Question:

What will be the acceleration due to gravity at a depth d where g is acceleration due to gravity on the surface of earth?

Updated On: Jul 7, 2022
  • $\frac{g}{\left[1+\frac{d}{R}\right]^2}$
  • g${\left[1-\frac{2d}{R}\right]}$
  • $\frac{g}{\left[1-\frac{d}{R}\right]^2}$
  • $g{\left[1-\frac{d}{R}\right]}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Acceleration due to gravity at the surface of the earth $g=\frac{GM}{R^2}=\frac{4}{3}\pi \rho GR$ ...(i) Acceleration due to gravity at depth d from the surface of earth $g'=\frac{4}{3}\pi \rho \pi G(R - d)$ ....(ii) From Eqs. (i) and (ii) $g'=g\left[1-\frac{d}{R}\right]$
Was this answer helpful?
0
0

Top Questions on Gravitation

View More Questions

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].