# 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

$(a)\;8\;hours \quad (b)\;12\;hours \quad (c)\;6\;hours \quad (d)\;24\;hours$

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.

edited Mar 14, 2014
The answer sud be 6hrs u r wrong