The mean radius of the earth is $R$, its angular speed on its own axis is $\omega $ and the acceleration due to gravity at earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be
The mean radius of the earth is $R$, its angular speed on its own axis is $\omega $ and the acceleration due to gravity at earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be
${R^2}g/\omega $
${R^2}{\omega ^2}/g$
$Rg/{\omega ^2}$
${R^2}g/{\omega ^2}$
The mean radius of the earth is $R$, its angular speed on its own axis is $\omega $ and the acceleration due to gravity at earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be
Orbital velocity ${v_0} = \sqrt {\frac{{GM}}{r}} = \sqrt {\frac{{g{R^2}}}{r}} $ and ${v_0} = r\omega $
This gives ${r^3} = \frac{{{R^2}g}}{{{\omega ^2}}}$
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