# Re: SPACE: Lunar Warfare

Mark Grant (mark@unicorn.com)
Wed, 15 Jan 1997 12:21:22 +0000

On Sun, 13 Jan 2002, Michael Lorrey wrote:

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

Well, once it reaches equlibrium it will emit 1200W/m^2, the amount of
energy it absorbs from sunlight (AFAIR). James is correct that that is a
lot more energy than would be returned from a radar signal. However, an IR
detector is going to be much smaller (say, 1m^2) than a radar receiver and
that will compensate to a small extent. Most important is that you're only
getting a power-of-two attenuation and not power-of-four.

Let's see... if the frontal area of the warhead is 0.5 m^2, then it will
emit 600W, which will be down to 600/(2*pi*(4*10^8)^2) ~ 6 * 10^-16 W/m^2
at the lunar surface if it's just left Earth. I don't know how much power
you'd need to detect the warhead, but if we assume a microwatt again then
we'll need to reduce the distance by around 10^5, which would bring the
detection range down to about 4-10km. That seems too small to me. Anyone
know more about IR sensors than I do and want to check my sums?

However, there are ways to get around this. For example, we could build a
sunshield perpendicular to the warhead's trajectory which would reflect
most of the IR away in directions that the lunar folx couldn't detect. We
could also pack the warhead with liquid helium to cool the casing down to
near-background temperatures before it reached detection range. The main
concern would be how far we could drop the temperature of the warhead
before the hardware would malfunction, and since emission has a
power-of-four factor again even halving the temperature would cut the
emission from 600W to about 30W.

Finally, of course, it wouldn't be against the background of space as it's
coming from Earth and would obviously be against the warm background of
Earth itself, at least until it neared the moon. This may make life easier
for you if I did cool the warhead down, of course.

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

Yes, but we don't really care because we've been assuming all along that
"the rock always gets through".

> Considering that the Nuclear winter effects are caused by the dust
> thrown up,

Didn't we cover this a year or so ago? Nuclear winter is far from proven
and in fact seems quite unlikely.

> 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)

The same nuclear winter simulations proved that millions of people were
going to starve to death in Asia after Iraqui troops set the Kuwait oil
fields alight. As this was totally untrue, I have little faith in their
models. Garbage in, garbage out, as one of my computer science tutors used
to keep telling us.

> I
> think your estimates are being inflated again simply to bolsteryour won
> position.

Hardly. Your mass-driver would take years to launch that many rocks. Even
if the nuclear winter theories are correct they assume that the explosions
all happen in one day, and the effects would be greatly reduced when

Anyway, James just showed that rocks of that size will be slowed so much
by the atmosphere that the question is immaterial. You're now down from
10,000 tons per rock to nearer 0.1 ton.

> 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.

Yes, I mentioned this a few days ago. But small nukes are nowhere near as
effective against deeply-buried structures, and the cost of launching
numerous small nukes to the moon will probably be much higher than launching
just a few large ones. Of course, if we launch 100 rather than 10 we'd have
a much greater chance of doing at least some damage to your colony.

As I think I mentioned earlier, the reason I originally picked 10MT was
just that I happened to have the figures for crater size handy. I don't
have them for smaller bombs.

> While I do not know how to calculate the actual
> or estimated equivalence, I think that this is a significant factor to
> consider.

I don't care, because all that really matters to me is the crater size and
James figures seemed to scale to roughly the same values as on Earth. The
only disadvantage is that I'd have a harder time destroying your
mass-driver. *However* just think what my EMP will do to your
multi-kilometer long string of superconducting coils? I may not even need
to hit it to destroy it. James?

Mark

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