Re: High speed collisions

On Dec 11, 7:06 am, "Carey Sublette" <carey...@xxxxxxxxx> wrote:

Could you offer an estimate of the effect of the colliders being made of
iron, how the larger charge would affect the interaction?

A quick c++ implementation of the Bethe equations for nuclear stopping
power (and a quick lookup of the average ionization energy of iron,
see Phys. Rev. A 39, 1033 (1989)) gives the following:

At a speed of 1-1e-16 c, the electronic stopping power is a loss of 3
TeV per meter. The variation of this quantity with speed is small, at
1-1e-9 c, it is 2 TeV/m, and at 0.999 c it is 1 TeV/m. So over nearly
the entire range of the particles, a nucleus with charge +26 will
loose 3 TeV/m.

Nuclear stopping is going to confound our attempts to make things
simple by looking at energy deposition from electronic stopping.
Assuming each nucleon has a radius of 1.2 fm, and any intersection
between the path of a target nucleus and a nucleus in the impactor
results in a nuclear interaction, I get a mean free path of a little
less than 20 cm in solid iron. Note that a nuclear interaction will
not stop the projectile nucleus and cause it to dump all its energy
into the target. Rather, it will fragment the target and projectile
nucleus into hundreds of nucleons, each with a charge of 1 or 2.
Since the rate of electronic stopping goes as the square of the
nuclear charge, this may very well result in a lower rate of energy
deposition for a while until those newly formed nucleons collide with
another nucleus. This is a problem for a Monte Carlo radiation
transport code like GEANT - from my experience with GEANT, this
probably means a day or so setting up the problem before we get an
answer out (which is too much work for me right now).

Also the area density of spacecraft might be 100 times higher than in the
example. Is the energy deposition rate per gram the same?

For hydrogen on hydrogen, we have each nucleus depositing about 35 MeV/
m. For iron on iron, it is more like 3 TeV/m per nucleus - a
significant difference.

The original poster did not say what the space craft hit, but it might be
much larger object such as a ejected comet kilometer across. This would stop
the spacecraft I expect.

If it is an iron comet (huh? what? an iron comet? where did that
come from?) a kilometer would give you 5,000 nuclear interaction
lengths (assuming a 20 cm nuclear mean free path). This should be
plenty to stop all of the spacecraft's nuclear particles.

For a spacecraft of iron impacting a comet of solid hydrogen, I get an
electronic stopping power of a bit less than 70 GeV/m and a nuclear
mean free path of a bit over 5 meters. This gives on the order of 200
nuclear interaction lengths in 1 km - if each of these halves the
energy of the projectile nucleon, this still gives a km of H_2 ice
completely stopping the nuclear particles. Muons might still get
through, as will neutrinos.

So lets take our cubic km of iron "comet" and slam a 100 ton (1e5 kg)
ultrarelativistic (beta = 1-1e-16, gamma=71,000,000) spacecraft into
it. Assume all energy is deposited in the target. The target has a
mass of about 8e12 kg, lets call it 1e13 kg. The net momentum of the
system is (spacecraft's mass)*beta*gamma*c = 2e21 kg m/s. The center
of mass of the system (comet + spacecraft) will have a beta*gamma of
(momentum)/[(comet's mass)*c], assuming comet's mass + spacecraft's
mass ~ comet's mass, which all works out to a speed of a bit less than
0.6 c.

The energy of the system is 1e13 kg *c^2 + 1e5 kg * gamma * c^2 ~
1.7e13 kg *c^2 ~ 1.5e30 J. There are (6.0e23 particles/mol)*(1e13
kg)/(0.056 kg/mol) = 1.1e38 iron nuclei. Each of these gets 1.4e-8 J,
or about 90 GeV. An individual iron nucleus has a rest energy of 52
GeV. This means our cloud of debris will be expanding at about 0.8
c. Also, since an iron nucleus has a binding energy of about 0.5 GeV,
the debris will be mostly protons (and maybe helium, I'll ignore the
helium and concentrate on the protons) with an energy of 1.6 GeV each
(note that only about 0.7 GeV of that is kinetic energy, the rest is
the rest mass of the proton) in the debris rest frame.

In the rest frame of the debris, earth is about 4e16 meters away. At
this distance, it will be struck by (1.1e38 nuclei * 56 protons/
nucleus)/[4 pi (4e16 meters)^2] = 310,000 iron nuclei per square meter
over a period of a few years, each with an average kinetic energy of a
couple of GeV. The first cosmic ray impacts will be the highest
energy, and the energy will taper off as time goes by. However, our
planet's magnetic field is likely to stop most of these. Space
weather monitoring instruments may well pick up something weird, if
the plasma is not stopped at the heliopause.

Luke



.

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