Re: SPACE: Lunar Warfare

James Rogers (
Mon, 13 Jan 1997 02:08:23 -0800

At 05:11 PM 1/12/97 +0000, Mark Grant wrote:
>> 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.
>> 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

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.

>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.

>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.
>> Your assertion that a 1 ton rock falling from lunar space to earth would
>> only impact with 10 tons of force is also completely laughable. You've
>> obviously never taken a physics course.

Check your math and your units.

Your 10 tons of force figure is even more laughable since "tons" are not a
unit of force. As for nuclear equivalence, 1-ton falling rock =
0.01-kiloton nuclear yield = 10 tons of TNT nuclear equivalence (which are
units of energy). To get into energies at the bottom of the nuclear range
you would need a rock the size of a building.

>Actually, I have a degree in the subject, which is why your figures looked
>dubious and I checked them rather than taking them at face value. My
>calculations are publically available above; please either prove me wrong
>or stop claiming that you're right. The whole problem with your scenario
>is that your calculations are out by a factor of 1000, which invalidates
>the whole scheme.
>> Force equals mass times
>> acceleration ( for which (end velocity^2)/2 can be substituted). Using
>> gravity, a 1 ton rock accelerates at an average of .5 G for over 150,000
>> miles (assuming a negligible velocity near the L1 point where Earth and
>> lunar gravity forces are equal). Falling falling falling.
>And if you do the integration the answer is around 11km/s. As I used in
>the calculations above.

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.

-James Rogers