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For a satellite orbiting close to the surface of earth the period of revolution is 84 minute. The time period of another satellite orbiting at a height three times the radius of earth from its surface will be

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(84) 2 \sqrt{2} minute

(84) 8 minute

(84) 3 \sqrt{3} minute

(84) 3 minute

answer is C.

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Detailed Solution

T = 2 π \sqrt{\frac{{(R + h)}^{3}}{GM}}

$T \propto {(R + h)}^{3 / 2}$

$\frac{T_{1}}{T_{2}} = \frac{{(R + 2R)}^{3 / 2}}{R^{3 / 2}} = \frac{3 \sqrt{3} R^{3 / 2}}{R^{3 / 2}} = 3 \sqrt{3}$

$T_{2} = 84 3 \sqrt{3} minute$

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