# Why is the strong nuclear force > electrostatic repulsion?

In a nucleus there is a gravitational force between the nucleons and also electrostatic repulsion between the protons, and since electrostatic repulson >> gravitational attraction, it follows that there must be an additional attractive force acting on the nucleons or else there is nothing stopping them from flying apart.

So if we let the gravitational and electrostatic forces be $g$ and $e$, respectively, and denote the additional attractive force by $x$, then we would need $g+x=e$ (because the attractive and repulsive forces must balance).

This gives $x=e-g<e$, implying that the additional attractive force must be less than electrostatic repulsion. Since the strong nuclear force must be part of $x$, we then have strong nuclear force $\le x<e$.

But in my revision guide, it says that the strong nuclear force is more than electrostatic repulsion, which seems counter-intuitive according to the above.

• "then we would need $g+x=e$ (because the attractive and repulsive forces must balance)" You are thinking of nucleons like little billiard balls and that model isn't really appropriate in this intrinsically quantum realm. That equality does not hold. – dmckee May 7 '15 at 18:57
• @dmckee in the Feynman lectures, Feynman said that the nuclear force can be approximated as $F=(1/r^2)exp(-r)$. It seems to me that although it falls off really fast, it has infinite range. So how come it has short range? – Omar Nagib Jul 13 '15 at 17:01