Question

Derive the formula of variation of small g with depth that is:
$g ' = g(1 - \frac{d}{r} )$
where, g' is g at depth d.​

Answer 1

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.

Answer 2

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

$g = \frac{gm}{ {r}^{2} }$

Here g in gm/r² is G (capital)

Let density be d1 and volume equal to

$\dfrac{4}{3} \pi \: r {}^{3}$

So,

$as \: mass \: = density \: \times volume \\ \\ mass \: = d \: \times \frac{4}{3} \pi \: r {}^{3} \\ \\ \\ thus \: \: \: \: \: \: \: g = g \times d \times \frac{4}{3} \pi \: r {}^{} \: (1)$

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)

$gd \: = \frac{gm}{ {(r - d)}^{2} }$

$density \: = \frac{mass}{volume} \\ \: mass \: = density \: \times volume$

Volume of sphere with radius R- d is equal to

$\dfrac{4}{3} \pi \: (r - d) {}^{3}$

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

$\dfrac{gd}{g} = \dfrac{r - d}{r } \\ \\ = 1 - \frac{d}{r} \\ \\ gd = g(1 - \frac{d}{r} )$

Therefore, gd = g(1 - d/r) is equation of acceleration due to gravity at depth d

Hence, proved the equation

Extra information

1. At centre of earth d = r so the gd will become zero. Thus if body of mass m is taken to centre it's weight will become zero but mass never become zero
2. g is maximum at surface of earth
3. g goes on decreasing above the earth and below surface of earth