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(D) For a geostationary satellite,the orbital period $T$ is equal to the rotational period of the earth,which is $T = 2\pi/\omega$.According to Kepler

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$R^2g/\omega^2$

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(D) For a geostationary satellite,the orbital period $T$ is equal to the rotational period of the earth,which is $T = 2\pi/\omega$.According to Kepler's third law,the square of the orbital period is proportional to the cube of the orbital radius $r$,given by $T^2 = (4\pi^2/GM)r^3$.We know that the acceleration due to gravity at the surface is $g = GM/R^2$,which implies $GM = gR^2$.Substituting $GM$ into the Kepler's law equation: $(2\pi/\omega)^2 = (4\pi^2 / gR^2)r^3$.Simplifying this: $4\pi^2/\omega^2 = (4\pi^2 / gR^2)r^3$.Canceling $4\pi^2$ from both sides: $1/\omega^2 = r^3 / (gR^2)$.Therefore,$r^3 = gR^2/\omega^2$.

The mean radius of the earth is $R$,its angular speed on its own axis is $\omega$,and the acceleration due to gravity at the earth's surface is $g$. The cube of the radius of the orbit of a geostationary satellite will be

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