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Can a planet, star or otherwise have a magnetic field that is stronger or have more range than its gravity?

Muze
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    interesting question! – uhoh May 30 '19 at 04:40
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    Magnetar? "The magnetic field of a magnetar would be lethal even at a distance of 1000 km due to the strong magnetic field distorting the electron clouds of the subject's constituent atoms, rendering the chemistry of life impossible": https://en.wikipedia.org/wiki/Magnetar – jamesqf May 30 '19 at 16:44
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    Magnetic field and force have different units/dimensions and cannot be compared directly. – ProfRob May 31 '19 at 07:37
  • At what point does a black hole with charge and angular momentum become (super-) extremal? – JollyJoker May 31 '19 at 07:57
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    @Jamesqf Proton precession ...? – Russell McMahon May 31 '19 at 11:37
  • @uhoh remember that question of using a diamagnetic sail off a Sun spot? This is related to that, but off topic here. – Muze May 31 '19 at 19:11
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    While both electromagnetism and gravitational fields have infinite range, magnetic fields generally fall off as 1 / r^3 rather than 1 / r^2 for gravity. – J. O'Brien Antognini May 31 '19 at 23:03
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    It would seem to me that for a pair of ordinary permanent magnets, their gravitational interaction is negligible compared to their magnetic interaction. So if "or otherwise" is to allow for small objects at small distances, the answer would appear to be positive. – Marc van Leeuwen Jun 01 '19 at 06:57
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    @MarcvanLeeuwen Heck, is there anything stopping someone from just scaling up a bar magnet into a giant planet-sized bar magnet? – nick012000 Jun 01 '19 at 09:32

3 Answers3

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It depends on what object it's acting on. There are many objects, including stars, that have magnetic fields where Lorentz forces on charged particles like electrons and protons are stronger than the gravitational force on them.

Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough moving electron even here on Earth will receive a larger magnetic force than a force of gravity. This is how the Earth's magnetic field is able to contain charged particles in the Van Allen belts that its gravity could not contain.

uhoh
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Ken G
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    Excellent! +1 I totally forgot about the Lorentz force experienced by charged particles and just did plain old static magnetic force versus gravitational force. – uhoh May 30 '19 at 04:55
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    Nice answer to a somehow ill posed question – Alchimista May 30 '19 at 09:35
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    +1 also for pointing out the big difference. Gravity is unaffected by (discrete || <<< c) speed while Lorentz force is. – Mindwin Remember Monica May 30 '19 at 12:38
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    @Alchimista That is the Pearls >>> Sand the Stack runs on. Scoop the bottom of the ocean. Questions are like sand but the scooped section may contain pearls somewhere. A question can be measured by the quality of the answers it sparked. – Mindwin Remember Monica May 30 '19 at 12:40
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    @Mindwin Thank You very much. I actually put much thought in the question. How would you word it? – Muze May 30 '19 at 23:08
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    @muze I don't know. The question's purpose is trifold: to solve your doubts as the OP, spark experts to post their knowledge and work as a repository of knowledge for future users. The actual wording of the question is important to understand what is being asked and to help index the search engines, but beyond that feel free to phrase it as you like. I see no problems with this one. I can understand the problem and it is elegant in its simplicity. – Mindwin Remember Monica May 31 '19 at 15:03
  • @Mindwin I am glad I can draw new minds like your own and thank you and nice answer to Ken G. – Muze May 31 '19 at 18:19
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    @muze thanks for the compliment, but I am not exactly new. =) P.S.: Keep doing the awesome. – Mindwin Remember Monica Jun 01 '19 at 11:56
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Let's look at the proper magnetic force (as opposed to the Lorentz force on a moving, charged object described in @KenG's answer) on a specimen $S$ of magnetized material with mass $M_S$ as a way to try to compare. Let's arbitrarily assume it has a fixed, permanent magnetic moment $m_S$. We can't use iron because it will saturate too easily.

Then let's look at how the forces scale differently with distance

$$\mathbf{F_G} = -\frac{G M_S M}{r^2}\mathbf{\hat{r}} \tag{1}$$

$$\mathbf{F_B} =\nabla (\mathbf{m_S} \cdot \mathbf{B}(\mathbf{r})) \tag{2}$$

If we reduce these to scalar equations at a radius $R$ (assume $\mathbf{m_S}$ and $\mathbf{B}$ are parallel) assume all forces are attractive, and evaluate the potentials and their gradients on the equator of the body at it's physical radius $R$. Since the magnetic force on our dipole specimen drops off faster than the gravitational force, we have to evaluate the two at the closest physically possible distance:

$$F_G = \frac{G M_S M}{R^2} \tag{3}$$

$$F_B = \frac{3 m_S B_{r=R}}{R} \tag{4}$$

where our specimen is a distance $R$ from our field source, and it's moment $m_S$ is a magnetization of 1 Tesla times the volume of a 1 kg rare earth magnet, about 0.000125 cubic meters.

All MKS units, all rough, ballpark numbers with emphasis on strongest magnetic fields

Body             R (m)      M (kg)    B(r=R) (T)    F_G  (N)    F_B (N)    F_B/F_G
Earth            6.4E+06    6.0E+24   5.0E-05       9.8E+00     2.9E-15    3.0E-16
Jupiter          7.1E+07    1.9E+27   4.2E-04       2.5E+01     2.2E-15    8.8E-17
Neutron Star     1.0E+04    4.0E+30   5.0E+10       2.7E+12     1.9E+03    7.0E-10
Magnetar         1.0E+04    4.0E+30   2.0E+11       2.7E+12     7.6E+03    2.8E-09

So even for a Magnetar (see also 1, 2) a kind of neutron star with a very strong magnetic field), the magnetic force on our 1kg specimen of permanent magnet is only 3 parts per billion as strong as the gravitational force.

You might see a much more favorable ratio if you compared two subatomic particles at short ranges (e.g. 1E-15 meters) but for astronomical objects, gravity seems to win smartly.

uhoh
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  • I don't think your expression for magnetic force is correct. For a magnetic material it should depend on $B^2$. And if you are putting in $G$ and using SI units, then where is the $\mu_0/4\pi$? – ProfRob May 31 '19 at 07:45
  • @RobJeffries the word "magnetic" is an artifact from a previous version and I'll change it to "magnetized". The next sentence states that it's a permanent magnet with magnetic moment $\mathbf{m_S}$ (1 kg, density of about 8000 kg/m^2, 1 Tesla magnetization) and later I mention that we can assume $\mathbf{m_S}$ and $B$ to be parallel (or antiparallel) It's of course preposterous to put a magnet near the surface of a neutron star (unless it's in a General Products hull). I just want to show that gravity wins by a landslide. – uhoh May 31 '19 at 07:48
  • Comments are not for extended discussion; this conversation has been moved to chat. – called2voyage Jun 01 '19 at 22:52
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It isn't impossible, but the short answer is "no".

A gravitational field will accelerate all matter and energy equally while a magnetic field will only accelerate moving electric charges (other magnets).

The force due to gravity is proportional to the inverse square of the distance, and the force due to magnetism asymptotically approaches the inverse cube of the distance. At some critical distance the gravitational force will become stronger than the magnetic force.

Unless most of the large body was magnetic, even over the magnetic poles the magnetic field would probably be too low to levitate a typical magnet in the large body's gravitational field.

anonymous
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