Question:

Escape velocity for 'a projectile at Earth's surface is v,. A body is projected from earth's surface with velocity $2v_e$. The velocity of the body when it is at infinite distance from the centre of the Earth is

Updated On: Jun 8, 2024
  • $v_e$
  • $2 \, v_e$
  • $\sqrt{2} \, v_e$
  • $\sqrt{3} \, v_e$
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The Correct Option is D

Solution and Explanation

Let $v$ be the speed of the body when it is at infinite distance from the centre of the Earth and u be speed of projection of the body from the earth's surface
According to law of conservation of mechanical energy,
$\frac{1}{2}mu^{2} - \frac{GMm}{R} = \frac{1}{2}mv^{2} - 0 $
or, $v^{2} =u^{2} - \frac{2GM}{R} $
$v^{2} = u^{2} -v^{2}_{e} \left( \because \:\: v_e = \sqrt{\frac{2GM}{R}} \right) $
$ v = \sqrt{\left(2v_{e}\right)^{2 } - v_{e}^{2}} $
$= \sqrt{3} v_{e}$
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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].