11
$\begingroup$

What is the most powerful magnetic field that could be generated by a planetary object? The measurement of 'powerful' will be field strength ($\mathbf{B}$-field) at the planet's surface, in teslas.

Restrictions

  • The planet may have any composition (gas giant, rocky, methane-ball, exotic)
  • The planet must be sub-stellar; that is, too small to start fusion.
  • The planet may be at any stage of development
  • The planet must generate this magnetic field itself, not induced from a nearby magnetar or something exotic.
  • The planet must be able to orbit a Sun-like star.
  • The planet must reasonably be able to form naturally
$\endgroup$
  • 1
    $\begingroup$ How do you determine planetary surface for a gs giant? $\endgroup$ – Gary Walker Sep 9 '18 at 23:03
  • 1
    $\begingroup$ @GaryWalker How are the surfaces of Jupiter and Saturn defined? This is a solved problem. $\endgroup$ – kingledion Sep 9 '18 at 23:33
  • $\begingroup$ If a magnetar has somehow been captured in an orbit by a much larger star, does it count as a planet? $\endgroup$ – Dubukay Sep 9 '18 at 23:39
  • $\begingroup$ @Dubukay This fails the 'sub-stellar' and 'must be able to orbit Sol' restrictions. $\endgroup$ – kingledion Sep 9 '18 at 23:47
9
$\begingroup$

TL;DR-- I think even under ideal circumstances, a field strength of about $0.4$ Tesla would be the limit.

So first off, ArtificialSoul is probably correct that no one here is going to be able to give you anything resembling a super accurate answer. But like most problems in fluid dynamics, If I rapidly wave my hands fast enough I can give an extreme estimate that should give you an idea of the order of magnitude that's possible.

Now, the actual planetary dynamo of the Earth is due to the coriolis force acting on convection currents of molten iron, twisting them into spiral flows. Then, a complex feedback mechanism between current carried by these flows, previously created magnetic fields, and resistive dissipation combine to produce a somewhat stable magnetic field. This is an absolute nightmare to model. So, instead, I'm instead going to approximate these flows as a current density $$\mathbf{J} = J_0s\hat{\phi}$$ Where we are in cylindrical coordinates and $s$ is the coordinate telling you how far you are from the z axis. The current will follow this form until the radius $R$, at which point it will become $0$. $R$ in this case is the radius of the zone where the iron is molten-- the outer core in the case of earth. I am going to neglect the fact that the inner core is solid, because less work = good.

The reason I chose the form I did for the current density is because it's the same as the current density of a charged sphere azimuthally rotating at some angular velocity. Again, the real picture is much more complex, but this will give us a good upper bound because it assumes all the flows of molten metal are working together to generate a magnetic field, whereas in reality, different regions will be counteracting each other and just making a general mathematical mess.

Now, we will use Biot-Savart to calculate the magnetic field along the z axis, because I would assume that's where the magnetic field would be strongest. More Importantly, however, it's much simpler. Now, Biot-Savart states $$\mathbf{B} = \frac{\mu_0}{4\pi}\iiint_{current}\frac{\mathbf{J}\times\mathbf{r'}}{(r')^3}d\tau'$$ I've made one more simplification, which is that I'm ignoring magnetization of the molten iron and I'm simply using the permeability of free space. This is reasonably accurate though since molten iron is well above the curie temperature and thus doesn't act very magnetic.

Plugging in all our nasty expressions in with the proper coordinates and simplifying considerably, we end up with the following expression (as long as $z>R$): $$\mathbf{B}(z\hat{z})=\hat{z}\frac{J_0\mu_0R^2}{2}\int_{-1}^{1}du \bigg[ \frac{1-u^2+2(\gamma-u)^2}{\sqrt{1-2u\gamma+\gamma^2}}-2(\gamma-u)\bigg]$$ where $$\gamma = \frac{z}{R}$$ This seems really awful, but it's actually not too bad-- the integral is simply a function of $z/R$ that decreases like $1/z^3$ as you move far away. For clarification $z$ is the coordinate along the z axis where $z=0$ at the center of the planet. From this point on, $z$ will refer to this coordinate at the surface of the planet, since that is where the magnetic field is strongest.

The key points to take away from this is that $\mathbf{B}$ grows with increasing $J_0$ and $R$ given fixed $\gamma$, and shrinks with increasing $\gamma$. So, to find an upper bound for magnetic field strength, we want $J_0$ and $R$ to be as big as possible, and $z/R = 1$. I'm simply going to reverse engineer $J_0$ from Earth's magnetic field to get a rough order of magnitude estimate, which produces the result $J_0 = 1.4\times10^{-10} A/m^3$ (as always, this is a very rough estimate).

As for $R$, the biggest planet thus far discovered has a radius of about $1.25\times 10^8 m$, so we will use that value. Finally, the integral attains a maximum of $0.2666$ when $\gamma = 1$. Note that this would imply the entire planet is molten and carrying a current-- for planets that have a solid layer, $\gamma$ would be greater than one.

Putting all these values together, we get an upper bound of $$|\mathbf{B}| \leq 0.38 T$$ at the surface of the planet. This is assuming $J_0$ is roughly the same for any planet, which likely isn't true-- faster rotation would imply higher $J_0$, and complex feedback mechanisms between the magnetic field and the convection currents could cause significant deviation from this value. But regardless, this should give you decent idea of the most extreme magnetic fields a planet can have that retain a shred of scientific plausibility.

I don't have time right now to add more detail of my calculations, but if you're interested I can probably edit them in later. Hopefully this helps!

$\endgroup$
6
$\begingroup$

I think this question is not sufficiently answerable by current science.

We know that turbulent molten metal currents (as in earths core) has the ability to create and sustain magnetic fields.

We also know that there are certain particles that can enhance this effect.

With the added ingredient of suspended magnetic particles, the material has the ability to generate and manipulate magnetic fields up to five times stronger than those generated by liquid metal on its own, say researchers from Yale University.

In general, Magnetohydrodynamics is not a well understood field in astronomical scales. So it is unlikely anyone can calculate you any numbers.

There are a few other mechanisms that can give planets magnetic fields.

How about finding data then?

Well, I've found a little bit. Jupiter has a magnetosphere larger than our sun. The equatorial field strength is 776.6µT - which is 25 times stronger than earths.

All the other data I found was significantly lower than that. In the realm of 1nT to 1µT equatorial field strength.

I did not manage to find any data on bodies that are not within our solar system, though. Only about the planets and their moons within it.


Addendum:

Look at el duderinos answer. He makes a very good upper bounds estimation.

So there is a good side to all of this:
Nobody can disprove your magnetic field strength. So you are about as free as el duderino described. :D

$\endgroup$
1
$\begingroup$

Neutron stars have field strengths up to 65 TT (Tera-Tesla), how is a neutron star a planet?

Well:

  • a neutron star, or other stellar remnant, could in fact be part of a new star system partly composed of material from it's own formation, as well as other supernova debris. Said second generation system being of sufficient total mass and travelling in an appropriate direction to capture the old remnant as part of it's accretion disk.

  • in which case it may end up in orbit of a new star as a planet of exotic but perfectly natural origin.

  • the new star that eventually forms need be no bigger than Sol.

  • neutron stars produce their own magnetic fields because of their high rotational speed and exotic composition.

Possibly it violates the "sub-stellar" clause since it is heavy enough to undergo fusion if it was composed of normal matter but a neutron star by definition does not undergo fusion so I feel it's still a valid candidate.

Sorry if this doesn't quite match what you had in mind but I've always liked the idea of a stellar remnant as a planet, I mainly think in terms of a black dwarf, the concept works for any leftover chunk of a star big enough to be re-integrated at the second or third generation.

$\endgroup$

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.