Geophysical Effects of Supernova Neutrino Pulse While neutrinos are generally totally harmless, a supernova produces enough of them at once that the neutrino flux would kill you if you could stand close enough (and not be killed by other means first). See XKCD What-If: Lethal Neutrinos for back-of-the-envelope calculations, and this previous question("Death by neutrinos - polonium go home").
But, say there's a relatively Earthlike (i.e., similar mass, roughly similar bulk composition, don't really care about climate / habitability) planet nearby. Just how far away would it have to be to avoid destruction, and what effect would the neutrino flux have at that distance? I.e., could it deposit enough energy / create enough radioactive isotopes to re-start volcanic activity on an otherwise inactive planet?
 A: OK, let's try a back of the envelope...
http://www.slac.stanford.edu/econf/C0805263/Slides/Budge.pdf states that the total anti-neutrino count from a supernova is on the order of 1e58 and the neutrino energy is on the order of 40MeV. Let's imagine all of these neutrinos would go trough earth, which is the worst case scenario. 
The nuclear cross section of 40MeV neutrinos is given by http://cupp.oulu.fi/neutrino/nd-cross.html as 9.3e-48m^2(En/1MeV)^2, which would lead to an effective cross section of roughly sigma=1.5e-44m^2 at 40MeV. 
With the total exposed area of Earth being 3.14*6.5e6m^2=1.3e+14m^2, we get an integrated neutrino flux of Jint=1e58/1.3e14m^2=7.7e43 particles/m^2. 
The total number of atoms in the planet is around N=1.3*e50, so the total number of reactions is Jint*N*sigma=(1e58/1.3e14)*1.3e50*1.5e-44=1.5e50. 
This means a fraction of 1e-8 of the neutrinos are absorbed, and basically every nucleus gets hit about once. (And I think I forgot a fudge factor of 1/2 for the spherical shape of the planet). 
I would call this a rather unhealthy outcome, as the whole planet would basically become one giant ball of radioactive plasma.
Now, if we raise the distance, the total number of neutrinos hitting us goes down with 4pi(d/r)^2. Let's make that multiplier a factor of 1e-8, which would still wipe out all of life, but only deposit about the amount of energy in a chemical reaction per atom. Now the planet would melt, or stay barely solid, but would have to be at a distance of roughly 8000 earth radii from the source. That's about a third of an astronomical unit. 
To prevent complete destruction of the biosphere (let's give, at least, the cockroaches a chance), we need another factor of 1000, or so, which now means we have to be 0.3AU*sqrt(1000)=10AU from the supernova, just to survive the neutrinos.
And now I apologize, if I am completely wrong about all of this... I haven't been a student for a long, long time. 
