Question 331: Gravitation

Question

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 '$$g$$')

Accepted Answer

$$\begin{aligned}
& \operatorname{mr} \omega^2=\frac{G M m}{r^2} \
& \omega^2=\frac{G M}{r^3}=\frac{g R^2}{r^3} \quad \ldots .\left(\because g=\frac{G M}{R^2}ight)
\end{aligned}$$
$$	herefore \quad$$ Radius of the orbit of the satellite is:
$$r=\left(\frac{g R^2}{\omega^2}ight)^{\frac{1}{3}}$$

Answer

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

Answer

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

Answer

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

Answer

$$\frac{\mathrm{gR}^2}{\omega^2}$$

Gravitation

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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