Question:
Assuming earth to be a sphere of a uniform density, the value of $g$ in a mine $100\, km$ below the earth surface will be
Assuming earth to be a sphere of a uniform density, the value of $g$ in a mine $100\, km$ below the earth surface will be
Updated On: Jul 27, 2022
- $ 3.66\,m/s^{2}$
- $ 5.66\,m/s^{2}$
- $ 7.64\,m/s^{2}$
- $ 9.66\,m/s^{2}$
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The Correct Option is D
Solution and Explanation
Here $:$ depth $(d)=100\, km =10^{5} m$
Radius $R=6380\, km =6.38 \times 10^{6} m$
Using the relation
$g'=g\left(1-\frac{d}{R}\right)=9.8\left(1-\frac{10^{5}}{6.38 \times 10^{6}}\right)$
$=\frac{9.8 \times 62.8}{63.8}=9.65\, m / s ^{2}$
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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].