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In Randall Munroe's What If? He is calculating the Lethal Neutrinos dose.

If you observed a supernova from 1 AU away—and you somehow avoided being being incinerated, vaporized, and converted to some type of exotic plasma—even the flood of ghostly neutrinos would be dense enough to kill you.

How do I stay alive to be killed by neutrinos?

Can I pick a large supernova or some other cosmic event, and hide behind a nearby neutron star?

hpekristiansen
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How do I stay alive to be killed by neutrinos? You wouldn't. The point is being made that even the beam of neutrinos with a supernova at one astronomical unit distance would be intense enough that enough of them would interact with the matter of your body to be lethal. So even the neutrinos would get you if all the other stuff - notably $\gamma$s didn't. However, you'd likely be plasma long before even a handful of neutrinos interacted with you in such a case.

The article is meant to give a physicist, who knows how weakly neutrinos interact, some feeling for the unearthly intensity of all kinds of supernova radiation. In line with the article's idea, a feather in low Earth orbit orthogonal or head-on with an astronaut's own orbit would be lethal, too.

As for hiding behind a neutron star, from this source I glean a cross section of $10^{-45}{\rm m^2}$ for the neutrino-proton scattering cross-section:

The case for neutrino is very representative in this case: neutrino-proton cross section for typical solar neutrinos of 1 MeV is around $10^{-41} {\rm cm^2}$, despite of a very larger proton “size”.

Let's think of a neutron star 10km in diameter, and you sit at the opposite side of it's equator to the supernova. Given a nuclear density of $6\times10^{17}{\rm kg\,m^{-3}}$, a cubic metre of neutron star holds about $6\times10^{17}\times 6\times 10^{23}\times 10^3 = 3.6\times 10^{44}$ neutrons. This means that the one square metre cross section cubic metre presents an effective collision cross-section, given the figure above, of about $0.36{\rm m^2}$, or a probability of interaction with one neutrino of about 0.36 (somewhat less, as I'm simply adding probabilities linearly). So the eleven kilometer wide neutron star would begin by shielding you from neutrinos very well (it would absorb almost all of them). However, absorption means energy dumping, and, given neutrinos account for only a small amount of the energy shed by a supernova, I'm not sure how long even a neutron star at one Au distance would last. I suspect that it too would be vaporized, and, if it weren't it would swiftly heat to a level where its $\gamma$ radiation would be lethal, but I'll have to leave that question to someone else.

Selene Routley
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    "However, you'd likely be plasma long before even a handful of neutrinos interacted with you in such a case. " Nah. The neutrinos get out before anything else. Even the photons are repeatedly scattered. Just how much earlier the neutrinos get out is a matter of considerable debate and something that might let us measure the absolute mass scale of the neutrinos. – dmckee --- ex-moderator kitten Dec 01 '13 at 02:50
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    @dmckee Interesting - not sure whether to delete the offending sentence or leave your contradiction to stand as the most effective way of communicating the facts. I'd like to incorporate your comment (attributed of course) into my answer sometime if that's OK. – Selene Routley Dec 01 '13 at 02:55
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    @dmckee that's so cool I'm giving it its own question. – Emilio Pisanty Dec 01 '13 at 03:49
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    "neutrinos account for only a small amount of the energy shed by a supernova" - I thought the $\nu$ luminosity was $\sim 100$ times the optical luminosity. –  Dec 01 '13 at 09:08
  • @ChrisWhite You'd be in a much better position with your background than I to know - my impression (as a nonspecialist) was the other way around: do you not know for sure? If not, I'll check this one. – Selene Routley Dec 01 '13 at 09:17
  • Your argument is valid. Your calculations are very wrong. You can only add cross sections like that in the zero limit (I believe you know that). – hpekristiansen Dec 02 '13 at 09:35
  • Why will a sufficiently small neutron star not work? – hpekristiansen Dec 02 '13 at 09:35
  • @Hans-PeterE.Kristiansen Well, yes of course: I'm simply estimating orders of magnitude: let's say then we add the cross sections up till the probability is 0.1, or 0.01 or whatever (so that the addition is valid). We can then get the mean penetration depth of a neutrino. What my very rough calculations show is that the penetration depth is of the order of metres. So, whichever way you do the calculation, effectively all of the neutrinos will be absorbed by the neutron star which is of the order of ten thousand metres thick. What will happen after that I don't know; as I say ... – Selene Routley Dec 02 '13 at 10:02
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    @Hans-PeterE.Kristiansen we need more of a specialist to answer this: but that neutron star is absorbing energy at a colossal rate so its a fair bet things are going to end in tears for anyone sheltering in the neutron star's lee. – Selene Routley Dec 02 '13 at 10:03