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Instead of placing many magnetic satellites close to Mars to artificially generate a magnetic field to protect Martian colonists from harming radiation, we could place a magnetic satellite at Lagrangian point L1. The magnetic field does not need to be as strong to deflect the solar wind enough, since the deflection takes place over millions of km before reaching the planet. What would this require to be practical?

Nathan Tuggy
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Richard Beaudry
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    And I just did a quick calculation: sun wind = 400km/s, L1 is about 1000000km from Mars, so you need a magnetic field that induce a perpendicular speed of 1.4km/s. Lets assume that the interaction between the magnet and the ion particle takes place within 10km, you just need 1.7E-6 Tesla to deviate it! We can easilly make 20T magnets powered by solar panel. So the effect will affect a greater cross-section. It seems too easy. Could someone double check my calculation? – Richard Beaudry Mar 06 '16 at 04:40
  • Unfortunately a satellite at L1 isn't going to be enough. As Mark Adler indicates in this answer, the solar proton flux is pretty much omnidirectional at Mars. – Hobbes Mar 06 '16 at 19:33
  • What about protection from other cosmic radiation? Just curious. – Mikey Mar 02 '17 at 23:13
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    Dr Jim Green of Nasa's Planetary science division has proposed positioning a magnetic dipole shield at the Mars L1 LaGrange point to create an artificial magnetosphere to protect against solar wind and radiation. https://phys.org/news/2017-03-nasa-magnetic-shield-mars-atmosphere.html – johnM Jul 06 '17 at 06:06

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This answers the question on how to block ions coming directly from the Sun from hitting Mars by a satellite stationed at L1. It does not cover the fact whether such a shield is effective in reducing the radiation level on Mars surface. See e.g. here for more details on radiation.

First, let's have a look at the magnetic field needed. Inside a magnetic field (gray area), the particles follow a circular trajectory (green). The radius depends on magnetic field strength and particle momentum while the total deflection angle also depends on the length of the field. Trajectory

Using a distance estimate of 1 million kilometer between Mars and L1, and a Mars diameter of 6000 km we see that we need to deviate the solar wind by about 0.3°. Further assuming a thickness of the magnetic field of 10 km we see that the radius of the path of particles in the magnetic field needs to be about 1500 km. Now, we can use the formula for the magnetic rigidity $$B= \frac{m v}{q r}$$ and plug in the known values of mass and charge of alpha particles, we get a magnetic field strength of 5 nT - or about 10000 times weaker than the Earth magnetic field.

Sounds simple enough? Let's have a look at pure geometry. In the drawing all values are given in thousands of kilometers - not to scale for obvious reasons.

Shield size - geometry

Sun (yellow) diameter is about 1.4 million kilometer and we need to fully cover it. A shield (blue) positioned 1 million kilometers from Mars (red) that covers the sun completely has a diameter of: $$R = \frac{R_{sun}d_{L1}}{d_{sun}} = 6100~\rm{km} $$ But, we want to cover the whole surface of Mars (green shield), so we need to add another 3000 km in each direction, yielding a total diameter of 12000 km. That is equivalent to build a ring around the equator of Earth!

Unfortunately it is not enough to build a single coil with this diameter - the magnetic field needs to be perpendicular to the solar wind and the intended direction of deflection. That is, pointing away from the viewer, into the screen.

Dual coil setup

A single coil of reasonable size is not able to produce such a field that extends over thousands of kilometers along the axis of the coil. A simple way to provide it would be a pair of Helmholtz coils (orange). Again assuming a diameter of 12000 km (which is already beyond the validity of simple formulas as the distance between the two coils needs to be smaller than their diameter) and a strand of copper with a cross-section of 1 square meter, the whole setup would need a reasonable 10 MW of power but weigh about a billion tons, or in other words, the world-wide copper production of 50 years launched to space.

Note that it is not strictly necessary to use a cross-section of 1 square meter. We can always trade mass versus power needed due to increased resistance of a thinner copper "wire": Half the cross-section requires double the power.

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asdfex
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  • Thanks Asdfex to answer. Why do you need strand of copper with a cross-section of 1 square meter? – Richard Beaudry Mar 09 '16 at 02:59
  • Thanks Asdfex to answer. Why do you need strand of copper with a cross-section of 1 square meter? And, it does not require to be uniform magnetic field as in a pair Helmholtz coils. All we need to deviate the alpha particles is that at all point of the cross-section, B > 5nT and perpendicular to its speed. Also, in my first message, I set to 10km interaction distance because the particle speed is quite high. Strong magnetic field can propagate farther than that in solar system plasma. The following book mention radius of 120km with 5 E6 Amp *m/kg. – Richard Beaudry Mar 09 '16 at 03:38
  • https://books.google.ca/books?id=x8LgBwAAQBAJ&pg=PA87&lpg=PA87&dq=interstellar+spaceship+magnetic+braking&source=bl&ots=-CYKViBmTc&sig=UXAOlVyCzGB-q8TlC26APh1KWqg&hl=fr&sa=X&ved=0ahUKEwiRjL__0LLLAhXCGT4KHXBWD44Q6AEIRzAF#v=onepage&q=interstellar%20spaceship%20magnetic%20braking&f=false – Richard Beaudry Mar 09 '16 at 03:38
  • So to avoid the satellite to become a magnetic sail, it needs to rotate such that forces cancels out over a complete rotation. Than, let line up 3 such satellites 240km apart, the 2 at each end are rotating and centrifugal force is compensated by magnetic force. Such that they stay at the same constant distance. – Richard Beaudry Mar 09 '16 at 03:47
  • Depending on the magnetic field generated, we can at least protect a large area on Mars surface. We can synchronize the rotation such that every time the martian settlement is facing the sun(so at noon for their time) their shielding effect is maximum. – Richard Beaudry Mar 09 '16 at 03:54
  • I added few sentences on cross-section and why to use a pair of coils. Note that we are not talking about the classical Helmholtz-setup but just two coils placed far apart. The field produced by them is far from homogeneous and far less than 5 nT as the interaction length is on the order of 10000 km and not just 10. I adjusted for this in my rough calculation of the coil setup but didn't mention it. I can't read the book you linked to, Google doesn't allow me to read these pages. @uhoh, yes I can try to make some pictures during the next days. – asdfex Mar 09 '16 at 22:59
  • Very nice calculations and thanks for sharing. But I hate to be the krytic. I think the calculations are bloated and we might need much smaller scheld. The diagram above for the Sun and Mars would be correct if sun and Mars were Flat bodies located at small distance. When you consider that Sun is round (spherical) then it becomes obvious that with a distance the solar wind intensity decreases. You can say the same for the Sun's brightness. If we follow the flat sun model above then we can see that its intensity would be uniform independent of distance. Just for example consider the visible Sun – Joy of woodworking hobby tony9 Mar 02 '17 at 19:22
  • diameter on Earth surface which is smaller than the Moon. So the calculations of needed Coil size could be more than three orders of magnitude bigger then we need, making it feasible. Can someone do the calculations again, taking into account that solar wind intensity drops with distance? – Joy of woodworking hobby tony9 Mar 02 '17 at 19:22
  • Intensity drops with distance, but this has nothing to do with the Sun being spherical or circular. I just drew it that way to make the graphics simpler. Also, the size of the magnetic field does not depend on the intensity, just on the energy of the individual particle, which doesn't change with distance. – asdfex Mar 03 '17 at 09:00
  • Instead of a single, 12000km diameter coil, couldn't there be many small coils orbiting around L1? – matthias_buehlmann Apr 02 '22 at 10:48
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The very idea was proposed earlier this month at the Planetary Science Vision 2050 Workshop.

The talk was titled: "A Future Mars Environment for Science and Exploration", J. Green, R. Bamford, et. al.

They combined heliophysics simulation tools with a Mars global climate model to do the simulation. Their abstract suggests placing a 1-2 Tesla dipole at L1 is indeed sufficient to protect and rebuild a Martian atmosphere. However, the scale is not indicated. The full talk specifies 50,000nT measured at one Earth-radius--which is comparable to regenerating Earth's magnetic field and clearly a challenge.

Few details were given on how to generate the artificial magnetosphere field, but in was noted that the field lines must remain parallel with the interplanetary field to function well as a shield.

enter image description here

You can see the full talk here (starts at 1:36:10).

Martin McCormick
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    This is not an answer to the question! The linked material may contain relevant information, but in that case the post must include it. You can edit your post by clicking at the 'edit' button. – SE - stop firing the good guys Mar 30 '17 at 07:02
  • How big of a 1 Tesla "dipole"? What size? If it's defined in units of Tesla, it should be a coil with some area specified. A 1 Tesla coil with a 1 liter volume will not protect a planet! However, you may want to consider answering the question Is NASA doing research on “mini-magnetospheres” to protect crew from radiation in space? which is more "behavioral" than quantitative! – uhoh Mar 30 '17 at 13:14
  • Exactly. Their simulations show they need 50,000nT measured at 1 Earth-radius (roughly same as Earth's magnetic field). So this is a huge, Earth-scale field with comparable strength. – Martin McCormick Mar 30 '17 at 16:30
  • Thanks Martin, very instructive presentation. I'm glad that someone is simulating it to study the global effect. It's a baby step in the herculus path for terraforming Mars.

    Thanks all for considering the challenge. The question is now how to make a stable bipole with that strenght in stable orbite at L1. It is a challenge because of magnetic sailing effect.

    Best regards

    R.B.

     – Richard Beaudry Mar 31 '17 at 23:36
  • @asdfex Do you have thoughts on this? (Sorry for the 2+ year bump ;p) – T3db0t Feb 18 '19 at 17:23
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It is not necessary for the artificial magnetosphere to block the radiation. It is sufficient to merely deflect the ions by 0.006 radians in any direction. Maybe 0.01 radians to be safe. The L1 point is a million km from Mars, and Mars has a diameter of 6000 km.

This is a much smaller field than would be required to enclose Mars in a magnetotail.

Perhaps worth considering building one of these for Earth L1 too. A shame to destroy the Northern Lights, but then we don't want to lose our power grid. Maybe spin it up on demand when a storm is detected.

Ken Seehart
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