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Atmospheric neutrinos correspond to neutrinos produced by the interaction of cosmic rays in the Earth atmosphere. The Earth atmosphere is at most $800$ km=$8 \cdot 10^5$ meters.

So how could the "baseline" (=distance of flight) of the atmospheric neutrinos be as much as $10^7$ meters, as stated in the Table of properties of atmospheric neutrinos in the famous (most cited book in particle physics) Particle Data Group review:

http://pdg.lbl.gov/2019/reviews/rpp2019-rev-neutrino-mixing.pdf

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And bonus question: how could solar neutrinos have a baseline (it is stated $10^{10}$ meters) higher than 150 millions km= order of $10^8$ meters: that looks impossible?

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Neutrinos produced in Earth's atmosphere have two opportunities to interact with detectors on Earth's surface: once on their way down from the sky, and again when they emerge unscathed on the other side of the planet. So the baseline is more like the diameter of Earth, which is about $12.7×10^6$ meters.

Famously, the IceCube detector in Antarctica looks down to see neutrinos from the north celestial sphere.

Re your bonus question: "million kilometers" is a stupid unit, and everyone makes off-by-thousand exponent-counting errors when using it. An astronomical unit is $1.5×10^{11}$ meters.

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  • $\begingroup$ Thank you, but it seems that your explanation does not work. The atmosphere is 8e5 meters. If you add 12742 km=12.7e5 of the diameter, it will give around 20e5=around 2e6 << 1e7. But in the table, the distance quoted is up to 10^7 meters. There is a similar problem for solar neutrinos : distance Earth-Sun is 150 millions km=1.5e8 meters, while the baseline for solar neutrinos is quoted as 10^10 meters. $\endgroup$
    – Mathieu Krisztian
    Commented Dec 21, 2019 at 21:17
  • $\begingroup$ My edit hopefully clarifies things. $\endgroup$
    – rob ♦
    Commented Dec 21, 2019 at 21:19
  • $\begingroup$ Thank you so much @rob . I made a mistake in my answer : 12742km is 12.6e6 : so indeed your explanation works very well !! And I made also a stupic computation for the Sun : 150 millions kimeters is 150e6 * 1e3=1.5e11. So it works !! But why do they report 10^10 only ? $\endgroup$
    – Mathieu Krisztian
    Commented Dec 21, 2019 at 21:24
  • $\begingroup$ That one I don't know. It might be worth its own question. $\endgroup$
    – rob ♦
    Commented Dec 21, 2019 at 21:27
  • $\begingroup$ ok thank you. I close this question with your answer as the official answer, and I open a new question for solar neutrino. $\endgroup$
    – Mathieu Krisztian
    Commented Dec 21, 2019 at 21:28

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