In a satellite, if the time of revolution is $ T $, then kinetic energy is proportional to

Question

In a satellite, if the time of revolution is $ T $, then kinetic energy is proportional to, $ \frac{1}{T} $, $ \frac{1}{T^{2}} $, $ \frac{1}{T^{3}} $, $ T^{-\frac{2}{3}} $

MARKS App · Accepted Answer

Suppose mass of satellite is M and radius of satellite is r . Velocity of satellite, v = G M r , K E ∝ v 2 ∝ 1 r and time period is T = 2 π r 3 G M ∴ T 2 ∝ r 3 , So, K E ∝ T - 2 3 .

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