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.cb BLACK HOLE ROCKET SHIPS


	This started with the fact, recently emphasized by the cosmologists,
that although gravitation is a very weak force, most of the energy in the
universe is gravitational, because gravitation is an entirely attractive
force.  As Wheeler point out, we could get the rest mass of an object
as energy by lowering it into a black hole.  However, lowering something
into a black hole isn't feasible, because the rope would break, and
the platform would have to be in orbit anyway so as not to fall in, and
then it wouldn't be possible to lower anything.

	With two black holes in mutual orbit it is much easier to extract
energy.  The simplest case to compute is when one of them is very much
smaller than the other.  Then a test object moves in the field of the
larger (call it ⊗M1) until it is much closer to ⊗M2 than the radius
of %2M2%1's orbit about ⊗M1.  The simplification occurs in both the
Newtonian and the general relativity cases, but I don't know how to
do relativistic calculations yet, and the Newtonian calculation is
very instructive.

	We assume that our test object ⊗M3 is very much lighter than ⊗M2. 
We can see that the details of its hyperbolic orbit about ⊗M2 don't matter;
⊗M3 orbits in the field of ⊗M1 and then has an elastic collision with ⊗M2 and
then orbits about ⊗M1 again.  This is true, because conservation of
energy and momentum leave only one free parameter - the angle at which
⊗M3 bounces off ⊗M2.  This can be anything from 0 degrees
in case of a distant encounter
to 180 degrees in the case when ⊗M3 comes very close to ⊗M2.  Moreover,
⊗M3 can bounce off in any direction depending on which side it approaches
⊗M2 from.  If we are shooting ⊗M3 into the system, we can control where
it meets ⊗M2 by when we shoot and we can control the miss distance and
the direction of the encounter by small changes in the initial direction
of ⊗M3.  It would seem that the considerations of this paragraph apply
in both the Newtonian and the Einsteinian cases.  If ⊗M2 is not
a point mass, then there is a limit on how close ⊗M3 can come
which prevents us from getting deflections close to 180 degrees -
which, alas, are the most useful and interesting.

	Suppose that ⊗M2 is in circular orbit about ⊗M1 with radius ⊗R 
and velocity ⊗V.  Centrifugal and gravitational forces must balance, so

!!a1:	%2(G M1 M2)/R%52%2 = M2 V%52%2/R%1,

and

!!a2:	%2V = sqrt(G M1/R)%1.

If ⊗M3 comes from infinity with initial velocity ⊗V3, it gains kinetic
energy at the expense of its potential energy, so that its velocity
⊗v at the time it reaches a distance ⊗R from ⊗M1 satisfies

!!a3:	
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.begin verbatim
John McCarthy
Artificial Intelligence Laboratory
Computer Science Department
Stanford University
Stanford, California 94305

ARPANET: MCCARTHY@SU-AI
.end

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