# What happens to graviational force when separation between 2 objects is very small?

I recently learned about gravitational force and found out the equation for gravitational force on a object by an object according to Newton's law of universal gravitation.

$$F = \frac{Gm_1m_2}{r^2}.$$

$$r^2$$ denotes the square of separation between two objects. So the question is what happens when this separation is very small. For example when someone is standing on ground, separation between him and Earth is very low (zero).

Does gravitational force $$F$$ becomes ∞ then?

• For future readers, the example I took is very inaccurate to the question. Ignore the example. – adp7 Mar 6 at 18:09

The $$r$$ value in this equation represents the separation of the two bodies' centers of mass.

So, when you're standing on the surface of the earth, then the value of $$r$$ is equal to $$r_E$$, the radius of the earth, which is $$6378$$km or $$6.378\times10^{3}$$m.

To answer your more general question, the force between two masses does, indeed, increase as separation decreases, but as separation gets smaller and smaller, other forces start to dominate.

The electrostatic forces between molecules and atoms are millions of times stronger than the gravitational forces between them.

The strong nuclear forces between nucleons are even stronger still.

Still, if you have enough mass, gravity can overcome all of these forces. When a large enough star collapses to an ever-smaller point, you get a black hole. While Physics tends to try to avoid talking about infinities, the nature of a black hole is such that you can, in theory, get arbitrarily close to its center of mass.

However, at this point, you are lost to the universe and what happens inside a black hole, stays inside a black hole. Physicists are not really sure what happens there.

• thank you. This clears me the idea. – adp7 Mar 5 at 16:21

Although the question is interesting, the example you took seems to be a spin off. I will address the physical example first to show why gravitational force doesn't really blow up when really close:

As you know distance is measured between the centers of the masses. Can this distance be very close to zero?

Actually no because mass would become zero as well! Think of it. Mass exists in matter and matter has volume, hence to have significant mass you at least require some volume (even if you take the densest material in the universe). Now lets take your example of a man standing on earth. Since the distance is measured from the centers, here the distance is actually slightly greater than the radius of earth (6300km)! Okay, so now you decide to reduce this distance and dig a tunnel towards the center of earth, what happens then? The mass of earth that contributes to the gravitational force changes and is lesser than the actual mass!

The effective mass of earth is given by (assume earth to be spherical and the density to be same everywhere on earth): $$m_{eff}^{earth}=M_{earth}.\frac{r^3}{R^3_{earth}}$$ where $$r$$ is distance between the man and the center of earth. Clearly as $$r\to0$$, $$m_{eff}^{earth}\to0$$ faster than the denominator ($$r^2$$). Hence physically is not possible to make the gravitational force blow up by decreasing the distance.

Now, why is your question interesting?

It is interesting because this is exactly the dilemma that physicists face when they want to study the small scale/distance effects of gravity (they just cant go to small scales without making the gravitational force negligible), in fact the dilemma is so huge that we still dont have a microscopic theory for gravitation. The main reason for this being what you pointed out together with the fact that at small scales other fundamental forces hugely dominates any gravitational effects, making the gravitational force negligible.

• The idea that gravitational force can't be made to blow up by decreasing the distance is a very important physical notion. – YouFoundMe Mar 5 at 16:55