Question

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

• A. ${\left(\frac{2R^{2}g}{{\omega }^2}\right)}^{1/3}$
  
• B. ${\left(\frac{R^{2}g}{{\omega }^2}\right)}^{1/3}$
  
• C. ${\left(\frac{R^{2}g}{3{\omega }^2}\right)}^{1/3}$
  
• D. ${\left(\frac{R^{2}g}{5{\omega }^2}\right)}^{1/3}$
  

Answer 1

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


Time period of rotation of earth$=\frac{2\pi }{\omega }$
(Duration of one day)

Geostationary satellite has same time period, $T=\frac{2\pi }{\omega }$.

Let $r$ be the radius of orbit of satellite.

So, the time period of satellite $=\frac{2\pi r^{3/2}}{\sqrt{G{M}_{earth}}}$

As, $g=\frac{G{M}_{earth}}{R^2}$

$\Rightarrow T=\frac{2\pi r^{3/2}}{R\sqrt{g}}$

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

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

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

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

Hence, option (b) is the correct answer.