Mars has only the ancient remnants of a magnetic field. What little chunks of field it does have (imprinted into magnetic rocks) are regional in scale and do nothing at all for radiation shielding (I once calculated this). Additionally, over a long enough timescale (tens of millions of years), the solar wind will erode the atmosphere of a terraformed Mars. So, let’s just get on with replacing the lost magnetic field.
To do this, we aren’t going to do something silly like restart the core. We’re going to rely on a tried and true existing technology: cryogenic superconductors. Just put a superconducting ring around the equator of Mars. Turns out, this wouldn’t even cost that much to build. I will be using Magnesium Diboride (MgB2) because it’s cheap, has pretty good performance (critical temperature of 39K, critical current of like 105 A/cm2 at 5K), and both Boron and Magnesium are known to exist fairly commonly on Mars. (Magnesium is super common, and Boron has been found in clays at >150ppm concentrations–according to this paper–and probably exists in far higher concentrations somewhere since Mars once had a quite active water cycle. I’ll assume that by the time the residents of Mars want to do this sort of thing, the costs of extracting these minerals will be similar to that of Earth (maybe a poor assumption, but why would you do this unless there are like millions of people on Mars, at a minimum?). We’ll also assume that because of the ridiculous scale being operated at, and because MgB2 is a pretty easy superconductor to make (just need to heat the mixture of Magnesium and Boron powders), the cost of actually building the ring will be some single-digit multiple of the raw costs of the material.
Okay. So just how big is the Earth’s magnetic field? We’ll use its total energy (when approximated as current on a sphere) to estimate what we’ll need as far as current in the ring. According to this, the Earth’s magnetic field stores about 1026 erg, or 1019J (roughly half the energy the world uses in a year). The energy stored in an inductor is just:
(according to Wikipedia)
Where L is inductance and I is the current in the ring.
To calculate the inductance L of a ring of radius R with wire radius a and number of turns N, we use the approximation:
Since N=1 and we’ll conservatively (very conservatively) say the relative permeabilityÂ Â , and since the current I is related to the critical current density Jc such that: , we can write the equation as:
If we let a=.42m, R=rMars, and Jc = 105 A/cm2:
https://www.google.com/webhp?#q=.5*r_Mars*(mu_0)*(ln(8*r_Mars/(.42m))-2)*(10^5A/cm^2*pi*(.42m)^2)^2 = 1.047*10^19 Joules, when we only needed 10^19 J to equal the same energy as Earth’s magnetic field.
Given the density of MgB2 is 2.57g/cc (source), the mass of the superconductor is:
https://www.google.com/search?q=2*pi*r_Mars*pi*(42cm)^2*2.57g/cc or about 3*1010kg, 30 million tons, almost have of which is boron. The Earth mines about 4 million tons of Boron a year, so the Earth produces enough boron to build that thing in about 4 years (we’ll mine this on Mars, of course). Pretty reasonable, considering we’re doing some pretty hardcore terraforming, here.
Given a price of about 10USD/kg for Boron (just spitballing here, since Ferroboron is half boron by molarity and 15-20% boron by mass and is 1-2USD/kg… of course, boron ore is much cheaper) and like 2USD/kg for magnesium metal (just look up the spot prices for Mg and Ferroboron), so about USD6/kg of bulk MgB2.
This whole thing would cost about 180B USD in raw materials but would store about 3 trillion kWh for a ridiculously low price per kWh of storage (like 6 cents/kWh! For storage that can be reused!). Of course, there is also insulation and cooling, plus some method to inject power into the ring.
NOTE: Some Japanese researchers recently (last 2-3 years I think?) published a paper about the thing I am proposing here. I can’t find the paper now, but I assume they did I better job than I did. Also, I don’t really buy into Jim Greene’s L1 magnetosphere, since the solar wind does actually shoot straight out from the sun but actually follows the spirals of the Interplanetary Magnetic Field, so I’m pretty sure solar wind would hit Mars because the shadow of a big magnetosphere at L1 would miss Mars.
Latest posts by Chris Stelter (see all)
- How much mass can we put in orbit before running into atmospheric constraints? - July 19, 2020
- Adding an Earth-sized magnetic field to Mars - June 18, 2020
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