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A geostationary satellite is orbiting the earth at a height of 6R above the surface of the earth. The time period of another satellite at a height of 2.5R from the surface of the earth is

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6hr

8hr

12hr

6 \sqrt{2} hr

answer is D.

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Detailed Solution

R_{1} = R + 6 R = {7R   R}_{2} = R + 2.5 R = 3.5 R

$T_{1} = 24 hrs$

${(\frac{24}{T_{2}})}^{2} = (\frac{7R}{35 R})$

$\frac{24}{T_{2}} = \sqrt{8}$

$T_{2} = \frac{24}{\sqrt{8}} = \frac{24}{2 \sqrt{2}} = 6 \sqrt{2} hrs$

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