Periodic time of a satellite revolving above Earth’s surface at a height equal to $R$, radius of Earth, is [$g$ is acceleration due to gravity at Earth’s surface]

Question

Dr. Abubacker · Accepted Answer

(b) $T = 2\pi\sqrt{\frac{(R + h)^{3}}{gR^{2}}} = 2\pi\sqrt{\frac{(2R)^{3}}{gR^{2}}} = 4\sqrt{2}\pi\sqrt{\frac{R}{g}}$

Q. Periodic time of a satellite revolving above Earth’s surface at a height equal to $R$, radius of Earth, is [$g$ is acceleration due to gravity at Earth’s surface]

Explanation

Reviewed by Dr. Abubacker

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