AP EAMCET · PHYSICS · Gravitation
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 \(2^{\mathrm{n}}\) times the initial distance. Then the value of ' \(n\) ' is
- A \(\frac{2}{3}\)
- B \(\frac{3}{2}\)
- C \(\frac{1}{3}\)
- D \(\frac{4}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{2}{3}\)
Step-by-step Solution
Detailed explanation
\( m r \omega^2 = \frac{G M m}{r^2} \) \( r^3 = \frac{G M}{\omega^2} \) \( r \propto \omega^{-2/3} \) \( r' = r \left( \frac{\omega/2}{\omega} \right)^{-2/3} \) \( r' = r \left( \frac{1}{2} \right)^{-2/3} \) \( r' = r (2)^{2/3} \) \( n = \frac{2}{3} \)
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