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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A geo stationary satellite orbits around the earth in a circular orbit of radius $36000km$. Then the time period of a spy satellite orbiting a few Kilometers above the earth's surface $(R_{earth}=6400 Km)$ will be approximately

\[(a)\;\frac{1}{2}\;hour \quad (b)\;2\;hour \quad (c)\;1\;hour \quad (d)\;4\;hour \]

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1 Answer

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We know time period of geostationary satellite is $24 hrs$
Also $T^2 \;\alpha\; R^3$
$\large\frac{T_2}{T_1}=\bigg (\large\frac{R_2}{R_1}\bigg)^{\large\frac{3}{2}}$
$\large\frac{T_2}{24}=\bigg (\large\frac{6400}{36000}\bigg)^{\large\frac{3}{2}}$
$T_2=24 \times \bigg (\large\frac{6400}{36000}\bigg)^{\large\frac{3}{2}}$
Hence b is the correct answer.


answered Aug 26, 2013 by meena.p
edited Feb 17, 2014 by meena.p

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