From: neufeld@helios.physics.utoronto.ca (Christopher Neufeld)
Subject: Re: Invading planets
Organization: University of Toronto Physics/Astronomy/CITA
Date: Sun, 24 Jan 1993 00:29:57 GMT

In article <1js764INN6q7@uxa.ecn.bgu.edu> mgcbo@uxa.ecn.bgu.edu (Charles B. Owen) writes:
>
>Everyone is very much in love with dropping or throwing rocks in space.
>Remember, though, that to drop that asteroid so that it can destroy a
>city, it must have enough kinetic energy to destroy a city and that
>energy has to come from somewhere.  It will not come from just falling
>into the Earth's gravity well.  Text's such as Lucifer's Hammer assume
>an object on an unrelated, but intercepting course.  That is not the
>same as "dropping a rock".  Now many may say that we will just "divert"
>a rock from the asteroid belt.  The energy required to do so will be
>simular to that expended by the rock reentry.  So, you are not
>accomplishing much.
>
   Wrong on several counts. The delta-v required to bring a rock from the
asteroid belt to the Earth is at or below 3 kilometres per second. This
rock then hits the ground at greater than escape velocity, somewhere over
twelve kilometres per second.
   The three kilometre (or much, much less for some closer orbiting
bodies) per second delta-v can be delivered over a relatively large
period of time, while the energy released by the impact of the asteroid
is liberated all at once.
   Let's take a hypothetical near-earth asteroid. It's ten kilometres in
diameter, iron-nickel, and in an orbit that just happens to be running it
close to the earth a year or two from now. A delta-v of perhaps one
kilometre per second, perhaps even less, has to be applied to this rock.
A formidable task, but not one which is obviously beyond the reach of a
civilization capable of mounting a war in space. This rock now hits the
target city of my choice. I have a lot of choice, by taking advantage of
the gravitational force of the earth in the final approach I can hit
almost any latitude target quite easily, though I might be forced to
restrict my choices to the hemisphere facing the rock when they meet,
since I might not be able to delay the meeting by as much as a half day.
Well, this rock hits, releasing about 2E20 joules of energy as heat.
That's 52 gigatons of TNT. It's as much energy as hits the whole earth
from sunlight in twenty minutes.
   This is overkill.
   Now, take that same rock, and break it into kilometre-sized bodies. I
have a thousand of these, each with one thousandth of the energy. 52
megatons of TNT is a city-buster, so that one rock, properly used, can
destroy one thousand city-sized targets on the ground.

>Anyway, this planet has been able to produce 60,000 nuclear weapons.
>Why worry about dropping rocks.
>
   I presume we're still talking about a space-based war context, such as
between two planets, or invaders to the system against natives on a
planet there. Rocks are available locally (saving on transportation
costs) and require only low-tech manufacturing. They are durable, immune
to electromagnetic countermeasures, and can deliver their energy even if
struck head-on by an anti-missile weapon.  These all make them more
convenient than nuclear weapons in the context of this thread. Further,
in a war between colonials and a home planet, one might presume a
government which prevents the easy construction of nuclear munitions by
the colonies. You can outlaw atomic bombs, but you can't really outlaw
rocks.

-- 
 Christopher Neufeld....Just a graduate student  | Everyone talks about
 neufeld@helios.physics.utoronto.ca    Ad astra  | apathy, but no one does
 utzoo.utoronto.ca!generic!cneufeld              | anything about it.
 "Don't edit reality for the sake of simplicity" |