Derive the formula of variation of small g with depth that is:
[tex]g ' = g(1 - \frac{d}{r} )[/tex]
where, g' is g at depth d.
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Derive the formula of variation of small g with depth that is:
[tex]g ' = g(1 - \frac{d}{r} )[/tex]
where, g' is g at depth d.
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Explanation:-
Let the radius of the earth be R and mass of the earth be M and density be ρ.
Mass of earth:-
=> Mass = Volume × Desnsity
=> M = 4/3πR³ρ ---(1)
We know that:-
=> g = GM/R²
=> g = [G×4/3πR³ρ]/R²
=> g = 4/3πGRρ ---(2)
If the value for acceleration due to gravity at the surface of the earth is 'g' and a depth 'd' from the earth's surface is g', then :-
=> g' = 4/3πG(R-d)ρ ---(3)
On dividing eq.3 by eq.2 , we get:-
=> g'/g = [4/3πG(R-d)ρ]/[4/3πGRρ]
=> g'/g = R-d/R
=> g' = g× R-d/R
=> g' = g(1 - d/R)
Thus, value of g decreases by a factor
(1 - d/R) as we go down below the surface of earth.
Some Extra Information:-
When a body is dropped from a certain height above the ground, it begins to fall towards the earth under gravity. The acceleration produced in the body due to gravity is called acceleration due to gravity . It is denoted by g. It's value close to the Earth's surface is 9.8m/s².
Verified answer
Solution:
Refer above attachment for understanding the derivation below
Consider the outer circle as earth and let mass be M and radius be R of earth
Point O is centre of earth and point M at depth d below is at distance of R-d from centre of earth. Consider acceleration due to gravity at depth d below surface of earth as gd at point M
As we know
[tex]g = \frac{gm}{ {r}^{2} } [/tex]
Here g in gm/r² is G (capital)
Let density be d1 and volume equal to
[tex] \dfrac{4}{3} \pi \: r {}^{3} [/tex]
So,
[tex]as \: mass \: = density \: \times volume \\ \\ mass \: = d \: \times \frac{4}{3} \pi \: r {}^{3} \\ \\ \\ thus \: \: \: \: \: \: \: g = g \times d \times \frac{4}{3} \pi \: r {}^{} \: (1)[/tex]
Here in above (1) we have substituted the value of m and cancelled r so it became only r in equation
So for gd ( acceleration due to gravity at depth d)
[tex]gd \: = \frac{gm}{ {(r - d)}^{2} } [/tex]
[tex]density \: = \frac{mass}{volume} \\ \: mass \: = density \: \times volume[/tex]
Volume of sphere with radius R- d is equal to
[tex] \dfrac{4}{3} \pi \: (r - d) {}^{3} [/tex]
and density be d1
Then,
Mass = Volume × Density
= 4/3 π ( r-d) ³ × d1
So,
gd = g × 4/3 π (r-d)³ ×d1 ÷ (r-d) ²
gd = g× 4/3 π (r -d) d1 equation (2)
Dividing equation 2 by equation 1 we get
[tex] \dfrac{gd}{g} = \dfrac{r - d}{r } \\ \\ = 1 - \frac{d}{r} \\ \\ gd = g(1 - \frac{d}{r} )[/tex]
Therefore, gd = g(1 - d/r) is equation of acceleration due to gravity at depth d
Hence, proved the equation
Extra information