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The time period of a satellite, revolving above earth’s surface at a height equal to R will be $(Given g = π^{2} m / s^{2}, R = radius of earth)$

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$\sqrt{32 R}$

$\sqrt{4 R}$

$\sqrt{2 R}$

$\sqrt{8 R}$

answer is A.

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Detailed Solution

$T = \frac{2 π}{\sqrt{GM}} r^{3 / 2} = \frac{2 π}{\sqrt{GM}} (R + h)^{3 / 2} Here h = R, GM = {gR}^{2} ∴ T = \frac{2 π}{\sqrt{{gR}^{2}}} (2 R)^{3 / 2} = \frac{2 π}{πR} \sqrt{8} R^{3 / 2} = \sqrt{32 R}$

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