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From: David.Smith@cmu-cs-ius.arpa
Newsgroups: net.space
Subject: Re: Orbital plane change
Message-ID: <12598@sri-arpa.UUCP>
Date: Mon, 23-Apr-84 12:18:43 EST
Article-I.D.: sri-arpa.12598
Posted: Mon Apr 23 12:18:43 1984
Date-Received: Sat, 28-Apr-84 10:39:58 EST
Lines: 37

In my original message about using aerodynamic lift for orbital plane
changes, I neglected an important fact.  If L/D is 4, then you get
four times the effective thrust for an equivalent rate of fuel expenditure.
But this does not mean that the fuel requirement is one-quarter what it
would have been.  Rather, it means that for equivalent delta-V, the required
mass ratio is the fourth root of what it would have been.  This follows from
the fact that deltaV = Vexhaust * ln(mass ratio).

Let's take the problem of transferring between the orbits of two space
stations, each with orbital inclination of 28.5 degrees (latitude of
Cape Canaveral), but launched 12 hours apart, so that the tug has to make a
57 degree plane change.  As I recall, that came out to about a 16,400 mph
delta-V.  And let's be old fashioned and use chemical rockets.

The specific impulse of hydrogen-oxygen rockets is often listed as 425
"seconds."  This is 425 seconds of kilogram-force per kilogram of propellant.
Make that 425 * 9.8 = 4165 newton-seconds per kilogram.

Use the formula of f = ma = m dv/dt, and consider that instead of
accelerating a finite block of exhaust over some time, we kick out an
infinitesimal of mass at instantaneous exhaust velocity.  This changes the
force formula to f = Ve dm/dt.  Combine this with 4165 n-s/kg, and we get
Ve = 4165 meters per second, or 9319 mph.

If we want to get that 16400 mph with pure propulsion, then the mass ratio
must be exp(16400/9319) = 5.81.  Using the L/D=4 maneuver, the mass ratio
must be 5.81**(1/4), or 1.55.  The aerodynamic plane change cuts the
required propellant by a factor of 8.7.

This analysis ignores other factors, such as:
	mass of wings
	reduced mass of tanks
	fuel to lower and raise perigee
	fuel and structure that would have been required to put 8.7 times
		the propellant into orbit (either from the earth or the moon)

					-- David Smith