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A geo-stationary satellite is orbiting the earth at height 6R from the surface of the earth. The time period of another satellite revolving around the earth at a height 2.5 R from the earth's surface is

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$12 \sqrt{2} hr$

$12 hr$

$6 \sqrt{2} hr$

$6 hr$

answer is C.

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Detailed Solution

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{R_{1}^{3}}{R_{2}^{3}}} = \sqrt{\frac{{(7 R)}^{3}}{{(\frac{7}{2} R)}^{3}}}$

$\Rightarrow T_{2} = \frac{24}{2 \sqrt{2}} = 6 \sqrt{2} hr$

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