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Unlike Earth Mars

  • does not have an atmosphere with oodles of Carbon Dioxide and water-vapour.
  • lacks a planet-wide magnetic field

Even alone these two factors make a significant difference to the amount of radiation incident on the Martian Surface.

Astronauts in Earth orbit space-missions are subjected to cosmic rays, as also to charged particles expelled by Sol. Wikipedia writes to say

Exposures on the ISS average 150 mSv per year, although frequent crew rotations minimize individual risk

The Mars One program website sums up the exposure to say

The 210-day trip results in radiation exposure of the crew of 386 +/- 61 mSv. On the surface, they will be exposed to about 11 mSv per year during their excursions on the surface of Mars. This means that the settlers will be able to spend about sixty years on Mars before reaching their career limit, with respect to ESA standards

  • Does this figure of 11mSv per annum over a duration of 60 years incorporate the cumulative nature of such exposure?
  • What is the projected death-rate displacement for colonists to Mars relative to the corresponding age-specific rate on Earth?
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You should quickly find sources converge on a coefficient to convert dose into cancers. Modern science has quantified this rather well.

http://rationalwiki.org/wiki/Linear_no-threshold

LNT estimates that the risk of premature death from radiation-induced cancer is around 5% per sievert or 0.5% per 100 mSv of exposure.

The units are easy to mess up, so I'll rewrite the above coefficient:

0.00005 cancers / mSv

There are more variables you could put into this. I mean, our linear (as in the LNT model) assumption is, in effect a model. You could replace that model with a sophisticated computer code that takes in all kinds of adjustments, but I'm not interested in that accuracy here. One very notable correction is fractionated versus non-fractionated dose. Nuclear industry workers who get a large dose through a single work operation get something like twice the corresponding cancer risk, as opposed to getting that same dose distributed over a longer period of time. For Mars colonization, I would imagine the dose is pretty constant. But anyway, moving on:

(0.00005 cancers / mSv) x (11 mSv per year) = 0.00055 cancers / year

I don't know how long you'd expect to live on Mars. Let's say 40 years.

(0.00055 cancers / year) x (40 years) = 0.022 cancers --> 2.2% chance of getting cancer

Had this number been much closer to 1.0, then different statistical equations would be needed. For reference, these days the "normal" chance of someone getting cancer and dying from it is something like 23%.

This answers half your question, and I'm going to leave it here. I don't know if I believe that a year on Mars will result in 11 mSv. With some very cursory looking online about the subject, I'm starting to get a picture that it depends almost entirely on assumptions about how the colonists spend their time. Assume that they're under a rock all day, and the dose could drop to near zero. This isn't physically impossible. Just put them under 10 tons/m^2 of dirt. Finished. Radiation requirement met. On Earth this would be workable, even if unpleasant. Design of the facilities on Mars would be subject to... more constraints.

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  • $\begingroup$ Note that his quote said the 11mSv was during excursions. Thus I think your answer of putting them under dirt is exactly what's intended. Not to mention that the dirt provides protection from smaller bits falling from the sky. $\endgroup$ – Loren Pechtel Nov 21 '13 at 20:17
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    $\begingroup$ @LorenPechtel So the correct figure would be to add the dose during excursions plus the dose while in the habitat modules? That would likely put it closer to the other figures I looked at, which were closer to twice that. My point about the dirt is just that you can tune assumptions however you want. $\endgroup$ – AlanSE Nov 21 '13 at 20:31
  • $\begingroup$ Alan, SouthWest Research Institute has published the RAD data: marsdaily.com/reports/… - 0.67 milliSieverts per day on Martian surface on average. $\endgroup$ – Deer Hunter Dec 13 '13 at 11:29

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