Question

If the angular velocity of a planet about its axis is halved, the distance of the stationary satellite of this planet from the centre of the planet becomes $2^n$ times the initial distance. Then the value of $n$ is

• A. $\dfrac{2}{3}$
• B. $2$
• C. $\dfrac{1}{3}$
• D. $\dfrac{4}{3}$

Hint:

For stationary satellites, $R \propto \omega^{-2/3}$ helps relate radius to angular velocity.

Answer 1

The Correct Option is A

Stationary satellite condition: $\omega^2 R^3 = GM$
$\Rightarrow R \propto \omega^{-2/3}$
If $\omega$ is halved: $\omega' = \dfrac{1}{2} \omega$
Then $R' = R \cdot \left(\dfrac{1}{2}\right)^{-2/3} = R \cdot 2^{2/3}$
$\Rightarrow n = \dfrac{2}{3}$