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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Gravitation
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A satellite is seen after each 8 hours over equator at a place on earth when its sence of rotation is opposite to the earth. The time interval after which it can be seen at the same place when sense of rotation of earth and satellite is same , will be

\[(a)\;8\;hours \quad (b)\;12\;hours \quad (c)\;6\;hours \quad (d)\;24\;hours \]
Can you answer this question?
 
 

1 Answer

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Let $T_1$ and $T_2$ be time period of earth and satellite respectively then we have $8=\large\frac{2 \pi}{w_1+w_2} = \frac {2 \pi}{\large\frac{2 \pi}{T_1}+\large\frac{2 \pi}{T_2}}$
When earth and satellite are rotating in opposite sense.
$T_1=24 \;hrs$
$8=\large\frac{1}{\Large\frac{1}{24}+\frac{1}{T_2}}$
$T_2=12 hours$
When the earth and satellite are rotating in same sense.
$t=\large\frac{2 \pi }{\Large\frac{2 \pi}{T_1 }- \frac{2 \pi }{T_2}}$
$t=\large\frac{1}{\Large\frac{1}{24}-\frac{1}{12}}$
$t=24\; hours$
Hence d is the correct answer. 

 

answered Aug 24, 2013 by meena.p
edited Mar 14, 2014 by thagee.vedartham
 

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