3
$\begingroup$

Let's say we have two planets. Both are around the same mass and radius as Earth, and are 1 year away from colliding with each other, each traveling at 20 km/s. At first. Planet 1 has a sufficient enough imbalance of protons and electrons to have a net positive charge of 1 megacoulomb, and Planet 2 has a net negative charge of 1 megacoulomb. How will their collision differ from two different planets of the same mass, radius and velocity that have both a net charge of 0? And if 1 megacoulomb each isn't enough to cause a noticeable difference, then what will?

$\endgroup$
2
  • $\begingroup$ Kudos to the super-being that can aim planets so accurately across such distances despite all the intermediate perturbations. $\endgroup$ – user535733 Dec 14 '20 at 19:23
  • $\begingroup$ This honestly seems better-suited to Physics.SE. $\endgroup$ – jdunlop Dec 14 '20 at 21:24
5
$\begingroup$

Your planets won't be approaching each other nearly fast enough for anything to happen other than forming a bunch of hydrogen atoms. When a proton and electron come near each other, they start orbiting each other and you get atomic hydrogen. You'd have to shoot them at each other at relativistic speeds in order for them to merge and become a neutron and an electron neutrino.

So, the only way for anything interesting to happen is to throw these planets at each other at relativistic speeds. The difference in electric charge could affect their speeds towards each other near the end, but a net difference of two megacoulombs across an entire planet isn't enough to affect their speeds at all. A megacoulomb isn't even that much charge, it's only about 2x the surface charge on Earth right now.

How much charge WILL cause a difference? More than enough to be realistic, even in a sci-fi setting.

$\endgroup$
3
$\begingroup$

1 Megacoulomb each?

so.... absolutely normal planets? As other people have noted 1 Megacoulomb on a planetary scale is... absolutely negligible. It is 1.04 surplus electrons per Kilogram of mass of the Earth.
Coulomb = 6.241 x 10^18 charges.
MegaCoulomb = 6.241 x 10^24
Mass of Earth = 5.972 × 10^24 kg

The electrostatic charge between two planets would only get "interesting" when the force between the two planets approach the same force as gravity, or when the energies in the charge approach their gravitational binding energy(enough energy to disassemble the planet). Ironically, you do not need a second planet to demonstrate either of these, as each planet will start to flee itself.

This happens when the charge on the planet is about 4.5e14 Coulomb (450 million megacoulomb) At this point a 1kg chunk of the planet's surface will experience a repulsive force of 9.8N from the electrostatic charge of the planet, causing it to exactly counter gravity. As it raises up, the static repulsion decreases, but the gravity decreases at exactly the same rate. The planet acts as if it in not subject to gravity!

As for an arc between the two planets.. This is only really possible when they are very close. Lightning cannot arc through a vacuum, at best electrons emitted by the one pole will be pulled in a cloud to the other pole. The rate of electrons available for this is governed not by voltage but by the thermionic emission rate of the pole. So very little action until they get to within less than 2 radii from each other, and start sucking matter between the planets. (whether by Gravity or Electrostatic attraction). At this point the electric charge between the two will rapidly neutralize.

For the abovementioned case where electric charge is very close to gravitational binding energy.... The two planets will disappear in a rapidly expanding cloud resembling a stellar nova, as the combined energy of the electric discharge and the kinetic energy of collision is much higher than their binding energy.

$\endgroup$
3
$\begingroup$

If the net charges are limited to 1 megaCoulomb, there will be no effects on either planet. The ratio of electrostatic force to gravitational force is $$\frac{F_e}{F_g}=\frac{kQ_1Q_2}{GM_1M_2}\approx4\times10^{-18}$$ meaning that the electric attraction will be almost 18 orders of magnitude lower than the gravitational force. You can also run the numbers on each individual planet and find, similarly, that it's not possible for a charge of 1 megaCoulomb to tear a planet apart; the gravitational binding energy is just way too high.

For the two to even be within two orders of magnitude (i.e. $F_e/F_g>10^{-2}$), each planet would have to hold a charge of $\sim10^{14}$ Coulombs. That would require a charge imbalance of either $\sim10^{33}$ extra protons or electrons on each planet, which is still tiny compared to the mass of a planet. This seems quite unlikely to happen - I don't know of any mechanism that could strip away that much charge. I'd assume it would require ionizing atoms in the atmosphere and then allowing electrons to escape hydrodynamically or through some sort of handwavy magnetic fields, but freeing bound electrons at the requisite level seems difficult. (I'm also tempted to say "something something stellar winds", but that's really no better an explanation for ionization.$^{\dagger}$)

Were that to happen, I suspect you'd see a difference once the two atmospheres were able to touch, as suggested by PcMan. If much of the extra charge was distributed in the upper atmosphere, and the two planets were approaching each other at a somewhat realistic speed of $\sim20$ km/s, then there would be a period on the order of a couple of seconds between the atmospheres beginning to merge and the surfaces colliding. In that brief timeframe, it's quite likely that you'd see lightning and similar extreme electrical effects in the atmosphere.

At this point, assuming that the $\sim10^{33}$ extra protons on one planet combine with the $\sim10^{33}$ extra electrons on the other, there would be intense emission from recombination. If each electron were to immediately fall to the ground state, we'd see $$10^{33}\times13.6\;\text{eV}\approx10^{15}\;\text{Joules}$$ which Wolfram Alpha says is about five times the energy released by the impact forming Meteor Crater in Arizona.$^{\ddagger}$

This is significant in everyday life, but insignificant compared to the energy released by two colliding planets. You'd also see a significant amount of energy coming from the electrostatic potential energy of each planet - but again, this wouldn't be comparable to the energy released by the collision. Remember, of course, that this requires some absurd process to create the charge imbalance in the first place.


$^{\dagger}$ That's a slight lie - I can't rule out a process involving stellar winds. I also can't rule out something using ultraviolet ionizing radiation from the parent star, which is important for planets orbiting M dwarfs. On the other hand, this radiation typically doesn't result in such dramatic charge imbalances!

$^{\ddagger}$ In reality, this won't happen instantaneously - or even directly to the ground state - but instead quite slowly. Recombination to the ground state would release a photon of energy 13.6 eV, which could go on to ionize another atom, so recombination to excited states would be the relevant processes, just as in H II regions. As the recombination timescale is much longer than the ~1 second between when the atmospheres touch and the planets collide, the atmospheres would quickly be heated up, which in turn would lower the recombination coefficients, which usually go as $T^{-3/4}$. So we'd see a burst of recombination, followed by a sharp decrease, and depending on what happens to the atmosphere following the collision, it could still take some time for the majority of the electrons and protons to recombine.

$\endgroup$
2
$\begingroup$

The issue here is local charge distribution vs. global charge neutrality. Planets, and astrophysical bodies in general, must have approximate global charge neutrality. This is because the immense repulsive nature of the electrostatic force from net charge accumulation on a microscopic scale, where as the gravitational force is attractive, with effects on the macroscopic scale.

For example, in the seminal work of Eddington the approximate net charge of the Sun was determined to be on the order of 77 C. Recent considerations have reiterated this conclusion A&A 372, 913–915 (2001). This rules out large global charges on the order of Mega Coulombs. For planetary bodies, this is especially compounded by the low energies (in comparison to stellar media) within planetary bodies. The Earth for instance is electrically neutral with large charge imbalances between the surface and various layers of the atmosphere. Large global net charge imbalances cannot accumulate in the interior of solid and liquid objects. Attempts to calculate the total net charge of the Earth arrive at values much less than that of the Sun, for example one such estimate arrived at a value near $|\Delta Q| \approx$ 1 C J. Biemond viXra (2010) . Whereas the local charge distributions occur on the order of 10$^5$ C in the atmosphere.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.