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Q.

A sattelite is seen after each 8h over the equator at a place on the earth when its sense of rotation is opposite to earth The time interval after which it can be seen at the same place when the sense of rotation of the earth and the satellite is the same will be

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a

8h

b

6h

c

24h

d

12h

answer is A.

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Detailed Solution

$\large I \to \omega = {\omega _1} + {\omega _2}$
$\large \frac{1}{T} = \frac{1}{{{T_1}}} + \frac{1}{{{T_2}}};\frac{1}{8} = \frac{1}{{24}} + \frac{1}{{{T_2}}} \Rightarrow {T_2} = 12h$
$\large II \to \frac{1}{{{T_0}}} = \frac{1}{{{T_2}}} - \frac{1}{{{T_0}}} = \frac{1}{{12}} - \frac{1}{{24}} = \frac{1}{{24}} \Rightarrow {T_0} = 24h$

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