Question:

The earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is :-

Updated On: Jul 5, 2022
  • 44563
  • $\sqrt 2 $
  • $ \frac{1}{\sqrt 2} $
  • 44564
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The Correct Option is C

Solution and Explanation

According to question and by using COME $-\frac{ GMm }{ R + R }+\frac{1}{2} m ( fv )^{2}=0+0$ $\Rightarrow fv =\sqrt{\frac{ GM }{ R }}$ but $ v =\sqrt{\frac{2 GM }{ R }}$ Therefore $f \sqrt{\frac{2 GM }{ R }}=\sqrt{\frac{ GM }{ R }}$ $ \Rightarrow f =\frac{1}{\sqrt{2}}$
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Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

  • 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].