Question

If the intensity of the gravitational field at all places inside the Earth is presumed to be constant, then the relation between the density of the Earth ($\rho $) and the distance (r) from the centre of the Earth will be?
(A) $\rho \propto r$
(B) $\rho \propto \dfrac{1}{r}$
(C) $\rho \propto \sqrt r $
(D) $\rho \propto \dfrac{1}{{\sqrt r }}$

Answer 1

Hint:
The gravitational intensity depends on the Earth and the distance at which it is being calculated. The mass is also related to the volume and the density of the Earth.
Formula used: $E = \dfrac{{GM}}{{{r^2}}}$, where E is the gravitational intensity, M is the mass and r is the distance at which the intensity is being calculated. G is the universal gravitational constant.

Complete step by step answer:
The intensity of the gravitational field at any point uses the force experienced by any object placed at any point due to the Earth’s gravitation. In this question, we are asked to determine the relation between the density of the Earth and the distance of this object from the centre of the Earth provided the gravitational intensity remains constant.
We know that the gravitational intensity is given as:
$\Rightarrow E = \dfrac{{GM}}{{{r^2}}}$
We also know that the mass depends on the density as:
$\Rightarrow M = \rho V$
Here, $\rho $ is the density of the Earth, and V is the volume encapsulated in the given distance r. This volume will be the volume of a sphere of radius r. Hence,
$\Rightarrow V = \dfrac{4}{3}\pi {r^3}$
The gravitational intensity will become:
$\Rightarrow E = \dfrac{{G\rho V}}{{{r^2}}} = \dfrac{{G\rho \dfrac{4}{3}\pi {r^3}}}{{{r^2}}}$
Solving it further gives us:
$\Rightarrow E = \dfrac{{G\rho 4\pi {r^3}}}{{3{r^2}}} = 4\pi \dfrac{{G\rho r}}{3}$
Now, since this intensity is constant, and so is G, we can rearrange the above equation to get a relation as:
$\rho \propto \dfrac{1}{r}$.
Thus, the correct answer is option (B).

Note:
We saw that the gravitational field intensity depends inversely on the radius of the sphere, so where is it maximum in context of the whole Earth. Since the Earth is an oblate spheroid, it bulges out in the middle which is at the equator. That is why the poles are a little closer to the centre of gravity. Thus, the intensity of gravity is the maximum at the poles as compared to the equator on the surface of the Earth.