The radius of the orbit of a geostationary sa | vedclass

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(A) For a geostationary satellite,the gravitational force provides the necessary centripetal force for circular motion.$m r \omega^2 = \frac{G M m}{r^

Vedclass · Accepted Answer

$\left(\frac{gR^2}{\omega^2}\right)^{\frac{1}{3}}$

Vedclass · Answer

(A) For a geostationary satellite,the gravitational force provides the necessary centripetal force for circular motion.$m r \omega^2 = \frac{G M m}{r^2}$Where $m$ is the mass of the satellite,$r$ is the orbital radius,$M$ is the mass of the Earth,and $\omega$ is the angular velocity.Simplifying the equation: $\omega^2 = \frac{G M}{r^3}$.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 $\omega^2$:$\omega^2 = \frac{g R^2}{r^3}$Rearranging to solve for $r$:$r^3 = \frac{g R^2}{\omega^2}$$r = \left(\frac{g R^2}{\omega^2}\right)^{\frac{1}{3}}$

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

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