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MASS production of nuclear devices.



 
 
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  #1  
Old April 15th 04, 12:35 PM
Carey Sublette
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Default MASS production of nuclear devices.


"Roger Moore" wrote in message
...
Ian Stirling writes:

Where would be the bottlenecks if in 5 years I wanted a million 10Mt
devices a year?

...

If the goal is to destroy an asteroid before it hits the Earth, a million
10 Mt weapons is probably not the way to do it. It would probably be a
lot more efficient to build exceptionally high yield, 3 or more stage
bombs instead. They could use cheap, readily available LiD as their
primary fusion fuel, and theoretically could be scaled to almost
arbitrarily high yield with proper design. For a ~10 km diameter
asteroid, it might be more practical to blast it to pieces with a single
absurdly high yield nuke (and then deal with any remaining large chunks
that are still on a collision orbit) rather than trying to redirect its
orbit.


You really don't want to fragment a large impactor, the results are
unpredictable
and potentially worse than the originally body (multiple impact devastate
more
area than one big one). You just want to nudge it into an non-collision
path.

Since momentum is proportional to
mass*velocity, while kinetic energy is proportional to mass*velocity^2,
a low velocity high mass exhaust provides a much larger momentum
transfer for a given amount of kinetic energy. The asteroid is entirely
expendable, so using a large quantity of its mass as the propellant
could greatly reduce the size and/or number of nuclear explosives
required for a given deflection.

The impactor that formed the KT boundary was about 10 km wide and
weighed around 10^15 kg. If interception was carried out 1 year before
impact, then a 1 meter/sec velocity change should be sufficient to turn
a collision into a miss (over a year this could create something like a
30,000 km orbital deflection), so a 10^15 kg-m/sec momentum transfer
will suffice.

If a 100 megaton (4.2 x 10^17 J) explosion were to deposit its energy in
a layer with a mass density of 0.5g/cm^2 over an area of 50 km^2 (1/6 of
the asteroid surface) then the layer would mass 2.5 x 10^8 kg, and have
a kinetic energy of 1.7 x 10^9 J/kg, and would expand outward at a
velocity of 58 km/sec, providing an impulse of 1.45 x 10^13 kg-m/sec. 69
such explosions would provide the necessary total impulse.

A 100 Mt bomb design has actually been created and tested (at a reduced
50 Mt test yield). This device weighed about 30 tonnes, but a weight of
15-20 tonnes is possible with known technology so a total mass transfer
of 1000-1400 tonnes would be required. If the system could be built and
deployed at a cost of $10,000/kg of delivered payload then deflecting
such a catastrophic menace would only cost $14 billion (much less than
the insurance payout from the Northridge earthquake).


Carey Sublette



  #2  
Old April 29th 04, 11:35 PM
Carey Sublette
external usenet poster
 
Posts: n/a
Default MASS production of nuclear devices.


"John Schilling" wrote in message
...
"Carey Sublette" writes:

[Asteroid diversion]

You really don't want to fragment a large impactor, the results are
unpredictable and potentially worse than the originally body (multiple
impacts devastate more area than one big one). You just want to nudge
it into an non-collision path.


[...]

The impactor that formed the KT boundary was about 10 km wide and
weighed around 10^15 kg. If interception was carried out 1 year before
impact, then a 1 meter/sec velocity change should be sufficient to turn
a collision into a miss (over a year this could create something like a
30,000 km orbital deflection), so a 10^15 kg-m/sec momentum transfer
will suffice.


If a 100 megaton (4.2 x 10^17 J) explosion were to deposit its energy in
a layer with a mass density of 0.5g/cm^2 over an area of 50 km^2 (1/6 of
the asteroid surface) then the layer would mass 2.5 x 10^8 kg, and have
a kinetic energy of 1.7 x 10^9 J/kg, and would expand outward at a
velocity of 58 km/sec, providing an impulse of 1.45 x 10^13 kg-m/sec. 69
such explosions would provide the necessary total impulse.


Not bad, but slightly pessimistic. I think you underestimated the
thickness of the ablated layer.


Could be. It is the softness number in the computation and is just a
representative BOTEC
(back-of-the-envelope-calculation) to show feasibility and I was trying to
avoid unwarranted optimism.

OTOH - I'm not sure the Ahrens and Harris number is necessarily better
(unless someone at a weapons lab provided them a well grounded estimate).


I have in my archives a reference to: "Deflection and Fragmentation of
Near-Earth Asteroids", Thomas J. Ahrens and Alan W. Harris, 1995, on
this subject. Somehow forgot to record the full publication details,
and I don't have the actual paper in hand. But the key result, was
a determination that a one-megaton detonation with optimum standoff
would, by X-ray ablation, impart an impulse of 5 x 10^11 kg-m/sec
on a large (1E12 ton) asteroid. This was fairly linear with yield
over a respectable range so long as you tweaked the standoff properly,
so 100 MT should give 5E13 N-s, about a factor of three better than
your estimate.

Or maybe a hundred megatons would be beyond the range of linear scaling,
so you could be right.


I think scaling should hold. Scaling problems arise when the ratio of yield
to asteroid size becomes too large. This is not a problem in this case.

The point about not fragmenting the asteroid is quite valid, and quite
important. However, in the case of dinosaur-killer scale asteroids,
gravity is our friend. Crudely speaking, if the momentum imparted
to the asteroid is small compared to that required to accelerate its
mass to its own escape velocity, the bulk of the asteroid will remain
*gravitationally* bound even if it is mechanically fragmented. If you
blast it into a pile of rubble, or if that's all it ever was to begin
with, the bits of rubble will mostly fall back into one loose mass on
a slightly displaced trajectory in a timescale of hours.


There are two considerations that limit the problem of scattered fragments
htting Earth, gravitational binding means they will coalesce if the fragment
velocity is too low. The other is if the fragment velocity is high enough to
escape they will end up spread over such a large volume that collision with
Earth is very unlikely.

But this becomes something hard to evaluate satisfactorily. A very large
impactor could actually be composed of many discrete pieces of varying size.
If one was trying a surface burst to maximize momentum it detonate over a
smallish fragment that would get ejected at greater than escape velocity.

But another problem with the idea of surface bursts is the problem of
creating a deadly debris cloud for following charges in the case when
multiple detonations are required. An ablation stand-off approach minimizes
this problem also.


Carey Sublette

 




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