The mean radius of earth is R, its angular speed on its own axis is $\omega$ and the acceleration due to gravity at earth's surface is g. What will be the radius of the orbit of a geostationary satellite?
(1992)
$(\frac{R^2g}{\omega^2})^{\frac{1}{3}}$
$(\frac{Rg}{\omega^2})^{\frac{1}{3}}$
$(\frac{R^2\omega^2}{g})^{\frac{1}{3}}$
$(\frac{R^2g}{\omega})^{\frac{1}{3}}$
Solution:
Gravitational force provides the centripetal force: $$\frac{GMm}{r^2} = m\omega^2r$$.nSince $$g = \frac{GM}{R^2}$$, we can write $$GM = gR^2$$.nSubstituting this gives $$\frac{gR^2}{r^2} = \omega^2r \Rightarrow r^3 = \frac{gR^2}{\omega^2} \Rightarrow r = (\frac{R^2g}{\omega^2})^{\frac{1}{3}}$$.
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