A satellite can be in a geostationary orbit around the earth at a distance $ r $ from the center. If the angular velocity of earth about its axis doubles, a satellite can now be in a geostationary orbit around the earth is its distance from the center is
(A) $ \dfrac{r}{2} $
(B) $ \dfrac{r}{2\sqrt{2}} $
(C) $ \dfrac{r}{{{\left( 4 \right)}^{\dfrac{1}{3}}}} $
(D) $ \dfrac{r}{{{\left( 2 \right)}^{\dfrac{1}{3}}}} $

Hint: For the geostationary satellites, the angular velocity of the geostationary satellites is equal to the angular velocity of the Earth. By using Kepler’s third law, we can find the relation between the angular velocity and the distance from the center.

Complete answer:
A geostationary satellite can be defined as the satellite having the orbital periodic time of $ 24\;hours $ which is similar to the Earth.
Hence, it moves with the same angular velocity as the earth. Due to this, the satellite appears stationary when viewed from the earth, hence it is known as the geostationary satellite.
Now, we know that the gravitational force on the satellite is balanced by the centripetal force on the satellite.
 $ \dfrac{m{{v}^{2}}}{r}=\dfrac{GMm}{{{r}^{2}}} $ , where $ m $ is the mass of the satellite, $ M $ is the mass of earth, $ r $ is the distance from the center of the earth, and $ v $ is the velocity of the satellite
 $ \therefore {{v}^{2}}=\dfrac{GM}{r} $
Now, we know the relation between linear velocity and angular velocity can be written as,
 $ v=\dfrac{2\pi r}{T}=\omega r $
Substituting this value in the equation.
 $ \therefore {{\left( \omega r \right)}^{2}}=\dfrac{GM}{r} $
 $ \therefore {{\omega }^{2}}=\dfrac{GM}{{{r}^{3}}} $
From this the relationship between the angular velocity and the distance is
 $ \therefore {{\omega }^{2}}\propto \dfrac{1}{{{r}^{3}}} $
For the given case, $ {{\omega }_{2}}=2{{\omega }_{1}} $
By taking the ratio,
 $ \therefore \dfrac{{{\omega }_{1}}^{2}}{{{\omega }_{2}}^{2}}=\dfrac{{{r}_{2}}^{3}}{{{r}_{1}}^{3}} $
Substituting the given data,
 $ \therefore \dfrac{{{\omega }_{1}}^{2}}{\left( 2{{\omega }_{1}}^{2} \right)}=\dfrac{{{r}_{2}}^{3}}{{{r}_{1}}^{3}} $
 $ \dfrac{{{r}_{2}}}{{{r}_{1}}}={{\left( \dfrac{1}{{{2}^{2}}} \right)}^{\dfrac{1}{3}}} $
Simplifying the powers,
 $ {{r}_{2}}=\dfrac{r}{{{4}^{\dfrac{1}{3}}}} $
Hence, the correct answer is Option $ (C) $ .

Note:
The relation derived here between time period and distance from the center of the earth is known as Kepler’s third law, which states “The square of the time period of the revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit.”