Given radius of Earth $‘R’$ and length of a day $‘T’$ the height of a geostationary satellite is [$G-$Gravitational Constant, $M-$Mass of Earth]
Given radius of Earth $‘R’$ and length of a day $‘T’$ the height of a geostationary satellite is [$G-$Gravitational Constant, $M-$Mass of Earth]
${\left( {\frac{{4{\pi ^2}GM}}{{{T^2}}}} \right)^{1/3}}$
${\left( {\frac{{4\pi GM}}{{{R^2}}}} \right)^{1/3}} - R$
${\left( {\frac{{GM{T^2}}}{{4{\pi ^2}}}} \right)^{1/3}} - R$
${\left( {\frac{{GM{T^2}}}{{4{\pi ^2}}}} \right)^{1/3}} + R$
Given radius of Earth $‘R’$ and length of a day $‘T’$ the height of a geostationary satellite is [$G-$Gravitational Constant, $M-$Mass of Earth]
$T = 2\pi \sqrt {\frac{{{r^3}}}{{GM}}}$
$\Rightarrow \,{T^2} = \frac{{4{\pi ^2}}}{{GM}}{(R + h)^3}$
$ \Rightarrow \,\,R + h = {\left[ {\frac{{GM{T^2}}}{{4{\pi ^2}}}} \right]^{1/3}} $
$\Rightarrow \,h = {\left[ {\frac{{GM{T^2}}}{{4{\pi ^2}}}} \right]^{\frac{1}{3}}} - R$
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