Satellite Period and Orbital Speed – Rankers Physics
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Satellite Period and Orbital Speed

Assertion (A): Period of revolution of satellite in circular orbit around earth is inversely proportional to cube of its orbital speed. Reason (R): Period of revolution in uniform circular motion is given by \( T = \frac{2\pi r}{v} \) where \( r \) is radius of orbit and \( v \) is speed.
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Solution:

Assertion (A) is true. For a satellite in circular orbit, orbital speed \( v = \sqrt{\frac{GM}{r}} \) implying \( r \propto \frac{1}{v^2} \). The period is \( T = \frac{2\pi r}{v} \). Substituting \( r \), we get \( T \propto \frac{1/v^2}{v} \propto \frac{1}{v^3} \). Reason (R) is true. The formula \( T = \frac{2\pi r}{v} \) is the correct definition for the period of uniform circular motion. However, (R) is a kinematic definition and does not explain the dynamic relationship between \( T \) and \( v \) for a satellite, which requires considering gravity. Thus, (R) is not the correct explanation of (A).

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