Charged particles such as protons are associated with the strong nuclear force and coulombic interactions, but as they're tiny, we can only perceive them through static. If a proton gets as big as a soccer ball - either by accumulating millions of protons or becoming - a single solid sphere of positive charge (charge it holds proportional to its size), would it cause destruction equivalent to an atomic bomb?

  • $\begingroup$ What is the mass or energy of that "proton"? The big radius just makes it a big target in this probabilitistic world ;D $\endgroup$
    – user6760
    Commented Jun 25, 2020 at 16:45
  • $\begingroup$ xkcd had a What-If with a similar premise, only there it was a moon made of electrons. what-if.xkcd.com/140 $\endgroup$
    – kutschkem
    Commented Jun 26, 2020 at 6:55
  • $\begingroup$ You are missing the fact that a soccer ball of pure protons would have a very very very large mass. Look up neutron star. $\endgroup$
    – RoyC
    Commented Jun 26, 2020 at 10:16
  • $\begingroup$ I looked stuff up on XKCD too and mass of proton (which is equal to a neutron, also looked up neutron star) so my object is going to be very heavy, DENSE! probably will crush everything with its gravity :( $\endgroup$
    – McGucket
    Commented Jun 28, 2020 at 4:40

3 Answers 3


The energy contained within this soccer ball-sized object is called the electrostatic self-energy of the object. It's the energy required to assemble all the infinitesimal chunks of charge together, and can be calculated by classical electrostatics: $$U=\frac{3kQ^2}{5R}$$ where $Q$ is the charge and $R$ is the radius. We can also write it in terms of the charge density $\rho_e$, and we get a dependence of $U\propto\rho_e^2R^5$.

Now, it's tempting to just use the proton's charge density and scale up the radius with that $R^5$ proportionality. The trouble is, a normal proton doesn't have a definite size in the same sense that a soccer ball has a definite size. We can try to characterize it by what we call a charge radius, a quantity which is important in scattering experiments. We reach the charge radius by setting the electrostatic self-energy of a sphere with the proton's charge equal to the proton mass: $$\frac{3ke^2}{5r_c}=mc^2$$ Let's say we want to define some sort of charge density $\rho_e$ for the proton, and have our supersized proton have the same charge density. Measurements show that $r_c\approx0.84\times10^{-15}\;\text{m}$, so if we multiply the proton's charge density by a factor of $(R/r_c)^5$, I get a self-energy value of roughly $U\sim6\times10^{57}\;\text{J}$, which is about 12 or 13 orders of magnitude greater than the energy released in a typical supernova. It corresponds to the mass-energy of about $\sim10^{10}M_{\odot}$, or about 1% of the Milky Way.

It turns out that when you pack an energy equivalent to 1% of the Milky Way into a FIFA regulation-sized soccer ball, it becomes dense enough that the escape velocity at its surface is substantially higher than the speed of light. In other words, your charged "proton" would turn into a black hole.

Such an object would not make a good soccer ball.

Realistically, this sort of supersized proton will not have the same charge density as a proton. I don't think the strong nuclear force would be able to bind it together; if you tried, electrostatic forces would easily resist you once you packed enough protons worth of charge together. We already see this sort of effect resist efforts to make heavier and heavier nuclei in atoms; it would be even worse on larger scales.

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    $\begingroup$ I think once you exceed the power of supernova, all explosions should just be measured in how far away it has to be to explode with the same apparent luminosity as an h-bomb pressed up against your eye: what-if.xkcd.com/73 $\endgroup$
    – Nosajimiki
    Commented Jun 25, 2020 at 22:02
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    $\begingroup$ With 6*10^57 J of energy, wouldn't this not explode at all, but instead just turn the 600 AU radius around it into a black hole? $\endgroup$ Commented Jun 25, 2020 at 23:19
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    $\begingroup$ Upvoted for "Such an object would not make a good soccer ball" (it was already worth an upvote, but that made me sign in so I could vote) $\endgroup$ Commented Jun 26, 2020 at 1:33
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    $\begingroup$ So to answer the question, no. A "charged particle as big as a soccer ball" would not "create destruction equivalent to an atomic bomb". This would be worse. $\endgroup$ Commented Jun 26, 2020 at 6:54
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    $\begingroup$ @Nosajimiki : if we are referring to xkcd, this is also relevant: what-if.xkcd.com/140 . It's basically the same scenario, just a few orders of magnitude bigger. (there the particle is planet-sized and its energy equivalent is rivaling that of the entire visible universe) $\endgroup$
    – vsz
    Commented Jun 26, 2020 at 8:05

If a proton became the size of a soccer ball it would mess with the strong nuclear force so badly that the universe as we know it probably could not exist.

If sufficient protons were somehow put into a soccer ball to "fill it up" it would be necessary to some how suspend electromagnetism in that space temporarily.

Assuming that was possible, the moment electromagnetism was switched back on the repulsive forces would be so large that it probably would generate forces like a nuclear bomb, although in this case more of an intense radiation bomb as countless trillions of protons were blasted out in all directions at extremely high velocity ionising the air for miles around and creating a huge electrical disturbance as the charge dissipated.

In fact it would effectively be a giant exploding nucleus made of protons and huge amounts of energy would be released as in fission from changes in binding energy. Probably much larger than a standard thermonuclear device

  • $\begingroup$ Yes. But what about the former version of charged particle? Proton/positively charged particle as big as a soccer ball. Would it rip surrounding atoms? Extracting free electrons from them until it forms a stable systems changing properties of surrounding or something I haven't thought?? $\endgroup$
    – McGucket
    Commented Jun 25, 2020 at 14:13
  • $\begingroup$ @McGucket your super-proton would rip electrons off the surrounding atoms and at the same time severely push away surrounding protons. To an extent until the charge is roughly neutralized (speaking of orders of magnitudes here). If I trust the numbers from the other answer, "surrounding" atoms may refer to way more than the solar system as a whole $\endgroup$ Commented Jun 26, 2020 at 8:43
  • $\begingroup$ @McGucket re the single proton the size of a football. The strong nuclear force would not allow a proton the size of a football You would need to rewrite nuclear physics and the result would depend on how physics was rewritten. $\endgroup$
    – Slarty
    Commented Jun 26, 2020 at 13:32

To Explode or not to Explode. That is the question.

I believe HDE's original hypothesis about a resulting super explosion is most likely correct when modeling the event using the laws of relativity, but this got way to long to just be a comment.

A proton nucleus the size of a soccer-ball will have a total volume of ~5.58e42 protons. Since the atomic mass of a Proton is ~1.67e-27 kg, this gives you a total mass of 3.34e15kg.

According to the acceleration of gravity equation: g=G*M/(R+h)^2, this would result in a gravitational force at the surface of the soccer-ball equal to ~1842km/s. This is much less than the speed of light which means that your soccer-ball will be able to explode with all of its 6e^57 J of glory instead of collapsing.

But E=MC^2 right?

Yes, this is generally true, but one of the funny things about particle physics is that not every particle that has energy also has mass. Even though massless particles are effected by gravity, gravity is only created by those particles which have mass meaning the 6e57 J of electromagnetic force does not automatically convert to an equivalent mass.

Because electromagnetism exchanges energy through massless photons, your hypothetical scenario means you will have an object with the energy of a super massive black hole but the mass of a large comet.

The Theory of Relativity justifies the existence of mass-less particles by assuming that the asymmetry caused by photons is always balanced out by the interactions of both positive and negative forces, but Special Relativity only requires balance, not balance within a local frame of reference. So creating the ball to begin with means that somewhere else in the universe you've formed an opposite and equal electron mass, and that you've already (somehow) expended the necessary energy to pull these electrons and protons apart and compress the ball down to it's current size.


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