Question

The pressure caused by gravitational pull inside the earth at ‘a’ distance, a measure from the center, when mass and radius of earth are ‘m’ and R’ respectively, is:
A: ${\displaystyle \frac{8G{m}^2}{3\pi R^4}}\left(1+{\displaystyle \frac{a^2}{R^2}}\right)$
B: ${\displaystyle \frac{8G{m}^2}{3\pi R^2}}\left(1-{\displaystyle \frac{a^2}{R^2}}\right)$
C: ${\displaystyle \frac{3G{m}^2}{8\pi R^4}}\left(1-{\displaystyle \frac{a^2}{R^2}}\right)$
D: ${\displaystyle \frac{8G{m}^2}{3\pi R^4}}\left(1-{\displaystyle \frac{a}{R}}\right)$

Answer 1

Hint: The gravitational pull below the surface of the earth also varies with the distance(it varies linearly) from the center. The pressure will vary too from the centre to the surface of the earth.

Complete step by step answer:
For all bodies below the surface of the earth the pressure opposes the weight of the body.
Since the body is now below the surface the value of acceleration due to gravity also changes.
And pressure will be the sum of all small pressures opposing the small masses.
So since we know that pressure opposes the weight of the body we have;
$dp=\rho g_{r}dr$
Here dp is the small pressure opposing a small mass of length dr.
Also $g_r$ is the acceleration due to gravity as a function of r.
Now we know that for a boy under the surface of the earth the acceleration due to gravity is given by;
$g_r={\displaystyle \frac{Gmr}{R^3}}$ here, G is the universal gravitational constant, r is the distance from the center, m is the mass of the body and R is the radius of earth.
Also th density of the body will be;
$\rho ={\displaystyle \frac{m}{{\displaystyle \frac{4}{3}}\pi R^3}}$
Hence, $dp={\displaystyle \frac{3G{m}^{2}r}{4\pi R^6}}dr$
So, $P=\underset{a}{\overset{R}{\int }}{\displaystyle \frac{3G{m}^{2}r}{4\pi R^6}}dr$
Hence $P={\left[{\displaystyle \frac{3G{m}^2{r}^2}{8\pi R^6}}\right]}_a^R$
$P={\displaystyle \frac{3G{m}^2}{8\pi R^4}}\left(1-{\displaystyle \frac{a^2}{R^2}}\right)$
Hence option C is correct.

Note: The concepts of gravitation and electrostatics are quite similar, so while finding the electric field as we use Gauss Law, for gravitational field problems too we can use Gauss law equivalent to find for example the field inside the earth.