FT-Nukes

1 posts · Dec 4 2001

From: Bif Smith <bif@b...>

Date: Tue, 4 Dec 2001 01:01:16 -0000

Subject: FT-Nukes

Someone was asking about nukes in space, and saying are the MT missiles
x-ray laser head nukes or normal nukes. Well, heres some answers from
the HH discusion board (and 2 quotes by einstein on the end).

BIF

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Viewing message <38C8C080.21236E2@t-online.de>

From: Till Poser (pose.infora@t-online.de)
Subject: Re: AOV nit (Spoilers)
Newsgroups: alt.books.david-weber
View: Complete Thread (7 articles) | Original Format
Date: 2000/03/10

Dan Swartzendruber schrieb:
> S

Nukes in atmosphere have one very special energy-eater to fight against:
the atmosphere itself. On the other hand, the atmosphere is a major
destructive factor, once heated up.

The energy of the nuclear explosion gets converted, to a large degree, to
kinetic energy of the atmospheric molecules. This accounts for a e^r
dissipation factor (the exact calculation involves statistical mechanics,
which isn´t exactly my forte) due to the kinetic interactions within the gas
aggregate. This is also one reason why larger nukes within the atmosphere will
not have the same effect as the same amount of energy in smaller packets.

In space, the major effect of a nuclear explosion will be energy release,
because there is little for the radiation to "push against" and the plasma of
the nuke itself is largely irrelevant.

A typical thermonuclear bomb of current design works in three stages:

o the first stage is the primer, a small nuke with a yield of a few ktons. It
will generate to conditions for the next stage, namely create a very focussed
energy source to facilitate fusion. The primer will generally have the minimum
amount of Plutonium necessary to go critical (about 5
kg).

o the second stage is a shell of Deuterium-Tritium, or other mixture.
This mix gets energised by the primer. The individual nuclei now have the
kinetic energy to make fusion between the dirfferent constituents likely.
Besides the nuclear binding energy released in fusion, this stage creates an
enormous neutron flux.

o The third stage of the bomb is a shell of U_238. This shell turns into
a breeder reactor, when the neutron flux of the second stage hits it. The
  neutrons will be captured by the U_238 nuclei, turning them into
Plutonium, which in turn will go into fission when hit by another neutron.

For a nuke of 20 MTons yield, You have to have all three stages.

Why am I so verbose? The above reaction takes time, some milliseconds, to take
place. One of the finer arts of bomb making lies in the problem of keeping the
whole kaboodle together long enough for each reaction to take place and thus
yield energy. For the sake of argument, I will assume that the energy relase
will be roughly along a bell curve (I know it won´t, it will probably look a
bit like an mirrored truncuated Poisson-curve, but there will be a build
up stage, a productions stage and a dissipation stage, so let´s just go with
it.)

So lets assume a 20 Mton device, and a explosion time of 4 msec. Taking the
bell curve, the energy spread of the explosion will be 16, 34, 34, 16%
respectively. A 20 Mton device translates to a yield of 8,6*10^16 joules, and
as I have said before, in space there is no internal dissipation. At a
distance of 50 km, every m^2 will receive 2,7 Mjoule of energy (equivalent to
500 g of tnt, or 0,27
kjoule/cm^2. The momentum transfer, assuming absorption, will be largely
neglegible. The maximum energy flux will be around 100 kjoule/cm^2/sec.
The absorbed energy will cause the hull to warm up quite a bit, transferring
that energy to the inside, so things will get quite uncomfortable, but the
amount
of boil-off should be neglegible (less than 0,01 mm overall) and thus
the amount of explosive sublimation.

So, given the above parameters, a ship with minimum hull padding shouldn´t
blow up. Heat damage and possible radiation damage is a different game. I
assume that at that distance, the neutron flux will still be high, and so will
the gamma flux.

To do some serious mechanical damage, the transferred energy should be at
least a factor of 20 higher, which would mean a proximity of at most 11 km.
And that would probably hurt only unarmoured craft.

Hum. How say You?

Till Poser