Can a magnetic field of an object be stronger than its gravity?

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

interesting question!

Gravity and electromagnetism both have infinite range.

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

Magnetic field and force have different units/dimensions and cannot be compared directly.

At what point does a black hole with charge and angular momentum become (super-) extremal?

@Jamesqf Proton precession ...?

@uhoh remember that question of using a diamagnetic sail off a Sun spot? This is related to that, but off topic here.

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.

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.

@MarcvanLeeuwen Heck, is there anything stopping someone from just scaling up a bar magnet into a giant planet-sized bar magnet?

• uhoh Correct answer

3 years ago

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_GEarth            6.4E+06    6.0E+24   5.0E-05       9.8E+00     2.9E-15    3.0E-16Jupiter          7.1E+07    1.9E+27   4.2E-04       2.5E+01     2.2E-15    8.8E-17Neutron Star     1.0E+04    4.0E+30   5.0E+10       2.7E+12     1.9E+03    7.0E-10Magnetar         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.

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 $\mu_0/4\pi$?

@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.

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