Question:
Assuming that the earth is a sphere of radius $R_{E}$ with uniform density, the distance from its centre at which the acceleration due to gravity is equal to $ \frac{g}{3} $ ($g$ = the acceleration due to gravity on the surface of earth) is
Assuming that the earth is a sphere of radius $R_{E}$ with uniform density, the distance from its centre at which the acceleration due to gravity is equal to $ \frac{g}{3} $ ($g$ = the acceleration due to gravity on the surface of earth) is
Updated On: Jun 8, 2024
- $ \frac{{{R}_{E}}}{2} $
- $ 2\,\,\frac{{{R}_{E}}}{3} $
- $\frac{R_{E}}{3}$
- $ \frac{{{R}_{E}}}{4} $
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The Correct Option is C
Solution and Explanation
For an $h$ depth below from the surface
we know that,
Given: $g'=g'=g(1-h / R)$
$g / 3=g(1-h / R)$
$\Rightarrow h=2 R / 3$ (depth from top)
Hence, from centre
$h'=R-h=R_{E} / 3$
we know that,
Given: $g'=g'=g(1-h / R)$
$g / 3=g(1-h / R)$
$\Rightarrow h=2 R / 3$ (depth from top)
Hence, from centre
$h'=R-h=R_{E} / 3$
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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].