A) 6000 km
B) 9000 km
C) 12000 km
D) 15000 km
Correct Answer: B
Solution :
Time period of satellite \[{{T}^{2}}=\frac{4{{\pi }^{2}}}{Gm}{{r}^{3}}\] or \[{{T}^{2}}=\frac{4{{\pi }^{2}}}{g{{R}^{2}}}{{r}^{3}}\] \[[\because GM=g{{R}^{2}}]\] 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, \[{{T}_{1}}=3h,\,\]\[{{T}_{2}}=24h\,\](geostationary satellite) \[{{r}_{1}}=R,\]\[{{r}_{2}}=36000\,km\] \[\therefore \] \[\frac{3}{24}={{\left( \frac{R}{36000} \right)}^{3/2}}\] or \[R={{\left( \frac{1}{8} \right)}^{2/3}}\times 36000\] or \[R=\frac{1}{4}\times 36000=9000\,km\]
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