Regarding a manned shelter on the surface of Mars...

With all of this esoteric talk of ice shelters, why not simply create concrete (Marscrete) buildings? The ice probably provides better radiation protection, but surely the difference is not that great.

So, how thick do walls have to be?

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    $\begingroup$ There's no answer to this because there's no decided formula for Marscrete yet. It's being worked on, but until we know what the end composition will be we won't know how thick it would need to be. $\endgroup$
    – GdD
    Jan 5, 2017 at 8:51
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    $\begingroup$ Not precisely, but surely within 25% or so, wouldn't you think? I mean the debate is over sulphur percentage, mostly, I think. $\endgroup$ Jan 5, 2017 at 20:35
  • $\begingroup$ I think you need to set an upper energy limit to the radiation you wish to stop before this question has an answer. Meaning, to be hyperbolic, there is nothing thick enough to stop an infinitely energetic particle. If you only care about, say, particles up to 10 MeV, then one can get relatively accurate estimates of the minimum thickness for a whole zoo of materials (lots of different models do this). $\endgroup$ Jun 30, 2017 at 18:31

2 Answers 2


According to "A Case for Mars", adequate protection can be achieved by simply filling sand bags and placing them over the shelter. Thus only a few inches (7-10 cm) of Mars soil would be required to provide adequate protection. A more accurate guess can be made with lunar soil, which is similar enough to Mars to give at least a first order guess. The lunar soil required to provide adequate radiation protection is 46 cm (18 inches). Mars would likely require less due to the atmosphere, which will protect it somewhat.

A bit more detail about the proposal from "A Case for Mars". The assumed Cosmic Ray radiation received is about 6 rem/year. The unsheltered radiation is assumed to be 9 rem/year. It seems like thicker shielding would be required for a truly long duration settlement, but these values are considerably below what the radiation exposure is in deep space, which Mars soil won't protect against.

And as Mark pointed out, to achieve the full protection of the Earth's atmosphere, one would need 14.7 psi/ Mars Soil Density, or about 12 feet (3.3 m) (roughly) of packed soil. Of course, one could get away with much less, say, 30% (Same as an airplane). That would give a thickness of about 4 feet( 1.2m)

Bottom line, I would go for 46 cm for "good" protection, and about 10 cm for "adequate" protection.

  • $\begingroup$ That's interesting... Sounds like it's far less of a challenge than I thought. $\endgroup$ Jan 8, 2018 at 16:06
  • $\begingroup$ A few inches? A quick back of the envelope to see how much sand I would need to get 14.7 pounds per square inch on the shelter, i.e. to be mass-equivalent to Earth's atmosphere in protection, I get 150 inches, or over 12 feet thickness! I wonder about that Case for Mars calculation. $\endgroup$
    – Mark Adler
    Jan 8, 2018 at 17:00
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    $\begingroup$ @ChrisB.Behrens Air pressure (in equilibrium) is equal to the total weight of the air above that area. So to get the same amount of mass between you and space that our atmosphere affords here on Earth, you would need something like 12 feet of dirt over you on Mars. $\endgroup$
    – Mark Adler
    Jan 8, 2018 at 18:29
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    $\begingroup$ That's only a very, very rough assessment however, since different materials offer different protection against radiation depending on, for example, the Z of the atoms (water ice is better by the way), and you can also get away with a lot less than 14.7 pounds above every square inch and still have pretty good protection against radiation. $\endgroup$
    – Mark Adler
    Jan 8, 2018 at 18:31
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    $\begingroup$ One could maintain the atmospheric integrity with much less thick walls. $\endgroup$
    – PearsonArtPhoto
    Jan 8, 2018 at 20:52

Absolute Minimum: 5 Inches

The OSHA limit on radiation is 5 REM / year, and surface radiation on Mars is 20-30 REM / year.

Using normal concrete, you'd need about 13 cm / 5 inches to reduce 20 REM to 5.

Realistic: 2 Feet

5 REM / year is still a lot of radiation.

50 mR / year is probably a more reasonable long-term value. So we're talking a three order of magnitude reduction - which is roughly 63cm or 25 inches of standard concrete.

The Wrinkle

You can play with "stay times" in different rooms. For example, if you make a two story habitat, then the upper story protects the lower story - all the air, the floor, and any equipment in the upper story absorb radiation that would otherwise be a problem. Similarly, exterior rooms shield the interior rooms.

So if you put the places where the colonists spend the most time (bunks, kitchens, etc) on the interior, and put places where the spend less time (atmosphere control, storage, waste-water processing, etc.) on the exterior, you can get away with thinner walls, with the understanding that people should restrict their time in the exterior areas.

Note that total does matters, so time outside the habitat would have to play into this calculation too - it could get fairly complicated trying to compute the actual answer, since you'd have to predict how much time people would spend in each area, and how material might move around the colony over time. Spare parts get used eventually, right?


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