The radius of the orbit of a geostationary sa | vedclass

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(A) For a satellite to orbit the Earth,the gravitational force provides the necessary centripetal force.$m r \omega^{2} = \frac{G M m}{r^{2}}$Here,$r$

Vedclass · Accepted Answer

$\left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 3}$

Vedclass · Answer

(A) For a satellite to orbit the Earth,the gravitational force provides the necessary centripetal force.$m r \omega^{2} = \frac{G M m}{r^{2}}$Here,$r$ is the orbital radius,$M$ is the mass of the Earth,and $\omega$ is the angular velocity.Simplifying the equation,we get $r^{3} = \frac{G M}{\omega^{2}}$.We know that the acceleration due to gravity on the Earth's surface is $g = \frac{G M}{R^{2}}$,which implies $G M = g R^{2}$.Substituting $G M$ in the expression for $r^{3}$,we get $r^{3} = \frac{g R^{2}}{\omega^{2}}$.Therefore,the radius of the orbit is $r = \left(\frac{g R^{2}}{\omega^{2}}\right)^{1 / 3}$.

The radius of the orbit of a geostationary satellite is (mean radius of the earth is $R$,angular velocity about its axis is $\omega$,and acceleration due to gravity on the earth's surface is $g$).

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