The force repulsing two protons never goes away, meaning that there is a constant force pushing the protons apart forever. Firstly, where are these protons getting the energy to constant apply this force, and where is the nucleus getting the energy to continue to hold them together? Is it the binding energy? If so does it eventually run out of energy to hold them together?
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1$\begingroup$ You are talking of classical mechanics and electrodynamics which do not apply at the sizes of nuclei and protons. Quantum mechanics was postulated as a theory to model the behavior at that level, $\endgroup$– anna vCommented Jun 14, 2023 at 4:31
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2$\begingroup$ To add to anna's comment, the assumption that it takes energy to generate a force is false in classical mechanics, already. Take the example of two magnets. They can "stick together" indefinitely without "running out of energy". What requires continuous work is to perform a continuous displacement against a force, but no such displacement is happening in stable configurations. $\endgroup$– FlatterMannCommented Jun 14, 2023 at 4:35
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1 Answer
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There is another force between protons which is stronger that electromagnetism, but it has a finite range, so it disappears over length scales longer than a nucleus. This force is named, by people who apparently had zero imagination, the "strong nuclear force."
The electromagnetic interaction between two pointlike unit charges is represented by a potential
$$ V_\text{e.m.} = + \frac{\alpha_\text{e.m.} \hbar c}{r} $$
Here $r$ is the distance between the pointlike charges. The constant $\hbar c = 197\rm\,MeV\,fm$ relates the energy and length scales, and the dimensionless constant $\alpha_\text{e.m.} \approx 1/137$ represents the strength of the electromagnetic interaction.
The long-range part of the strong potential is the "Yukawa potential,"
$$ V_\pi = -\alpha_s \frac{\hbar c}{r} e^{-r/r_\pi} $$
where the length scale is related to the mass of the $\pi$ meson, $r_\pi = \frac{\hbar c}{m_\pi c^2} \approx 1.4\,\rm fm$. There are separate Yukawa potentials associated with the other mesons, but longest-range part of the potential comes from the low-mass pion. The scale for the coupling constant $\alpha_\pi$ is about 1/10, which is much larger than the electromagnetic coupling constant.
Combining these two potentials means that, for nearby separations $r \lesssim \text{few}\cdot r_\pi$, the overall proton-proton interaction is attractive. Well, the overall nucleon-nucleon interaction is attractive: you can't neglect the role of the neutron. However, when two protons are many $r_\pi$ apart from each other, the exponential in the Yukawa term totally kills the strong attraction, and only the electromagnetic repulsion remains. This is the basic reason that the periodic table ends: by the time you get to uranium and the heavy actinides, the protons on opposite ends of the nucleus start to have important competition between their strong attraction to the rest of the nucleus compared to their electrical repulsion.
(For a list of caveats about this approach, see the first several chapters of your favorite nuclear physics textbook.)
