The mean radius of the earth is $R$, its angular speed on its own axis is $\omega$, and the accelera

Question

The mean radius of the 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 linear orbital velocity of a geostationary satellite?

Professor PC Thomas Classes · Answer

(a) : For a satellite in a circular orbit, the gravitational force provides the centripetal force: $\frac{G M m}{r^2} = m \omega^2 r$. This gives the orbital radius $r = \left(\frac{GM}{\omega^2}\right)^{1/3}$. Since $g = \frac{GM}{R^2}$, we can write $GM = gR^2$. Thus, $r = \left(\frac{gR^2}{\omega^2}\right)^{1/3}$. The linear orbital velocity is $v = r\omega$. Substituting $r$, we get $v = \omega \left(\frac{gR^2}{\omega^2}\right)^{1/3} = (\omega^3)^{1/3} \left(\frac{gR^2}{\omega^2}\right)^{1/3} = (g R^2 \omega)^{1/3}$.