Re: SPACE: Lunar Warfare

The Low Willow (
Tue, 14 Jan 1997 20:57:53 -0800 (PST)

From: James Rogers <>

10Mt (nuclear) = 900kton (mass @ 11000 m/s) = ~60m diameter ferrous rock

Impact parameters, assuming 10Mt (energy, nuclear equivalence) impacts on earth:

CRATER DEPTH: Nuclear: 60 meters Rock: 250 meters
CRATER DIAMETER: Nuclear: 2000 meters Rock: 1500 meters

IMPORTANT NOTE: According to sources, a rock must have a mass >500 tons in
order to impact at the above assumed velocity. Rocks with less than 500
tons mass are slowed by atmospheric drag to velocities less than 1000 m/s.
A one ton rock will not work.

From: Michael Lorrey:

No he showed a massive difference in ground penetration with the
advantage to rocks. He also compared what would be needed for
equivalence in damage on earth for a 10MT nuke and a rock. I am
proposing using smaller rocks, preformed as conic atmospheric
Penetration isn't too relevant, as few Earth structures are underground
anyway. Terrestrial military might be in trouble, but this thread
originally started as a MAD scenario.

I don't know my drag. Would conic rocks surive? Warheads do, but do
they have special surfaces? And the shaping won't help the fact that 1
ton rocks don't have much energy; James got megaton equivalence using
very BIG rocks. Rocks big enough for Earth to use active defense
against -- to detect and break up into badly shaped fragments that burn
up in atmosphere or at least slow down.

Merry part,
-xx- Damien R. Sullivan X-) <*>

"It was no sin, only a failure. 'And even if my troop fell thence
vanquished, yet to have attempted a lofty enterprise is still a trophy.'"
-"Forty-two years in Holy Orders, you hear all the sins in the Lexicon.
But angelism! Now there's a genuine rarey."
-- Julian May, _The Adversary_