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From: MCGRATH%OZ.AI.MIT.EDU@MIT-XX.ARPA ("Jim McGrath")
Newsgroups: net.space
Subject: Accelerator Momentum Loss
Message-ID: <12177393347.39.MCGRATH@OZ.AI.MIT.EDU>
Date: Wed, 22-Jan-86 20:39:31 EST
Article-I.D.: OZ.12177393347.39.MCGRATH
Posted: Wed Jan 22 20:39:31 1986
Date-Received: Fri, 24-Jan-86 21:25:42 EST
Sender: daemon@ucbvax.BERKELEY.EDU
Reply-To: mcgrath%mit-oz@mit-mc.arpa
Organization: The ARPA Internet
Lines: 57



This is a followup to a previous message of mine concerning the loss
of momentum (and thus the orbital decay) experienced by an accelerator
placed in low earth orbit (LEO) and designed to accelerate sub-orbital
payloads to low earth orbit velocity.  Two obvious solutions suggest
themselves.  First is to install reaction motors to accelerate the
structure again.  Since they may be far more efficient than those used
in the earth to orbit phase (e.g. ion rockets), you still turn out a
winner.  The second is to decelerate payloads from beyond LEO to
sub-orbital speeds.  The problem here is that you do not want to be
required to lose mass from the space environment to the earth.
Moreover, it is unlikely that you could balance the traffic in the
early stages of use.

Another solution is possible.  If a charge is placed on the
accelerator, then it becomes a gigantic charged particle moving in the
earth's magnetic field.  We know that in this case a force is exerted
on the particle, a force that can be used to transfer angular momentum
from the field to the particle, thus making up for the momentum lost
by the payload acceleration.  Note that no mass is needed to
accomplish any of this, only a power plant which may not even be
physically coupled to the accelerator.

How much of a charge is needed?  The following is a very simple
analysis, correct only to the first approximation.  A lot of
simplifying assumptions are used.  But the gist should be correct.

Assume throughout that the accelerator is in a circular orbit
concentric and coplanar with the earth.  Let delta J = J (final) - J
(initial) [ J = angular momentum of accelerator ].  We know that this
is equal to M*(rf*vf - ri*vi), where M = mass of the accelerator, rf =
final radius of orbit, fi = initial radius of the orbit, vf = final
orbital velocity, vi = initial orbital velocity.  Now after the loss
of momentum we charge up the accelerator, resulting in a delta J equal
to rf*Q*vf*B*t, where rf = initial radius of the accelerator before
charging, Q = charge on the accelerator, vf = initial velocity of the
accelerator before charging, B = earth's magnetic field, and t = the
duration that the charge is present.  Note that since the velocity is
increasing, and radius decreasing, with time, this is not strictly
correct, but is true in the limit as the mass of the payload becomes a
very small fraction of the mass of the accelerator.

Setting the two equal, we have: M*(rf*vf - ri*vi) = rf*Q*vf*B*t.  Using
conservation of linear momentum we have ((M-m)/M)*vi = vf, where m =
mass of the payload.  Also, we know that ri = GMe/vi**2, and rf =
GMe/vf**2, where Me = mass of the earth.  So we can combine all this
into the following (if I manipulated right):

        Q*t = m/B.

Unfortunately, I do not have a good reference value for B.  Does
anyone out there?


Jim
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