Question:

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

Updated On: Jul 13, 2024
  • $ \frac{1}{T} $
  • $ \frac{1}{T^2} $
  • $ \frac{1}{T^3} $
  • $ T^{-2/3} $
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The Correct Option is D

Solution and Explanation

Velocity of satellite $v=\sqrt{\frac{G M}{r}}$ $\therefore KE \propto v^{2} \propto \frac{1}{r}$ and $T^{2} \propto r^{3}$ $\therefore KE \propto T^{-2 / 3}$
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Concepts Used:

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  • F ∝ (M1M2) . . . . (1)
  • (F ∝ 1/r2) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M1M2/r2

F = G × [M1M2]/r2 . . . . (7)

Or, f(r) = GM1M2/r2

The dimension formula of G is [M-1L3T-2].