Periodic time of a satellite revolving above Earth’s surface at a height equal to $R$, radius of Earth, is $R$ [$g$ is acceleration due to gravity at Earth’s surface]
Periodic time of a satellite revolving above Earth’s surface at a height equal to $R$, radius of Earth, is $R$ [$g$ is acceleration due to gravity at Earth’s surface]
$2\pi \sqrt {\frac{{2R}}{g}} $
$4\sqrt 2 \pi \sqrt {\frac{R}{g}} $
$2\pi \sqrt {\frac{R}{g}} $
$8\pi \sqrt {\frac{R}{g}} $
Periodic time of a satellite revolving above Earth’s surface at a height equal to $R$, radius of Earth, is $R$ [$g$ is acceleration due to gravity at Earth’s surface]
$T = 2\pi \sqrt {\frac{{{{(R + h)}^3}}}{{g{R^2}}}} $
$= 2\pi \sqrt {\frac{{{{(2R)}^3}}}{{g{R^2}}}}$
$ = 4\sqrt {2\pi } \sqrt {\frac{R}{g}} $
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