{2}}=24\,h $ (geostationarysatellite) $ {{r}_{1}}=R,\,{{r}_{2}}=36000\,km $ $ \therefore $ $ \frac{3}{24}={{\left( \frac{R}{36000} \right)}^{3/2}} $ or $ R={{\left( \frac{1}{8} \right)}^{2/3}}\times 36000 $ or $ R=\frac{1}{4}\times 36000=9000\,km $