A) 100%
B) 41.4%
C) 50%
D) 59.6%
Correct Answer: B
Solution :
The period of revolution of geostationary satellite is the same as that of the earth. Orbital velocity \[{{v}_{o}}=\sqrt{g{{R}_{e}}}\] Escape velocity \[{{v}_{e}}=\sqrt{2g{{R}_{e}}}\] where\[{{R}_{e}}\]is radius of earth. % increase \[=\frac{{{v}_{e}}-{{v}_{o}}}{{{v}_{o}}}\times 100\] % increase \[=\frac{\sqrt{2g{{R}_{e}}}-\sqrt{g{{R}_{e}}}}{\sqrt{g{{R}_{e}}}}\times 100\] \[=(\sqrt{2}-1)\times 100\] \[=(1.414-1)\times 100=41.4%\]
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