# Solving for $r=0$ in an inverse square law

This occurred to me in thinking about Coulomb's Law for a spherical insulator, but obviously can be asked of any inverse square law force calculation. For example, all things on the surface of he earth are touching the earth.

Classical Newtonian mechanics says $F=\dfrac{Gm_1m_2}{r^2}$ but in the case of myself, my house, my dog, and all my neighbors $r=0$. But, clearly when $r=0$, then $F\neq0$.

It isn't zero (we aren't floating) and it isn't infinite (we aren't in a singularity, I think). Clearly there must be a value of the force when $r = 0$. However, no simple algebraic operation seems to produce a sane value. So, either we use calculus (limits, L'Hopitals, etc) or I'm missing something stupid.

• The "r" in question is the distance from the center of the earth, not from the surface of the earth. When you're on the surface of the earth, you're still a good deal away from its center. Feb 26, 2016 at 4:53
• maybe you are missing the basic that division by zero isn't defined. Feb 26, 2016 at 5:46
• Feb 26, 2016 at 10:29

The force law $F=\frac{Gm_1m_2}{r^2}$ holds, in general, only for two point masses. As it happens, if a body is spherically symmetric and you're strictly outside that body, then as regards the gravitational force it exerts (though not necessarily the force it experiences), it can be replaced by a point mass (of the same mass) at its centre.

In particular, if your mass is $m$ then the Earth's mass $M$ exerts a gravitational force $F=\frac{GMm}{r^2}$ on you, as long as you're not underground, where $r$ is the distance between you and the centre of the Earth: in particular, $r$ is at least as long as the Earth's radius $R=6300\:\mathrm{km}$, a long way from zero.

If you do want to go underground, then the force you experience changes. In particular, it turns out that the gravitational attraction from the spherical shells of rock above you precisely cancel out (because some bits point one way and others point the other way), and you're only attracted by the volume of rock underneath you, which scales down as $r^3$. The upshot is that the gravitational force underground goes roughly linearly with the distance $r$ to the centre of the Earth, $F=\frac{GMm}{R^3}r$ (assuming constant density!), and there is no singularity at the centre.

If you do have point masses, on the other hand, then the force law $F=\frac{Gm_1m_2}{r^2}$ does hold, and at $r=0$ there is indeed a singularity, and there's no getting around that. This causes all sorts of headaches, particularly in electromagnetism (where the Coulomb attraction is identical, and where we needed to invent renormalization to deal with the singularities) but in many ways these are only present on the theory side.

Unless, of course, you've got a pair of true point masses in your desk drawer, in which case you've got a promising future in physics ahead of you.

The idea that a uniform spherical charge/mass distribution is equivalent to a point charge/mass works for points at the surface and outside the spherical distribution and so the force on a mass outside the spherical distribution is proportional to the reciprocal of the distance of the charge/mass from the centre of the distribution squared ($\propto \frac{1}{r^2}$).

For a charge/mass inside the spherical distribution the force on the charge/mass is proportional to the distance of the charge/mass from the centre of the distribution ($\propto r$).

If there were point masses they would have infinite gravity at r=0. In reality there are no point masses with radius=0, so you can never be infinitely close to something's center of mass. In the center of a real mass like the earth the gravity is 0 (because of Newton's shell theorem).

• Ok, so in this case how can it be said in Couloumb's law that a spherical charge distribution acts exactly like a point charge? More to the point, how shall I calculate F on the surface of a spherical conductor, if F inside is zero? Feb 26, 2016 at 5:01
• It acts exactly like a point charge outside of the charge distribution, not inside. Feb 26, 2016 at 8:46

In the context of classical Newtonian mechanics, a point mass is always an approximation for an small object with a finite volume. The inverse square law is only valid outside of the mass distribution of the point-like object ; as @Farcher pointed out, the law varies linearly with distance to the center inside of the mass distribution.

Thus you never get to reach the $r=0$ part of the inverse square law. However, in the context of General Relativity, there exists such singularity of space time that are called black holes. Correct me if I'm wrong, but as far as I know GR only deals with what happens outside of this particular singularity point where the physical description breaks down.

You aren't actually touching the ground you're standing on. You're hovering a few pico/nanometers above the ground, suspended by electrical fields.

If you get enough gravity, you can get closer, and closer, and eventually collapse everything into a black hole. Classically, we could say you now have "infinite" gravity at a point, but modern physics prefers to say we're just not sure what happens there.

Either way, nothing in normal life ever gets close enough to test the r=0 hypothesis. And with quantum physics, the presumption is the inverse radius laws break down when you get really, really close to something. In fact, if you ever see a physics equation that has an infinite output for some input value, it's pretty safe to assume the equation diverges from reality near that point.