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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?

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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.

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 = $N(0)H$

Values listed in that document are:

  • Mars: D = 0.020 kg/m3 x 11.1 km = 22 gm/cm2

  • Earth: D = 1.2 kg/m3 x 8.5 km = 1,020 gm/cm2

We can extract:

  • Mars:

    • N(0) = 0.020 kg/m3
    • H = 11.1 km
  • Earth:

    • N(0) = 1.2 kg/m3
    • H = 8.5 km

Doing again the math with all steps:

  • Mars:

ES = $0.020 \frac{kg}{m^3} * 11.1 km$ = $0.020 \frac{kg}{m^3} * 11100 m$ = $0.020 \frac{kg}{m^2} * 11100$ = $222\frac{kg}{m^2}$ = $222000\frac{g}{m^2}$ = $\frac{222000}{10000}\frac{g}{cm^2}$ = $22.2\frac{g}{cm^2}$

  • Earth:

ES = $1.2 \frac{kg}{m^3}* 8.5 km$ = $1.2 \frac{kg}{m^3}* 8500 m$ = $1.2 \frac{kg}{m^2}* 8500 $ = $10200 \frac{kg}{m^2} $ = $10200000 \frac{g}{m^2} $ = $\frac{10200000}{10000} \frac{g}{cm^2} $ = $1020 \frac{g}{cm^2} $

Other values for comparison:

  • onboard ISS: $ 10 \frac{g}{cm^2}$
  • Apollo space suite: $ 0.1 \frac{g}{cm^2}$

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).

($R_{Earth}$ = 6378 km, $R_{Mars}$ = 3374 km)

In tabular format:

N0 (kg/m3) H (km) ES (g/cm2) Received radiation (normalized to Earth)
Mars 0.020 11.1 22.2 x45
Earth 1.2 8.5 1020 1
Moon (in spacesuite) - - 0.1 x10200
ISS - - 10 x102
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  • $\begingroup$ 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, 2014 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, 2016 at 5:06
  • $\begingroup$ Voting down for atrocious formatting and mangling the units in a chaotic manner. This post is randomly switching back and forth from volume to area units (cm³ to cm²), using ambiguous non-SI compliant unit abbreviations ("gm" does not mean "gram", and is therefore not equivalent to "g"). I tried recovering this by suggesting an edit, but I gave up because I cannot even understand what the author's intention was for the most part. $\endgroup$
    – user47149
    May 17 at 20:58
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    $\begingroup$ @user47149 Reformatted $\endgroup$
    – jumpjack
    May 18 at 9:14
  • $\begingroup$ @jumpjack Switched votes, thanks. $\endgroup$
    – user47149
    May 18 at 9:51

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