Question:
Assuming earth to be sphere of uniform density what is the value of acceleration due to gravity at a point $100\, km$ below the earth surface (Given $R=6380\,\times 10^{3}\,m$)
Assuming earth to be sphere of uniform density what is the value of acceleration due to gravity at a point $100\, km$ below the earth surface (Given $R=6380\,\times 10^{3}\,m$)
Updated On: Jul 28, 2022
- $3.10\, m/s$
- $ 5.06\,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 $=100\, km =100 \times 10^{3} m$
Radius of earth $R=6380 \times 10^{3} m$
Acceleration due to gravity below the earth's surface is given by
$g,=g\left(1-\frac{d}{R}\right)$
$g'=9.8\left[1-\frac{100 \times 10^{3}}{6380 \times 10^{3}}\right]$
$=9.8\left[1-\frac{1}{63.8}\right]$
$=9.8 \times \frac{62.8}{63.8}=9.66\, m / s ^{2}$
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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].