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FT-Nukes

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:
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> I'm kind of surprised at something vis Cromarty's death.  When the
> missile targeted at Grayson One went off, it was "less than 50
> kilometers" from Honor's ship (which had to have been at least a few
> klicks from Grayson One (to clear the wedges).  My understanding is
that
> with our less than perfect knowledge of nukes in vacuum, they are far
> less effective than in atmosphere.  Here's what has me puzzled: if a
20
> MT warhead woudn't kill a ship 50 klicks away in atmosphere, why would
it
> do so in vacuum?  Or is there some reason the missile aimed at Grayson
> One went off much farther away than the one aimed at Queen Adrienne?

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

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