Researching this answer led to ³He in permanently shadowed lunar polar surfaces published in Icarus. The abstract is tantalizing but terse:


Because of their cryogenic temperatures, analysis indicates that permanently shadowed polar lunar craters may have substantially higher levels of ³He than sunlit lunar surfaces and are conservatively estimated to contain as much as 50 ppb or more.

Wikipedia's Helium-3; Solar nebula (primordial) abundance says:

One early estimate of the primordial ratio of ³He to ⁴He in the solar nebula has been the measurement of their ratio in the atmosphere of Jupiter, measured by the mass spectrometer of the Galileo atmospheric entry probe. This ratio is about 1:10,000, or 100 parts of ³He per million parts of ⁴He. This is roughly the same ratio of the isotopes as in lunar regolith, which contains 28 ppm helium-4 and 2.8 ppb helium-3 (which is at the lower end of actual sample measurements, which vary from about 1.4 to 15 ppb).

Question: Why exactly would "permanently shadowed polar lunar craters... have substantially higher levels of ³He than sunlit lunar surfaces?" What is it exactly about permanently shadowing surfaces from the Sun that is thought to allow them to accumulate up to 50 ppb of helium-3 compared to a lunar average of only 2.8 ppb?

Is it the far lower temperature, or the shielding from the solar wind, or something else?

Possibly relevant factoid, the boiling point of helium-3 is only about 3.2 Kelvin, much lower than the 4.2 Kelvin of helium-4.

  • $\begingroup$ Parts per billion are actually very low concentrations. $\endgroup$
    – user8269
    Aug 28, 2019 at 3:52
  • $\begingroup$ @PhilipNgai compared to what? $\endgroup$
    – uhoh
    Aug 28, 2019 at 3:52
  • $\begingroup$ Compared to any real source of energy. $\endgroup$
    – user8269
    Aug 28, 2019 at 3:54
  • $\begingroup$ @PhilipNgai I'm just interested in the dynamics of particles from the Sun interacting with lunar regolith. $\endgroup$
    – uhoh
    Aug 28, 2019 at 3:57
  • 1
    $\begingroup$ @uhoh I stand corrected; got a little too fast and loose with my terminology there. $\endgroup$
    – Roger
    Sep 26, 2019 at 15:51

1 Answer 1


That is fairly obvious from the factoid you state yourself. In regions, of a celestial body, that are not shielded by an atmosphere and that are never lit by the sun, the temperatature can go down very, very low - actually close to the temperature of the Cosmic Background Radiation (2.725 K). So in permanently shadowed lunar craters the Helium-3 simply has boiled away at a far lower rate than in sunlit regions.

  • 3
    $\begingroup$ -1 for unsourced, unsupported commentary. Helium-3 storage in regolith is a very complex process, it's not just stuck to the surface, it can be embedded inside the particles as well. So far you haven't convinced me if it's strictly thermal or if there are sputtering effects that come into play as well. Shielding from sunlight also means shielding from solar wind. Just repeating points in my question is not an answer to my question. Can you have a second look at what I've written after Question: and see if you can address that in more detail? Thanks! $\endgroup$
    – uhoh
    Aug 27, 2019 at 14:50
  • 1
    $\begingroup$ @uhoh Yes, can do that. Will be in a bit (few days, prolly) though, as I'm travelling. This year-2000 paper researchgate.net/publication/… has in its conclusion (and the section just before it) that 1. yes, there is a higher concentration of solar-wind hydrogen at the poles, especially in unlit areas, than elsewhere 2. sputtering concerns only the breakdown of water ice 3. He3 is also enhanced at the lunar poles. Anyways, I can dive deeper. $\endgroup$ Aug 27, 2019 at 15:08
  • $\begingroup$ @uhoh Also, this - admittedly speculative - paper 21sci-tech.com/Articles_2014/Moon_Chemistry.pdf cites the much more serious NASA paper by Schmitt, Henley et al: solarsystem.nasa.gov/studies/191/… which in turn, and indirectly points at Geochimica et Cosmochimica Acta, Supplements 1 and 2 (1970), dedicated to the results of Apollo 11. It'll be my pleasure to gut those for you :-) $\endgroup$ Aug 27, 2019 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.