In the book "The Case for Mars", cited in this answer, there is a study cited that states that 20-50 REM radiation occurs due to cosmic rays in deep space. It seems likely that the atmosphere of Mars would reduce that somewhat. By how much will the atmosphere of Mars reduce the radiation?


I found at SpaceMath ( http://spacemath.gsfc.nasa.gov ) a document stating that the density of a planetary atmosphere is defined by the exponential function:
$N(z) = N(0) e^{-z/H}$

where H is the scale-height of the gas. For the composition of Mars atmosphere, H = 11.1 km. The sea-level density for Mars is N(0) = 0.00020 g/cm3. (For Earth: H=8.5 km, N(0) = 0.012 g/cm3)

The amount of surface shielding for radiation arriving from 90° direction (vertical) is given by evaluating the integral of N(z), which results in:

Equivalent Shielding = $N(0)H$

Mars Equivalent Shielding = 0.020 kg/m3 x 11.1 km = 22 gm/cm2

(Earth Equivalent Shielding = 1.2 kg/m3 x 8.5 km = 1,020 gm/cm2)

(onboard ISS: 10 g/cm2)

(Apollo space suite: 0.1 g/cm2)

Assume R (Earth) = 6378 kilometers, R (Mars = 3374 km)

It's worth noting that a bare body on Mars' surface receives half the radiation of an astronaut onboard ISS.

It's also worth noting that to get equivalent shielding of the whole Earth atmosphere by using a water shielding, only 10 meters thickness is required (given that water density is 1000 kg m3, compared to 1.2 kg/m3 for Earth atmosphere).

  • $\begingroup$ The document you're referring to is this one: spacemath.gsfc.nasa.gov/Calculus/3Page24.pdf It only deals with matter density though, so the numbers aren't directly comparable because of reduced incident radiation within the Earth's magnetic field and different atmospheric composition. It's only good as first order approximation, for precise values we're still waiting for official measurements data from MSL RAD mars.jpl.nasa.gov/msl/mission/instruments/radiationdetectors/… with some preliminary results already out, e.g. spaceflight101.com/msl-rad-science-reports.html $\endgroup$ – TildalWave Apr 10 '14 at 20:03
  • $\begingroup$ Just by the way, the integral gives the wrong answer, and you don't need to integrate anyway. All you need is the surface pressure and the acceleration of gravity to get the mass of the column of air above. The pressure where you are is simply the weight of the atmosphere above you (when in hydrostatic equilibrium, which is a good assumption). Divide the weight per unit area (the pressure) by the acceleration of gravity to get the mass per unit area. $\endgroup$ – Mark Adler Mar 16 '16 at 5:06

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