I have a planet similar to Venus(similar size and atmosphere) that has floating continents in the atmosphere due to strong magnetic fields. If it is possible, what would the continents need to be made of, and how big could they be?

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    $\begingroup$ Related question: Can airborne floating/flying islands be scientifically possible?. Not quite a dupe, but close. $\endgroup$ Sep 12, 2019 at 6:27
  • $\begingroup$ Can you help with details on 1) Similar to Venus in terms of what? (size, material, atmosphere etc) 2) floating continents float on top of what? (magma, water, liquid gas etc)? $\endgroup$ Sep 12, 2019 at 8:08
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    $\begingroup$ Ow, my belt buckle is dragging me someplace! And I can't figure out where because my compass jumped out of my hand and flew away. $\endgroup$
    – user535733
    Sep 12, 2019 at 12:28
  • $\begingroup$ Are any other factors besides magnetism allowed, the lighter the material of your floating continents the less implausible it becomes, you might consider pumice rafts where the air pockets are filled with Helium or Hydrogen, Helium is probably best there if you want to avoid flammability issues, are any organisms allowed? perhaps some sort of 'coral polyp' analog that excretes Helium into pockets in its 'shell'. $\endgroup$
    – Pelinore
    Sep 12, 2019 at 13:36

3 Answers 3


To lift a continent the size of Australia on Venus, you'd need 13,682C of charge between continent and planet.

The only condition for a floating continent is that the force of the electromagnetic field is greater than the force of gravity. With this in mind,

$ F_{gravitational} < F_{electromagentic} $

and thus,

$ \frac{GMm}{r^2} < \frac{Q_1Q_2}{4\pi\epsilon_0r^2} {*} AreaOfSurface $

I've substituted in an electric field in place of a magnetic one (the effect is much the same- you get big amounts of repulsion). We assume the electrostatic force the planet exerts on the continent is the same in size from the continent to the planet. Now taking the mass of Venus to be $4.867x10^{24} kg$, the mass of Australia as $4.033x10^{16} kg$ and the area of Australia to be $7.692x10^{12} m^2$, we can rearrange the above equations to the following:

$6.67*10^{-11} * (4.867*10^{24}-4.033*10^{16})(4.033*10^{16}) < Q^2 * 7.692*10^{12} $

$ Q^2_{minimum} = \frac{1.440*10^{21}}{7.692*10^{12}} $

$ Q_{minimum} = 13,682 Coulombs $

Numerically, 13,682 Coulombs sounds OK. However, once you learn that a lightning strike is 15 Coulombs of charge, maintaining this kind of energy over the area you'd need to quickly becomes a task too great for modern technology to handle. Having this sort of energy as a planet-wide magnetic field would deep fry anything trying to live on the surface (consider for a start the impact this would have on electronics- how would anyone be able to live on this floating continent?).

If a floating continent is truly what your heart desires, maybe look into an artificial gravity source that pulls the continent away from the planet? Could be cool having a continent that's upside down relative to the rest of your world. It may not be that much more efficient, but the lack of magnetic fields would mean modern technology could still very much be embraced.

Happy worldbuilding!

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    $\begingroup$ This answer is worthy of a bounty. $\endgroup$ Sep 12, 2019 at 12:30
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    $\begingroup$ Your answer assumes levitation is due to electrostatic repulsion, but I think the question was about magnetic levitation, which is a separate effect that's due to the relative motion of charges in two objects and doesn't require that they have different net charges. There's an equation for magnetic levitation here though the paper from that lab here (pdf auto-download) says there is a "shape-dependent correction". $\endgroup$
    – Hypnosifl
    Sep 12, 2019 at 12:58
  • $\begingroup$ @Hypnosifl Thanks for pointing that out. As you say correctly, the equation you linked depends on the shapes of the object. I believe the upside-down triangle is an operator called the Laplacian, which relates the shape of an object to its second derivatives. Doing the 'true' calculations with geometry in mind is something I think I'll save for my PhD at Harvard. $\endgroup$
    – mcRobusta
    Sep 12, 2019 at 13:12

Here you find some examples of magnetic levitation. The only one that might suit your needs may be: 5. Repulsion between a magnet and a superconductor.

You can not levitate magnets in stable fields:

The stable levitation of magnets is forbidden by Earnshaw's theorem, which states that no stationary object made of magnets in a fixed configuration can be held in stable equilibrium by any combination of static magnetic or gravitational forces,. Earnshaw's theorem can be viewed as a consequence of the Maxwell equations, which do not allow the magnitude of a magnetic field in a free space to possess a maximum, as required for stable equilibrium.

Currently the only known way to do it is:

Diamagnets (which respond to magnetic fields with mild repulsion) are known to flout the theorem, as their negative susceptibility results in the requirement of a minimum rather than a maximum in the field's magnitude,. Nevertheless, levitation of a magnet without using superconductors is widely thought to be impossible. We find that the stable levitation of a magnet can be achieved using the feeble diamagnetism of materials that are normally perceived as being non-magnetic, so that even human fingers can keep a magnet hovering in mid-air without touching it.

Here is the example. It has a problem though:

One problem, though, is that if the magnetic field of the current flowing within the superconductor becomes large enough, the ceramic will drop out of superconductivity, even if it is cold. Large magnetic fields will destroy the superconducting state. So, there is always a balance between the temperature, the magnitude of the magnetic field due to the current, and the molecular structure in determining the suitability of the superconductor for a particular application.


This is not a complete answer because I haven't worked out how strong the magnetic field needs to be. One difficulty here is that a magnetic object cannot in general be stable in a static magnetic field. To keep this stable, you might want inside the continents giant superconducting sections. They could then use the Meissner effect or similar results which can result in stable positions. Edit see also this earlier question about this Can the Meissner effect explain very large floating structures?

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    $\begingroup$ The answer is very, very strong. Far beyond the strength of a planetary magnetosphere. There may be exceptional astronomical bodies capable of producing the required field strengths, but offhand I can't think of any. Artificial continents with superconductor strata might be plausible, though they might not be floating high. $\endgroup$
    – a4android
    Sep 12, 2019 at 8:54
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    $\begingroup$ "One difficulty here is that a magnetic object cannot in general be stable in a static magnetic field" This is true for ferromagnetism or paramagnetism according to Earnshaw's theorem, but diamagnetic levitation (which includes levitation of superconductors) isn't covered by this theorem and it can be stable, see here. From what I understand the difference is that ferromagnets and paramagnets in an external magnetic field try to align so they're attracted to the source, diamagnets align so they're repelled. $\endgroup$
    – Hypnosifl
    Sep 12, 2019 at 12:32

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