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Q.

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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a

8hr

b

$6 \sqrt{2} h r$

c

6hr

d

12hr

answer is D.

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Detailed Solution

$T = 2 π \sqrt{\frac{(R + h)^{3}}{GM}} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{R + h_{1}}{R + h_{2}})}^{3 / 2}$

$\Rightarrow \frac{24}{T_{2}} = {[\frac{R + 6 R}{R + 2.5 R}]}^{3 / 2}$ $∴ T_{2} = 6 \sqrt{2} hr$

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