From: Michael Lorrey (retroman@tpk.net)
Date: Thu Jan 16 1997 - 17:00:19 MST
The Low Willow wrote:
>
> More problems: when you shoot rocks at Earth, how long will it take them
> to get there? If you rely on Earth's gravity to supply most of the
> energy, but are launching from very far away, I'd think it will take a
> while, as most of the acceleration happens close to Earth. More time
> for Earth to defend (against big rocks... the bigger it is, the slower
> and easier the target)
slower? I don't think so. Any rock will accelerate the same regardless
of mass, as Galileo proved, and the bigger it is, the more momentum it
has so its harder to deflect from its course with a given repulsive
charge. The time from luna to earth is 1-3 days, depending on trajectory
taken and excess velocity when crossing the L1 point.
> or to launch a crushing counterattack.
Since luna does not NEED to attack earth, the onus will be on earth to
attack first. This puts earth gov't in a poor PR position, where they
will have to fabricate some terrorist bombing or something to rouse up
public sentiment to support their inhumane actions.
>
> Whereas if Earth wanted to attack first, I've already argued that they
> could hit Luna in 8 hours, by putting missiles and their fuel in Earth
> orbit first. Actually I was wrong; using a better mean distance (384400
> km) the time is more like 9.7 hours.
You were projecting the use of a propulsin system so efficient that it
could accelerate constantly at above one G from the earth to the moon?
You must really beleive in the Lorry Drive!!!!
And an easily detectable military mobilization in orbit will be seen to
be as much of a threat by any non allied nations on earth, raising
political cain on earth, in the UN, and in the media, which can be
easily used by loonies for propaganda purposes (i.e. Jane FOnda, El
Salvador, the Pentagon Papers, Iran-Contra). In addition, the logistical
costs of putting earth resources in orbit (since they won't be able to
get them from the moon) add additional expense to such an endeavor.
You don't really understand the magnitude of what you are saying. It
took $60 billion in equipment losses and expenses operating a trillion
dollars in equipment with the cooperation of over a dozen nations to
mount a successful 8 day campaign here ON EARTH to recover 100 square
miles of desert wasteland.
There is a HUGE differential in the costs that a defender must incur vs
the attacker to win a confrontation. Here on earth the differential can
be as much as 1 to 100. It is known by vets that any attack that only
outnumbers defenders by 10 to 1 has a very small chance of taking a hill
or other redoubt. The 10th Mountain Div. in WWII lost 90% of its troops
taking a small part of the Italian Alps back from the Nazis, who were in
a retreat (and thus had poor morale, yet succeeded in incurring heavy
losses), and required three tries to finally take the assigned
territory.
>
--
TANSTAAFL!!!
Michael Lorrey
------------------------------------------------------------
President retroman@tpk.net
Northstar Technologies Agent Lorrey@ThePentagon.com
Inventor of the Lorrey Drive Silo_1013@ThePentagon.com
Website: http://www.tpk.net/~retroman/
Now Featuring:
Mikey's Animatronic Factory
http://www.tpk.net/~retroman/animations.htm
My Own Nuclear Espionage Agency (MONEA)
MIKEYMAS(tm): The New Internet Holiday
Transhumans of New Hampshire (>HNH)
------------------------------------------------------------
Transhumanist, Inventor, Webmaster, Ski Guide, Entrepreneur,
Artist, Outdoorsman, Libertarian, Arms Exporter-see below.
------------------------------------------------------------
#!/usr/local/bin/perl-0777---export-a-crypto-system-sig-RC4-3-lines-PERL
@k=unpack('C*',pack('H*',shift));for(@t=@s=0..255){$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]}
This archive was generated by hypermail 2.1.5 : Fri Nov 01 2002 - 14:44:01 MST