# Is it possible to revive the Mars geodynamo? [closed]

I’ve done a bit of reading and have found multiple theories suggesting how the Mars geodynamo stopped fully working.

This site suggests that one possibility could be volcanic activity ejecting water and other necessary elements from the mantle significantly slowed the natural convection process.

Where as this site says that it also could have been massive asteroids colliding with the surface, heating the planet up to temperatures that would disrupt convection.

No one is quite certain what happened, all that is known is Mars doesn’t have much of a magnetic field at this point.

Today I saw that there’s an idea to create an artificial field that’ll act in place of a geodynamo’s naturally generated one.

My main concern for this is that the units creating the artificial field could be destroyed and the entire planet would become unstable afterwards.

It would seem if the magnetic field is supported by the planet itself, it would offer a higher degree of stability and less involvement for humans in the long run.

What’s the possibility for ‘restarting’ the Mars geodynamo or perhaps ramping it up enough to support a strong enough field? And if so, how would we be able to do it?

• Equivalent question on Space Exploration: space.stackexchange.com/questions/2423/… – user24157 Apr 23 '20 at 8:37
• @antispinwards That question is certainly related, but there's a lot of speculation and low quality info on that page... – PM 2Ring Apr 23 '20 at 9:54
• This question might be more on-topic on Space Exploration. – dalearn Apr 23 '20 at 12:34
• This question might be even more on topic on World Building. – David Hammen Apr 23 '20 at 13:27
• Unsurprisingly there are also questions on the subject of giving Mars a magnetic field on World Building. E.g. worldbuilding.stackexchange.com/questions/8832/… or worldbuilding.stackexchange.com/questions/30208/… – user24157 Apr 23 '20 at 13:58

Let's do a bit of math: magnetic energy density is at least $$u = \frac{1}{2}\frac{B^2}{\mu_0}$$. The magnetic field at the surface of Earth is of the order of 30µT which gives us a magnetic energy density in the order of $$10^{-3} J/m^3$$ Now we need to fill the entire volume of the planet and some surrounding space with this field in order to get some protection from the solar wind; let's assume a cube of $$(20000km)^3 = 8\cdot 10^{21}m^3$$. So that means that we need an energy in order of $$10^{18}W$$ to sustain a comparable field which equates to roughly 1 million state-of-the art GW nuclear power plants. We currently have significantly less than 1000 in the whole world combined and I have not considered losses of any kind (thus 100% conversion of power into magnetic field)

My rough calculation also does not include any additional power needed to sustain the field extent into space when subject to the magnetic pressure of the solar wind which will be needed in order to keep the atmosphere inside the magnetic field, especially in direction of the sun.

So please go and figure whether these scenarios are really realistic. Actually reviving any dynamo on Mars is an altogether different story and would need to understand how exactly it works/worked; it might entail to heat up the core which requires probably more energy and will pose even more severe technological challenges than a giant dipole magnet of planet size.

• Your calculations make no dimensional sense, they go from energy density to an enormous number for power without involving time at any point. No power input is required to maintain a static magnetic field. – Christopher James Huff Apr 26 '20 at 20:03
• @ChristopherJamesHuff But geodynamo's are understood to be electromagnets caused by convection of conductive fluid and not static magnetic fields. So it would require non-zero energy to mainint. I.e. the power from a planet's remnant heat is the energy source for geodynamos? – CognizantApe Feb 4 at 21:33
• @CognizantApe Power is required to maintain the electrical and fluid currents, not the resulting magnetic field. Those power requirements have basically nothing to do with the energy density*volume of the field divided by some unstated arbitrary amount of time. – Christopher James Huff Feb 4 at 21:40

I think a superconducting cable system around the Martian equator is far more easy. To get an Earthian $$\rm B$$, which is $$\approx 4 \mu T$$, from a Martian radius ($$\approx \rm{3000 km}$$), we would need $$I=\frac{2\pi r B}{\mu_0} \approx 680 \rm{MA}$$. It is about 6000 times of the ITER main coiling, elongated throught the whole Martian equator.

This might be even much smaller, depending on the magnetic permeability of the planet. While it currently does not have a magnetic field due to the lack of the convection in the mantle, it is likely magnetizable, because its iron content is probably higher than of the Earth(ref).

• The mere fact that this is "far more easy" speaks volumes. – James K Apr 23 '20 at 19:50
• @JamesK It can be done with today technology. Asteroids in the required size won't visit the Mars by us. – peterh Feb 4 at 23:15