Re: Life can eat ionizing radiation!

In article <1180181168.220615.286510@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<balpao@xxxxxxxxx> wrote:

I'm trying to find the energy contained in the cosmic rays, but since
now I've not found it on the network.

I mean... if you have a square meter collector in the direction of the
galactic centre (out of the eliopause, interstellas space), hoy many
watts, supposing a 100% efficiency?

The plot on Wikipedia is a good starting point, but you have to know
how to read it:
http://en.wikipedia.org/wiki/Cosmic_ray

This plot has a turnover at around 1 GeV Below that energy, it
flattens, above that energy it falls off. Thus the energy is more or
less concentrated in a ~1 GeV band around 1 GeV. (I could explain in
much greater length. You don't want me to.)

At 1 GeV, F = 1e3 (m^2 sr s GeV)^-1 . So the energy flux on a flat 1
square meter collector is 1e3 GeV / s .

This is about 160 nanoWatts per square meter. This is about 10^-10 of
what we get from solar energy near Earth. You have to go out to 1.5
light years from Earth before your solar panels drop to that. At which
time other bright stars will be taking up the slack.


Some caveats are that the low-energy turnover is partially due to
shielding by the Sun's heliosphere, so there is more energy below 1 GeV
than I am accounting for. Also, you don't have to point your collector
since cosmic rays are isotropic, due to being swirled about by magnetic
fields, and you get equal additional energy from cosmic rays coming in
the back of your detector. (But the calculation above isn't good to a
factor of 2 anyway, so that doesn't matter to this level of accuracy.)


But it's pretty thin gruel.

--
David M. Palmer dmpalmer@xxxxxxxxx (formerly @clark.net, @ematic.com)
.

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