Question:

The work done in shifting a particle of mass $m$ from the centre of the earth to the surface of the earth is (Where $M$ is the mass of the earth and $R$ is the radius of the earth)

Updated On: Jul 5, 2022
  • $- \frac{GMm}{R}$
  • $+ \frac{GMm}{R}$
  • $+ \frac{GMm}{2R}$
  • $- \frac{GMm}{2R}$
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The Correct Option is C

Solution and Explanation

Gravitational potential energy at the centre of the earth is $u_{i} = -\frac{3GMm}{2R} $ Gravitational potential energy at the surface of the earth is $u_{f} = -\frac{GMm}{R} $ Work done, $W = U_{f} - U_{i} $ $ = -\frac{GMm}{R} -\left(- \frac{3Gm}{2R}\right)$ $ W = - \frac{GMm}{R} + \frac{3GMm}{2R} $ $ = + \frac{GMm}{2R}$
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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].