The radius of the orbit of a geostationary satellite is (mean radius of earth is ' $R$ ', angular velocity…

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 ' $\mathrm{g}$ ')
  1. $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$
  2. $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{2}{3}}$
  3. $\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{2}}$
  4. $\frac{\mathrm{gR}^2}{\omega^2}$

Solution

$\mathrm{mr} \omega^2=\frac{\mathrm{GMm}}{\mathrm{r}^2}$ $\omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots\left(\because g=\frac{G M}{R^2}\right)$ $\therefore \quad$ Radius of the orbit of the satellite is: $\mathrm{r}=\left(\frac{\mathrm{gR}^2}{\omega^2}\right)^{\frac{1}{3}}$ ~

Asked in: MHT CET 2023 (11 May Shift 2)

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