Question

In the case of earth, mean radius is 'R', acceleration due to gravity on the surface is 'g', angular speed about its own axis is 'ω'. What will be the radius of the orbit of a geostationary satellite?

पर्याय

• ${\displaystyle {\left[\frac{{\omega }^2}{{\text{gR}}^2}\right]}^{\frac{1}{2}}}$

• ${\displaystyle {\left[\frac{{\omega }^2}{{\text{gR}}^2}\right]}^{\frac{1}{3}}}$

• ${\displaystyle {\left[\frac{{\text{gR}}^2}{{\omega }^2}\right]}^{\frac{1}{2}}}$

• ${\displaystyle {\left[\frac{{\text{gR}}^2}{{\omega }^2}\right]}^{\frac{1}{3}}}$

Answer 1

${\displaystyle {\left[\frac{{\text{gR}}^2}{{\omega }^2}\right]}^{\frac{1}{3}}}$

Explanation:

${\displaystyle \text{mr}{\omega }^2=\frac{\text{GMm}}{r^2}}$

${\displaystyle r{\omega }^2=\frac{\text{GM}}{r^2}}$

${\displaystyle r^3=\frac{\text{GM}}{{\omega }^2}=\frac{\text{GM}}{R^2}\times \frac{R^2}{{\omega }^2}=g\frac{R^2}{{\omega }^2}}$

${\displaystyle \therefore {\left[\frac{{\text{gR}}^2}{{\omega }^2}\right]}^{\frac{1}{3}}}$