The intention of the question is to provide an example of the weakness of gravity.

I imagine a horseshoe magnet located at the Earth's centre (remove the Earth), and a ferromagnetic moon. How intense would the magnetic field need to be to keep such a moon in orbit at the same distance that the moon is from the Earth?

  • $\begingroup$ You mean you want to replace Earth's gravity with magnet? You would need to produce the same acceleration as Earth's gravity at Moon's centre. $\endgroup$ May 23, 2014 at 12:40
  • $\begingroup$ Yes, exactly. How intense the magnetic field to account for Earth's gravity pull? $\endgroup$
    – harogaston
    May 23, 2014 at 12:44
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    $\begingroup$ I guess this would not work. Maybe you can reproduce the force at one instant, but dynamically there is a huge difference between the (monopole) gravitational field and a (dipole) magnetic field. $\endgroup$
    – Neuneck
    May 23, 2014 at 13:00
  • $\begingroup$ Well that was in my mind also, that is why I stated horseshoe magnet, I thought it might play the role. $\endgroup$
    – harogaston
    May 23, 2014 at 16:24

1 Answer 1


This is a more complicated problem than you may realize.

For electrostatic attraction the statement is nice and neat. You want the gravitational force $$ F_g = G\frac{M_\text{earth}M_\text{moon}}{r^2} $$ to be equal to an electrostatic force $$ F_e = \frac1{4\pi\epsilon_0}\frac{q_\text{earth}q_\text{moon}}{r^2}. $$ If you ask that the charge ratio $q_\text{moon}/q_\text{earth}$ be equal to the mass ratio $M_\text{moon}/M_\text{earth} \approx 1/80$, you find $$ q_\text{earth} = M_\text{earth}\sqrt{4\pi\epsilon_0 G} \approx 5\times10^{13}\,\mathrm C. $$ This is a lot of charge, but it's not a lot of charge. It's about $5\times10^8$ moles of fundamental charges, which is only 500 tons of extra protons, or a quarter-ton of extra electrons.

If you want to do a magnetic force, the problem is a lot thornier. The earth's magnetic field would be $$ \vec B = \frac{\mu_0}{4\pi} \frac1{r^3} \left( (\vec m_\text{earth} \cdot \hat r)\hat r - \vec m_\text{earth} \right) $$ where $\vec m_\text{earth}$ is the earth's dipole moment, $r$ is the distance from the dipole center, and $\hat r$ is a unit vector pointing away from the dipole center. In order to have a constant field over the orbit of our magnet-moon, it must orbit around the earth's magnetic equator. The force on the moon is $$ \vec F = \vec \nabla (\vec m_\text{moon}\cdot\vec B) = m_\text{moon} \vec\nabla |B| $$ where can make the second approximation only if we demand that the moon's dipole moment always be parallel to the local magnetic field. Luckily for us this is the way that the moon's dipole wants to align: two dipoles in the same plane want to orient antiparallel and come close to each other.

In this very restricted case, the total force is $$ \vec F = -\hat r\frac{3\mu_0}{4\pi} \frac{m_\text{earth}m_\text{moon}}{r^4}. $$ Setting the two dipole moments $m$ equal to each other and the force equal to the gravitational force, we find $$ m = r\sqrt{G M_\text{earth} M_\text{moon} 4\pi/3\mu_0} = 4\times10^{23}\,\mathrm{A\,m^2} $$ This is a huge dipole moment! Suppose you wanted to make this with an electromagnet that was the same size as the Earth. The cross-section of the Earth at the equator is $\pi R_\text{earth}^2 \approx 10^{14}\,\mathrm m^2$; to make this dipole moment you'd need to wrap a million turns of wire around the equator and run 4000 amps through each turn! Plus you'd need the same magnetic dipole moment at the moon.

I haven't addressed what I consider a detail in your question, that the moon would be an induced magnetic dipole, because ferromagnetism is another layer of messiness. I can tell you with confidence that you wouldn't be able to induce $m_\text{moon} = m_\text{earth}$, so you'd need an even bigger dipole at the earth to make the product $m_\text{moon}m_\text{earth}$ come out right. It might be that the strength of the induced dipole would also depend on the earth-moon separation, in which case it's quite possible that there would be no stable orbit at all.

It's also worth pointing out again that if we relax our assumptions that the lunar dipole is exactly antiparallel the the terrestrial dipole, and that the orbit takes place exactly in the plane of the equator, we lose the "nice" $1/r^4$ force. I have no idea whether this "nice" orbit is stable either.

What's happening here is that the magnetic force is a second-order effect of electromagnetism. It makes a huge difference that the force between two dipoles falls off like $1/r^4$ instead of $1/r^2$.

It occurs to me that it'd make more sense to compare this magnetic dipole to another astrophysical magnetic field, rather than to a laboratory field made of coils and currents. Typically the Earth's natural magnetic field is about 50 μT at the surface (about half a gauss). If the dynamo generating the "earth's" magnetic field were smaller than the radius of the earth, so that we could use the dipole field approximation at the surface, the field I've computed above would have strength $\sim 10^{3}$ T at one $R_\text{earth}$ from the center. This is essentially the same field strength as a magnetar: a magnetar may have a surface field of $10^8$–$10^{11}$ T, but they also typically have radii of $10^{-3}R_\text{earth}$.

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    $\begingroup$ Does the $1/r^4$ nature of the magnetic force create any issues with orbit stability? I vaguely remember hearing something related to that in one of my classes. Can't remember enough specifics to know if that was actually what my teacher meant. $\endgroup$ May 23, 2014 at 14:53
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    $\begingroup$ @Joshua I had the same memory. I think you only get closed elliptical orbits for forces proportional to $r^n$ where $n$ is $+1$ or $-2$; for other such forces you get pericenter precession, even if you only have two interacting bodies. The moon's perigee precesses like crazy because of interactions with the rest of the solar system, so I left that detail out. $\endgroup$
    – rob
    May 23, 2014 at 15:15
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    $\begingroup$ Thank you for your answer I can see you put some effort on in. I'll need to do some studying before I can fully understand it, so I'll keep the question open in the mean time. Anyways, I thought you may appreciate some feedback. $\endgroup$
    – harogaston
    May 23, 2014 at 16:33
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    $\begingroup$ @rob, for electrostatic attraction or repulsion, would ions work just as well (just a lot heavier)? Say, to counteract the Sun's increasing luminosity (and the effect of man's greenhouse gas emissions), could you effectively weaken solar gravity and move the Earth farther away by dumping X tons of positive ions on the Earth, and equal amount into the Sun, and discard the electrons somewhere in space? Would that be quickly neutralized or shielded in some way? Just a thought experiment... $\endgroup$
    – Phil Perry
    May 23, 2014 at 17:03
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    $\begingroup$ @PhilPerry The near-vacuum of the interplanetary medium is a weakly conducting plasma, so I think it's impossible to give any celestial object a permanent net charge. I could expand more in answer to a separate question. $\endgroup$
    – rob
    May 23, 2014 at 17:17

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