What is the Gravitational force between 2 bodies, when their centres of mass coincide? [duplicate]

According to Newton's Universal Law of Gravitation,

$F=G{\frac{Mm}{r^{2}}}$, where

$M$ and $m$ are the masses of the two bodies,

$F$ is the force of attraction between them,

$r$ is the distance between the centers of the 2 masses, and

$G$ is the Universal Gravitation Constant.

However, if the distance between the centers of the 2 bodies is $0$, i.e., their centres of mass coincide with each other, we would have $r = 0$.

Putting this in the formula would give,

$F = G\frac{Mm}{0^2}=G\frac{Mm}{0} = \infty$

Maybe, an example of this can be, when a terrestrial body is taken to the center of the Earth.

How can this be true? And if yes, how can this be practically explained? Isn't gravity supposed to Become $0$ at the earth's core?

• This has to be a duplicate. That said, $F=GMm/r^2$ applies to point masses, and by Newton's shell theorem / Gauss's law of gravitation, to objects outside of an object with a spherical mass distribution. Your question requires a six-fold integration of the force between all infinitesimal mass elements in object A to all of the infinitesimal mass elements in object B. – David Hammen Sep 19 '18 at 15:41

The gravitational force each feels would be zero (in the case of symmetric spheres).

This is because of the Shell Theorem. In short this means that if you're anywhere inside a spherically symmetrical shell you experience no net force from the shell on you. As a sphere is just a set of shells, then the net force is zero if you're at the center of the sphere.

The shell theorem also means that the net force you experience at some radius $r$ inside a sphere of radius $R$ is only from the mass inside your own radius, not all of the body.

Now if the bodies aren't quite symmetrical, things will be different as the contributions from shells won't quite cancel, but the net force will be small if the asymmetry is small.

If you're concerned about the problem that the distance between the centers of mass is zero and so $\frac 1 {r^2} \to \infty$, then note we can rewrite the acceleration due to gravity from a sphere of average density $\sigma$ as :

$$g = \frac {GM(r)}{r^2} = \frac {G4\pi r^3 \sigma}{3r^2}=\frac 4 3 G\pi \sigma r$$

So that you can see as $r\to 0$, the net acceleration felt is also zero and not infinite as you might think from the division by $r^2$.