Why does cosmic radiation diminish entirely, reaching zero, at sea level? enter image description here

If the atmosphere's thickness were random, it should not happen at sea level, but at a random altitude.

Is there a specific mechanism or interaction within the Earth's atmosphere that causes this complete nullification of cosmic radiation at sea level?

For example, maybe nitrogen is created by cosmic rays interacting with seawater or rocks, until the density stops it? (So, perhaps all planets with water oceans should have similar atmospheric densities).

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    $\begingroup$ My Geiger counter confirms a couple of times per minute that the cosmic ray flux at sea level does not go all the way down to zero. $\endgroup$
    – MTA
    Commented Jun 13 at 12:30
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    $\begingroup$ (speculation) I think there's a bit of Anthropic Principle at play here. Earth's atmospheric thickness depends on how much air there is, and how much gravity the Earth exerts on it to keep it close to the surface. It just so happens that, for Earth, these values are appropriate for most of cosmic radiation to not reach the surface — which is one of the reasons why life could develop and, ultimately, the reason why you're here asking this question :) IOW, if Earth had less air or gravity, cosmic radiation would reach the surface, and you wouldn't be asking this, because you wouldn't exist. $\endgroup$
    – walen
    Commented Jun 14 at 9:13
  • $\begingroup$ @MTA How do you know that that's due to cosmic rays, rather than some other source of background radiation? $\endgroup$ Commented Jun 14 at 18:11
  • $\begingroup$ Cosmic neutrinos are not at all stopped by the atmosphere. $\endgroup$ Commented Jun 14 at 20:59
  • $\begingroup$ @MarkMorganLloyd Process of elimination. There is no other known source of ground level background ionizing radiation that is continuous day and night, unchanging with wind or weather, independent of proximity to mineral deposits, etc. First studied in 1912 by Victor Hess and well known by discerning geeks everywhere since Dr Millikan of the famed oil drop experiment did extensive work in the field in the 1930s. $\endgroup$
    – MTA
    Commented Jun 14 at 21:00

1 Answer 1


It won't be (and isn't) zero. To see that, you want the plot on a logarithmic scale. Thus there isn't really a question to answer. The attenuation of cosmic rays will follow some exponential decay with airmass. Since most of the airmass is at low altitudes then that is where most of the attenuation takes place. The Earth's magnetic field is also a factor, which is why there is a latitude dependence.

Here's a similar diagram but with a logarithmic dose scale (Hwang et al. 2010).

Dose vs altitude

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    $\begingroup$ I would imagine there's also some latitude dependence due to oblateness of the atmosphere. $\endgroup$
    – Cadence
    Commented Jun 13 at 4:39
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    $\begingroup$ @wepajakeg A vertical slope on that figure means little or no attentuation, since the amount of cosmic rays isn't changing with altitude. Strong attenuation happens when the slope is closer to horizontal: as you go down in altitude, the number of cosmic rays you can detect goes down (a lot). $\endgroup$ Commented Jun 13 at 8:41
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    $\begingroup$ If I'm reading this correctly, the muon and photon curves (and, just barely, the electron and neutron curves) fold back on themselves at higher altitudes. Is there an obvious reason for this, or should I make it a new question? My own best guess is secondary radiation from proton (or unmentioned) collisions in the stratosphere. $\endgroup$ Commented Jun 13 at 14:26
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    $\begingroup$ @CristobolPolychronopolis Yes, it's secondary radiation, and yes, you should make a new question. $\endgroup$
    – zwol
    Commented Jun 13 at 15:04
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    $\begingroup$ @wepajakeg Your absolute graph shows nicely that the radiation is "almost zero" already at around 13,000 feet and probably "small enough" close to 20,000 feet which is more than a quarter of the strongly absorbing atmosphere. $\endgroup$ Commented Jun 13 at 23:22

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