Question:

The escape velocity of a body from the earth is $v_{e}$ . If the radius of earth contracts to $ \frac{1}{4}th $ of its value, keeping the mass of the earth constant, the escape velocity will be

Updated On: Jun 6, 2024
  • doubled
  • halved
  • tripled
  • unaltered
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The Correct Option is A

Solution and Explanation

Escape velocity $v_{e}=\sqrt{\frac{2 G M}{R}}$ If $R'=\frac{R}{4}$ $v_{e}'=2 \sqrt{\frac{2 G M}{R}}$ Since, $G$ and $M$ are constant, hence $v_{e}'=2 v_{e}$
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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].