Question

Derive an expression for acceleration due to gravity at depth below the earth’s surface.

Answer 1

The universal law of gravitation by Sir Isaac Newton gives the force of gravity between two attracting bodies. The gravitational force exerted by the earth can also be calculated using it. However, the shape of the planet is that of a geoid and not a perfect sphere, therefore, it does not exert equal force at all points on it and the force varies with depth and height of the object.

Complete step by step answer:
According to the universal law of gravitation, the force of gravity on any object is given as:
$\vec{F}=G\dfrac{mM}{{{r}^{2}}}$
Where,
$G=$ universal gravitational constant
$m=$ mass of body on which gravitational force is acting
$M=$ mass of body exerting the gravitational force
$r=$ distance between the centre of masses of two bodies
When we calculate gravity at the surface of the earth, $r=$ radius of the earth.
Also, force = mass $\times$ acceleration, $\Rightarrow \vec{F}=m\vec{a}$
Therefore, $m\vec{a}=G\dfrac{mM}{{{r}^{2}}}$
$\Rightarrow \left| {\vec{a}} \right|=\dfrac{GM}{{{r}^{2}}}$
$\Rightarrow g=\dfrac{GM}{{{r}^{2}}}$
Where, $g=$ acceleration due to gravity
Let $M=$ mass of earth and $r=$ radius of earth
And let ${{g}_{d}}=$ acceleration due to at depth, $d$from earth’s surface
We assume earth as a sphere of uniform density.
Hence, $\rho =$ the density of the earth
Let $B$ be a point inside the earth at depth $d$ from the earth’s surface
$\therefore OA-OB=d$
$\Rightarrow OB=r-d$
$\left( \because OA=r \right)$ -------- (1)
$\Rightarrow g=\dfrac{GM}{{{r}^{2}}}$ -------- (2)
Since, $density=\dfrac{mass}{volume}$
$\Rightarrow \rho =\dfrac{M}{\dfrac{4}{3}\pi {{r}^{3}}}$
$\Rightarrow M=\left( \rho \right)\dfrac{4}{3}\pi {{r}^{3}}$
Substituting value of mass of earth in equation (2);
$\Rightarrow g=\dfrac{G\left( \rho \right)\dfrac{4}{3}\pi {{r}^{3}}}{{{r}^{2}}}$
$\Rightarrow g=G\left( \rho \right)\dfrac{4}{3}\pi r$ ------- (3)
Now, ${{g}_{d}}=\dfrac{G\left( {{M}_{d}} \right)}{{{\left( OB \right)}^{2}}}$
Where,
${{M}_{d}}=$ mass of the earth (sphere) with radius OB
$\Rightarrow {{g}_{d}}=\dfrac{G\left( \dfrac{4}{3}\pi {{\left( OB \right)}^{3}}\rho \right)}{{{\left( OB \right)}^{2}}}$
$\Rightarrow {{g}_{d}}=\dfrac{4}{3}\pi G\left( OB \right)\rho$
From equation (1),
$\Rightarrow {{g}_{d}}=\dfrac{4}{3}\pi G\rho \left( r-d \right)$ -----(4)
Dividing equation (4) by equation (3):
$\Rightarrow \dfrac{{{g}_{d}}}{g}=\dfrac{\dfrac{4}{3}\pi G\rho \left( r-d \right)}{G\left( \rho \right)\dfrac{4}{3}\pi r}$
$\Rightarrow \dfrac{{{g}_{d}}}{g}=\dfrac{r-d}{r}$
$\Rightarrow {{g}_{d}}=g\left( \dfrac{r-d}{r} \right)$
$\therefore {{g}_{d}}=g\left( 1-\dfrac{d}{r} \right)$
Therefore, the acceleration due to gravity at depth $d$ below the earth’s surface is $g\left( 1-\dfrac{d}{r} \right)$.

Note: The force of gravity decreases as we go deeper into the earth and finally tends to become zero as we approach the centre of the earth if we assume uniform density of the earth. The reason it is zero is because there is equal mass surrounding the object in all directions so the gravity is pulling the object equally in all directions causing the net force on the object to be equal to zero.