Magnetizing Mars

... sorry, no


Superconductors do not make a stupid idea into a super idea.

Some talk about terraforming Mars, turning this dead ball of rock into a miniature Earth. Mars has no magnetic field, so an artificial atmosphere (somehow) would get stripped away by solar UV and the much weaker gravity. How would a terraformed Mars 2 compared to Earth and Mars 1 today?

units

Earth

Mars 1

Mars 2

Semimajor Axis

AU

1.0

1.52

1.52

Mass

kg

5.972e24

6.42e23

6.42e23

Radius

km

6371

3389.5

3389.5

Gravity

m/s²

9.81

3.71

3.71

Planet Density

kg/m³

5510

3930

3930

Escape Velocity

m/s

11200

5030

5030

Escape Energy

MJ/kg

63

12.6

12.6

Dipole moment

A-m²

7.94e22

~0

3.4e22

Iron Core radius (est)

km

3400

1800

1800

Atmosphere @surface:

Mass

kg

5.15e18

2.5e16

4e18

Temperature

K

288

210

288

Density

kg/m³

1.2

0.015

1.2

Pressure

kPa

101

0.64

101

Pressure Scale Height

km

8.5

11.5

22.8

Temperature Lapse Rate

K/km

6.5

2.2

?

Primary Components

N₂, O₂

CO₂

N₂, O₂

Molecular Weight

g/mol

30.5

44.0

30.5

Molecular Weight

kg

1.54e-24

2.23e-24

1.54e-24

Thermal Velocity √3kT/2m

m/s

485

345

485

(Vesc/Vthrm

533

213

108

CO₂ mass

kg

3.2e15

2.4e16

?

Conversion tables

The most disturbing problem with Mars 2 is that a hotter, lower-mass gas atoms are more likely to escape in the weak Mars gravity field ... (Vesc/Vthrm)² is halved compared to the colder, higher molecular weight current Mars 1 atmosphere.

But we'll ignore that, and also ignore how we can create enough greenhouse effect to heat Mars to Earth temperatures. Terraforming in science fiction novels is done with a few keystrokes.

The goal is making an Earth-equivalent magnetic field with an equatorial coil.

The shielding effect of the magnetic field is not due to surface field strength, but the dipole of the planetary magnetosphere compared to the solar wind pressure. Presume the solar wind density is inverse-square - (1/1.52)² or 0.433 less dense than at Earth, so the artificial Martian dipole may be 0.433*7.94e22 A-m² or 3.4e22 A-m².

The potentially magnetic core of Mars is small; it will reduce the reluctance of the path a bit, but the magnetic field we are concerned with goes far into space, and most of the field will be close to the equatorial coil, not the core of Mars.

The dipole field of a coil is the circumferential current times the area; the martian equatorial radius is R = 3396200 m, and the area of the coil is the \pi R^2 = 3.63e13 m², so the current is 940 mega-Amps. By Ampere's law for a round wire with radius r , the B field at the edge of the wire is B = \mu_0 I / 2 \pi r where \mu_0 = 4e-7 \pi , hence B = 2e-7 I / r or r = 2e-7 I / B . For a maximum field B $ at the edge of the wire of 20 Tesla, r = 2e-7 * 9.4e8 / 20 = 9.4 meters. Our superconducting wire (which may be a thin hollow cylinder of superconductor around a structural core) is about 19 meters (62 feet) in diameter. It might be oval shaped; however, the field will concentrate around the tightest radius, so the total surface area must be larger to limit the maximum field at the radius to 20 Tesla.

20 Tesla is the highest field ever generated in stacked magnetic fields in small lab experiments for a few days at very cold temperatures. In Real Life, The 4.3 km diameter Large Hadron Collider runs magnets at 7.7 Tesla in 96 tonnes of 1.9 Kelvin liquid helium; one small weld failure released as much energy as a high speed train collision, and took the collider offline for 14 months.

20 Tesla on the inside surface of the coil, almost that much on the outside; the magnetic pressure on the coil surface will be B^2/2 \mu_0 = 160 MPa, or the pressure at 1.6 kilometers deep in the ocean.

How much energy is stored in the field? Using an online calculator, the inductance is 27.5 Henries. The stored energy is ½ L I² = 0.5 * 27.5 * (9.4e8)² = 1.2e19 Joules or 3.4e6 GW-hrs or 2900 megatonnes explosive equivalent.

Solar insolation on Mars is 590 W/m², and a 40% efficient solar panel on the Martian equator might collect an average of 60 W/m² of electricity. An earth year is 31.56 Msec. so a square meter produces about 1.89e9 J per year. A solar array area of 1000 square kilometers (60 GW average, 230 GW peak noontime at Mars perigee), will produce enough energy to charge the coil in about 6 years.

However --- if even a tiny portion of the superconductor surface fails, that area will stop conducting and make I²R resistive heat instead. The superconductor system will need massive cooling systems, and a high thermal- and electrical- conductivity substrate to bypass the current and conduct away the resistive heat. If that fails ... a complete failure and a GYNORMOUS electrical arc, dissipating as much energy as 2900 megatonnes of nuclear weapons, starting near the disruption, with a shock wave launched around the planet in a few seconds. The incandescent arc will climb rapidly into the sky, like a nuclear fireball ringing the planet, probably emitting high energy x-rays.

This isn't terra-forming - it is hell-on-Mars-forming. Something that huge running for that many years must be perfect, forever.

Science fiction writers can type any numbers they want into their stories, and their science- and engineering- ignorant readers will accept them if the story is entertaining. However, entertainment won't terraform Mars.