Question:

Find ratio of acceleration due to gravity $g$ depth d and at height $h$, where $d = 2h$.

Updated On: Jun 20, 2022
  • $ 1:1 $
  • $ 1:2 $
  • $ 2:1 $
  • $ 1:4 $
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The Correct Option is A

Solution and Explanation

Acceleration due to gravity decreases both at altitude and depth.
If $g '$ is the acceleration due to gravity at a point, at a height h above the surface of earth, and $g$ be acceleration due to gravity on earths surface, then
$g'=g\left(1-\frac{2 h}{R}\right)\,\,\,...(i)$
If $g$'' be the acceleration due to gravity at a point at depth $d$, below the surface of earth, then
$g''=g\left(1-\frac{d}{R}\right)\,\,\,.... (ii)$
But $d=2 h$ (given)
$\therefore g''=g\left(1-\frac{2 h}{R}\right)\,\,\,...(iii) $
$\therefore \frac{g''}{g'}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1$
$ \therefore g'': g'=1: 1$
Note The value of acceleration due to gravity, also decreases due to rotation of earth.
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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].