The time period of a satellite,revolving abov | vedclass

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(C) The orbital radius of the satellite is $r = R + h = R + R = 2R$.The orbital velocity $v$ is given by $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R

Vedclass · Accepted Answer

$\sqrt{32 R}$

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(C) The orbital radius of the satellite is $r = R + h = R + R = 2R$.The orbital velocity $v$ is given by $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R}}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the velocity equation:$v = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}}$.The time period $T$ is given by $T = \frac{2\pi r}{v}$.Substituting $r = 2R$ and $v = \sqrt{\frac{gR}{2}}$:$T = \frac{2\pi(2R)}{\sqrt{\frac{gR}{2}}} = \frac{4\pi R}{\sqrt{\frac{gR}{2}}} = 4\pi R \sqrt{\frac{2}{gR}} = 4\pi \sqrt{\frac{2R}{g}}$.Given $g = \pi^2 \ m/s^2$,we substitute $\pi = \sqrt{g}$:$T = 4\sqrt{g} \sqrt{\frac{2R}{g}} = 4\sqrt{2R} = \sqrt{16 	imes 2R} = \sqrt{32R}$.

The time period of a satellite,revolving above earth's surface at a height equal to $R$ will be (Given $g = \pi^2 \ m/s^2$,$R =$ radius of earth).

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