Re: SPACE: Lunar Warfare

Michael Lorrey (
Sun, 13 Jan 2002 15:17:19 -0500

James Rogers wrote:
> At 05:11 PM 1/12/97 +0000, Mark Grant wrote:
> [...SNIP...]
> >> With boost phase, you do have an economic advantage in that with an
> >> opponent that has MIRVed missiles, you can take out multiple warheads
> >> with one hit, making each shot more cost effective.
> >
> >And because they're much easier to track when they're radiating MW of IR
> >from the rocket exhaust. On the moon you won't even have a reentry plume
> >to track them by.

WIth the background of space, anything only a few hours in space from
earth surface would radiate a lot of IR.

> >
> >> Such weapons based on the
> >> moon with no atmospheric problems would be able to hit any target with a
> >> lunar trajectory within 10,000 miles of leaving earth.
> >
> >How are you going to track a warhead with the cross-section of a marble
> >from 240,000 miles away!?!?!?!?!
> [...Radar calculations and such...SNIP...]
> >If you increase the scanning rate you're more likely to detect it, but
> >the detection range decreases because you put out less energy in each
> >scan. The crossover point where you're guaranteed to detect the warhead
> >before impact is at around 200m. That's about 33 milliseconds before
> >impact. It might even be close enough to the ground that I could
> >detonate the bomb anyway.
> >
> >Of course, I launched a thousand decoys with each warhead, so you have 33
> >microseconds to hit each one. Each one you miss is a 1/1000 chance of
> >losing your colony. How fast can your lasers retarget, and can they put
> >out enough power to destroy a warhead in 33 microseconds?
> This is all good and well, ASSUMING, that you are using radar sensing
> technology. However, current state-of-the-art ballistic missile defense
> sensors for coast, reentry, and terminal phases *do not* use radar
> technology. The current technology is also extremely capable of
> discriminating between dummy warheads and live ones. The sensor technology
> has an effective exoatmospheric (e.g. lunar) sensing range of thousands of
> miles.
> Taking this into consideration, I would say the odds of a viable defense
> aren't so slim anymore.
> [...Terminal guidance et al...SNIP...]
> >> Given our 10 kiloton clean ferrous warhead (weighing one ton), we
> >> could decimate the world with 110,000 one ton ferrous rocks.
> >
> >But this is the problem. I don't believe your figures, and still haven't
> >had time to find an official 'J to MT' conversion rate. The two figures I
> >have handy are that the explosion speed of TNT is about 5500m/s and that
> >something like 50g of antimatter is equivalent to 1MT. So that would give
> >figures of 0.5*5500*5500 = 1.5x10^7 or 0.1 * c^2 / 10^9 = 9x10^6 J per kg
> >of TNT. So let's say 10^7 J/kg.
> According to ASTM E-380-82 of Standards for Metric Practice, one ton TNT
> nuclear equivalent is 4.184^9 Joules. This number is not the amount of
> energy in a ton of TNT (which, BTW detonates at ~7000m/s in most military
> forms), but a standardized unit with a historical name. In fact, one ton
> TNT nuclear equivalent is defined as exactly 10^9 calories.

I found a ref ona NASA page stating that one Megaton was 4.2 x 10^22

> >Freefall velocity from the moon is around 11km/s, so a minimum energy
> >launch will give you 500 * 11000 * 11000 = 6 * 10^10 J. That's
> >equivalent to about 6 * 10^3 kg of TNT, or 6 tons. Hence, a thousand times
> >less than you claim. Sure you haven't got tons and kg confused?
> A one ton ferrous rock is pretty small. In fact, it amounts to a boulder
> roughly 0.5-0.7 meters in diameter. Therefore, it is not surprising that it
> only has an energy yield of 6*10^10 Joules.

Yes it is quite small, and could easily be formed into the same
stealthed shapes that Mark is puportin for nukes.

> >Even if that's true, you now have to increase the figure to 6,000,000, and
> >your total to 110,000,000. In my earlier message I showed that you can
> >launch 18,000 before forces arrive from Earth. Note, of course, that if my
> >warheads can't target you at 6km/s, then your rocks will certainly not be
> >able to target me at 11km/s, and decimating the military will still leave
> >90% of them just itching to wipe you out.
> >
> >The fundamental problem is this: launching rocks from the moon gives you
> >around a factor of four increase in energy from gravity. So you can put
> >four times as much energy into Earth as you can generate. In order to hit
> >the Earth with 60 MT worth of rocks you need to generate 15 MT equivalent
> >of energy. That's 15 * 10^7 * 10^9 J. With a 1GW reactor to power your
> >100% efficient mass-driver that will take you about five years.
> >'Decimating the planet' with 1,100 MT will be impossible, because you'd
> >need a century to launch that many rocks, in which time the population
> >would have increased by a factor of four.
> >

Considering that the Nuclear winter effects are caused by the dust
thrown up, and that simulations have shown that a limited exchange of
less than 10,000 kt would have significant environmental effects (this
limited exchange was considered to be a strictly military conflict) I
think your estimates are being inflated again simply to bolsteryour won

> Just some quick comments on nuclear effects in the lunar environment:
> I checked all my civil engineering books and others to get real figures and
> data. Assume a nuclear device with a yield of 1-Mt, which is a a few times
> higher than the actual average yield of the current arsenal. An airburst
> would have virtually no effect in the lunar environment, so we will assume a
> ground burst. Assuming a theoretically perfect energy yield for a 1Mt
> device, we have a total output of ~4*10^15 Joules. However, the vast
> majority of the energy is harmlessly radiated, since an atmosphere and
> surface living is required for these effects to have significance. The only
> significant destructive force is the ground shock. Note that much of the
> ground shock energy found in atmospheric environments is airslap induced
> ground shock, and is therefore irrelevant. I assumed a depth of 2-meters,
> since soil arching plays a significant role in shock resistance. After much
> nasty calculation based on the above values, you will be disappointed to
> learn that the safe distance from ground zero is between 500-1000 meters for
> a 1Mt surface burst, depending on soil conditions. The reason is apparently
> that virtually all the blast and cratering effects are consequences of
> having an atmosphere, and soil doesn't propagate shock waves very well. In
> fact, structures specifically hardened against nuclear blasts could survive
> quite near ground zero, since non-airslap induced ground shock is attenuated
> extremely rapidly.
> Note that this is not an exact science, and that there are many factors
> involved. I was simply demonstrating that nuclear weapons lose most of
> their effectiveness in subterranean lunar environments. Nuclear weapons are
> hundreds of times more effective on earth.

Thanks for a more expert opinion James. Based on a reference I found of
4.2 x 10^22 ergs per Megaton, I calculated 15 tons of TNT equiv. for the
1 ton rock, so while I was seriously wrong in my own estimate, Mark was
under by almost three times. I don't know why that number stuck in my
brain, I had read it somewhere...

Additionally, in terms of damage ability, large nukes have, even in
atmosphere, an attentuating effectivness proportionate with their yeild.
A 10 MT device will only have a damage area of between 4-8 times larger
than a 300 KT device, which is why the powers have gotten away from
using big nukes, as they are not as cost effective as using the same
amount of material in four or more devices.

For examply, a high altitude attack by 3-4 B-52s carrying conventional
2000 lb HE bombs can damage as much area as a 10-20 KT nuke, though
without the radiation effects. In fact, because the concussive
shockwaves are from dozens of sources, rather than one device, they
actually are more effective in their area of effectiveness.

Also, as James demonstrated, a big nuke htting a lunar target would have
much closer actual damage equivalence to a much smaller device hitting
an earth based target. While I do not know how to calculate the actual
or estimated equivalence, I think that this is a significant factor to

If we look at a more proportionate device, like the Oklahoma bombing,
that device, I beleive was estimated to be equivalent to a ton of TNT
(please correct me with more accurate data).


Michael Lorrey ------------------------------------------------------------ President Northstar Technologies Agent Inventor of the Lorrey Drive

Website: Now Featuring: My Own Nuclear Espionage Agency (MONEA) MIKEYMAS(tm): The New Internet Holiday Transhumans of New Hampshire (>HNH) ------------------------------------------------------------ Transhumanist, Inventor, Webmaster, Ski Guide, Entrepreneur, Artist, Outdoorsman, Libertarian, Arms Exporter-see below. ------------------------------------------------------------ #!/usr/local/bin/perl-0777---export-a-crypto-system-sig-RC4-3-lines-PERL @k=unpack('C*',pack('H*',shift));for(@t=@s=0..255){$y=($k[$_%@k]+$s[$x=$_ ]+$y)%256;&S}$x=$y=0;for(unpack('C*',<>)){$x++;$y=($s[$x%=256]+$y)%256; &S;print pack(C,$_^=$s[($s[$x]+$s[$y])%256])}sub S{@s[$x,$y]=@s[$y,$x]}