Question

The mean radius of earth is R, and its angular seed on its axis is $\omega$ What will be the radius of orbit of a geostationary satellite?

• A. ${\left(\frac{Rg}{{\omega }^2}\right)}^{1/3}$
• B. ${\left(\frac{R^{2}g}{{\omega }^2}\right)}^{1/3}$
• C. ${\left(\frac{R^{2}g}{\omega }\right)}^{1/3}$
• D. ${\left(\frac{R^2{\omega }^2}{g}\right)}^{1/3}$

Answer 1

The correct option is B ${\left(\frac{R^{2}g}{{\omega }^2}\right)}^{1/3}$

Satellite orbitary with a orbital velocity $v_o$ radius of orbit r.

$v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{g{R}^2}{r}}$

Time period $=\frac{2\pi r}{v_o}=\frac{2\pi r}{{\left(\frac{g{R}^2}{r}\right)}^{1/2}}=\frac{2\pi r^{3/2}}{\sqrt{g{R}^2}}$

As, $T=\frac{2\pi }{\omega }$

$\therefore \frac{2\pi }{\omega }=\frac{2\pi r^{3/2}}{\sqrt{g{R}^2}}$

$r^3/2=\frac{\sqrt{g{R}^2}}{\omega }$

$r^3=\frac{g{R}^2}{{\omega }^2}$

$r={\left(\frac{g{R}^2}{{\omega }^2}\right)}^{1/3}$.