Re: SPACE: Lunar Warfare

Mark Grant (
Sun, 12 Jan 1997 17:11:02 +0000

No insult taken, I just think you're wrong.

> 1) the colony would all be in one tight little place

Well, you'll be adding a lot of expense and complexity if you try to
spread your colony out too much. Anyway, a 10MT warhead will destroy any
habitation at a depth of six feet over an area of more than a square km.
The same could be achieved with a larger number of much smaller warheads.

This assumes, of course, that a groundburst explosion on the moon will
have the same underground effect as on Earth.

> 2) all habitats would be mapped out with correct public coordinates

What valid argument would your colony company have for keeping its
location secret? Wouldn't the governments of the world be rather unhappy
when they discovered just how hard you were working to prevent them from
controlling you?

> 3) that the warheads would be accurate (a big if at the velocities,
> course,and distance you are talking about if you know anything about
> such things)

Well, Apollo managed an accuracy of under a mile back in the seventies.
That's all a 10MT warhead would need in order to have a high probability
of destroying your colony. The mass-driver, of course, is a huge target so
it can easily be destroyed.

> Note: SDI failed not because of technical problems, but because of
> politics and money.

Yes, lack of an infinite money-pot to spend...

> 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!?!?!?!?!

Let's suppose you have a 1GW radar beam which scans all space between the
Earth and moon once a minute (and going faster would make detection
harder). That means your 1GW is going into an area of about 10^15 square
meters. My warhead has a radar cross-section of about 1 cm^2, so it reflects
10^9 * 10^-4 * 10^-15 W. That's 10^-10 W at a distance of about 400,000
km. Now, if I remember radar correctly, the reflection you receive will be
10^-10 / (4*pi*(4*10^8)^2) ~ 10^-28 W/m^2. Even if you set aside 1000 square
km as a receiver your signal will only be 10^-19 W. I don't believe that you
can do anything useful with such a small amount of power, and a realistic
value would be much lower.

Now let's assume that a microwatt is enough signal to detect something.
The received power will scale with the fourth power of distance, so you
need to improve the signal by 10^13, or reduce the distance by a factor of
>10^3. So your huge investment in radar technology would mean that you
could detect my warheads from 300km, around fifty seconds before impact.
If, more realistically, you can only build a 1000m^2 receiver then you'll
be down to 3km, or around 0.5 seconds to impact. Even worse, if you're
scanning the sky once per minute then you only have a 1% chance of detecting
it before it hits.

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?

Not to mention that your radar transmitter would be an wonderful beacon
for me to use in targetting my warheads... and if I can't do that I just
blast you with a 1GW radar from Earth and use that for terrain mapping
guidance in the last few seconds. I don't have to worry about the
power-of-four attenuation because my receiver is a thousand times closer
than the transmitter.

Note: You might be able to do something with radar receivers in orbit;
stealth works in a large part by scattering the radar reflections off in
a direction away from the radar source. Receivers in orbit might be able to
pick up some of that scattered radiation and help your tracking.

Finally, I could do something clever like braking around the back of
the moon to put the warhead on a low-angle ballistic trajectory towards
your colony. If you're only scanning the sky in the direction of Earth,
then you're history.

> This is controlled by a computer/radar/IR
> control system that has been demonstrated to be VERY effective.

But can it hit a marble in 0.000033 seconds? AFAIR the rate of fire is
only about 100 rounds per second, so a single gun certainly can't. If
you had one gun per incoming warhead and decoy you'd have 33 milliseconds,
giving you three shots at each target. Might work, but I don't think that
the US government would be willing to sell you a thousand of them ;-).

> The reports in the media that Patriot was wholly unsuccessful were
> untrue.

Well, I don't have my mailing list archives online, but I seem to remember
that the article was reporting on a military report to a Congressional
committee, so Congress might be interested to hear about it.

> if the majority of the installation is underground, they would never
> notice.

High-powered lasers and radar stations won't do you much good if they're
underground, and you don't need 1000m^2 of radar receiver for tracking
incoming transports with a radar cross-section the size of an office

> What in numbers of people? useless. in numbers of rocks launched by
> railgun? You would not need to decimate the entire surface, that
> assertion is just ludicrous.

But you were the one who first made that assertion. You said that you
could pulverize the Earth's surface before missiles from Earth hit you. I
proved that even with your optimistic figures that was far from the truth.

> The goal of battle is to destroy the
> enemy's ability to wage and support war.

And how do you do that with dumb rocks? How do you target military
installations from 250,000 miles away with no terminal guidance?

> Using estimates that the US and Russia
> have the combined ability to decimate the world ten times over,

That figure is, of course, nonsense. Realistic estimates show that even
wiping out the entire population of Britain (i.e. 'decimating them ten
times over') would take around 2,000 MT. That's 1/4 of your figures for
the total stockpile to kill off 1% of the world's population.

Even if your figures are correct, those 40,000 warheads were still not
considered to be enough to wipe out all military targets! This is why they
wanted even more. Of course you would only need to destroy those
installations that could hit you, but that would just leave the rest of
the military very pissed off.

> 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.

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?

> Also, as
> the military industrial complex of the world covers a much smaller
> percentage of it surface, I'll bet that a lunar colony could decimate
> the US Chinese and Russian military capability with less than 6000 rocks

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.

> (count the installations), and that is taking into account interception
> systems, giving 105 rocks to each target based on importance. Places
> like NORAD control in Colorado would get some extras just for fun.

10 ton impacts will have little effect on a base that's buried under
hundreds of feet of solid rock.

> Also, consider this: Luna could put any number of rocks up in orbit way
> ahead of time, as part of its minimg operation to supply the
> construction of solar sats and colonies.

True, but six million? You really need five year's backlog of launches in
space to supply construction? And what if one of those rocks is perturbed
enough by some passing body or gravitational change to fall to Earth by

> Keeping a decent backlog of
> rocks in orbit with little thruster packs would work as an excellent
> deterrent system that could not be sabotaged.

How much will six million 'little thruster packs' cost you? Lunar orbit to
Earth orbit will take around 1.5 km/s, so if you have solid rockets then
Mr = e^(1500/2500) ~ 1.5, and you'll need at least half a ton of current
rocket fuels. At current prices that's around $15,000 per thruster, giving
a total cost of at least $90 billion before we even think about
electronics and other hardware. You also have to launch all those
thrusters, which will take another two or three years.

So by now you've spent untold billions building your colony, and all your
mass-driver has done for the first eight years is stockpile rocks in orbit
in case you decide to attack Earth, *and* you've spent another $90 billion
on thrusters. Frankly I think your bank manager and creditors will come
after you long before the US government.

> 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.

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.

> no. safer for the inhabitants (as they will be less likely to be hit by
> collateral damage).

Which collateral damage? Your mass-driver is there to launch rocks for
construction of habitats, not as a weapon! You seem to think that you can
run your commercial colony on a war footing 24 hours a day, 365 days a

> Look at Area 51
> for a good example.

You mean that place that everyone knows about and of which aerial photos
have even been included with plastic kits of stealth planes?

[Commie Infiltrators]
> They didn't get every secret,

They got everything they needed to build a precise duplicate of the
American bombs. At one stage they were collecting so much classified
information from all over America (not just on nukes) that they had to
ship it out by the plane-load!

> You are also not
> understanding the dynamic of the people involved on the moon. ANyone
> forced off earth will feel toward earthgov as Cuban Expatriates in Miami
> feel toward Fidel Castro. How many of them would spy for old Fidel?

Uh, are we talking about completely different things here? First your
colony was a commercial venture to build space habitats, now it's some
kind of prison, presumably run from Earth. Which is it? If it's a company
then most people probably won't want to kill their relatives on Earth, and
if it's a prison then they won't be able to build the kind of hardware
that you're talking about.

> True, but fortunately, every credit agency in the world is an
> intelligence agency for hire. you can learn A LOT from a persons credit
> record.

And every credit agency in America cooperates with the US government as
part of the Witness Protection Program. Do you really think they don't do
the same for CIA agents?

> ALso, the kind of Americans likely to migrate to the moon are
> not likely to resemble candidates for CIA training. (not many Yale
> yuppies digging moon dirt)

Most of the people working on the moon would be engineers or scientists.
The original L5 studies (I presume you have read 'The High Frontier'?)
showed that their mass-driver could be continually fed with one bulldozer.
Doesn't leave much opportunity for unskilled dirt-diggers.

> No wonder we are having this argument, you haven't even studied the
> preparation material. What would your professors have said? Read it then
> come back.

I don't need to, I know the basic plot because it's been discussed so
many times before. I just want to know more about the specifics.


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