Question:

The value of acceleration due to gravity $'g'$, at earth's surface is $10\, m / s ^{2} .$ Its value at the centre of the earth which is assumed to be sphere of radius $' R'$ and uniform mass density, is

Updated On: May 11, 2024
  • $2.5\, R\, m / s ^{2}$
  • $5\, R\, m / s ^{2}$
  • $10\, R\, m / s ^{2}$
  • 0
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The Correct Option is D

Solution and Explanation

Acceleration due to gravity at a depth h from the earth�s surface $\quad\quad g'=g\left(1-\frac{h}{R}\right)=g\left(1-\frac{R}{R}\right)=g\left(1-1\right)=0.$
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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].