A geostationary satellite is revolving around | vedclass

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(D) For a geostationary satellite,the gravitational force provides the necessary centripetal force for its circular orbit.Let $r$ be the radius of the

Vedclass · Accepted Answer

$\left[\frac{R^2 g}{\omega^2}\right]^{1/3}$

Vedclass · Answer

(D) For a geostationary satellite,the gravitational force provides the necessary centripetal force for its circular orbit.Let $r$ be the radius of the orbit of the satellite and $M$ be the mass of the Earth.The condition for circular motion is: $m \omega^2 r = \frac{GMm}{r^2}$.Simplifying this,we get: $r^3 = \frac{GM}{\omega^2}$.We know that the acceleration due to gravity at the Earth's surface is given by $g = \frac{GM}{R^2}$,which implies $GM = gR^2$.Substituting $GM = gR^2$ into the expression for $r^3$,we get: $r^3 = \frac{gR^2}{\omega^2}$.Therefore,the radius of the orbit is $r = \left[\frac{R^2 g}{\omega^2}\right]^{1/3}$.

$A$ geostationary satellite is revolving around the Earth. If the radius of the Earth is $R$ and the angular speed of the Earth about its own axis is $\omega$,then the radius of the orbit of the geostationary satellite is ($g =$ acceleration due to gravity).

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