A satellite in an equatorial orbit has a time period of 6 hrs. At a certain instant, it is directly overhead an observer on the equator of the earth. It is directly overhead the observer again after a time T. The possible value of T is/are

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none of these

both (1) and (2)

4.8 hr

8 hr

answer is C.

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Detailed Solution

$ω^{'} = ω_{s} + ω_{E}$
or $ω^{'} = ω_{s} - ω_{E}$
$\begin{array}{l} = \frac{2 π}{6} + \frac{2 π}{24} = \frac{10 π}{24} \\ = \frac{2 π}{6} - \frac{2 π}{24} = \frac{6 π}{24} \\ ∴ T = \frac{2 π}{ω^{'}} = 4 .8 hour \\ ∴ T = \frac{2 π}{ω^{'}} = 8 hour \end{array}$

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