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From: zweig@brutus.cs.uiuc.edu (Johnny Zweig)
Newsgroups: sci.space.shuttle,sci.space
Subject: Re: Exhaust velocity
Message-ID: <1989Oct23.203306.24154@brutus.cs.uiuc.edu>
Date: 23 Oct 89 20:33:06 GMT
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conor@inmos.co.uk (Conor O'Neill) writes:

>In article <1989Oct18.174154.23242@utzoo.uucp> henry@utzoo.uucp (Henry Spencer) writes:
>>In general, correct.  For one thing, it's easier to build solid motors
>>in large sizes (i.e. high thrusts).  For another, the average molecular 
>>weight of the exhaust is higher, which is bad for getting maximum velocity
>>but good for getting maximum thrust.


>I've seen this said before (many times) but never with a simple explanation.
>My school physics seemed to imply that it is exhaust momentum which matters,
>not simply velocity nor molecular weight. Could someone please elaborate.

Note that the exhaust's momentum involves its velocity relative to the
spacecraft's motor. There are actually 2 kinds of thrust: Momentum Thrust
and Pressure Thrust. The first has to do with the momentum of the stuff
going out of the motor, the other has to do with the crose-sectional-area of
the nozzle and the difference between ambient pressure outside and the
exhaust gas pressure. (According to the "Space Handbook", available from the
US Gov't Printing Office).

I can't quite hack all the Isp, TC/m, and other mumbo-jumbo. I think it boils
down to the fact that the mass-flow (Kg/S) out the back of the motor gives
you part of your thrust, and the interaction between what comes out the back
"pressing against" the atmosphere is another. So if I have two rockets that
spit out the same mass of stuff per second, but one has higher nozzle
pressure, that one will give me somewhat more thrust off the launchpad (once
up in space there should be no difference since there's no atmosphere).

I would also be interested in a more accurate explanation, if someone who can
avoid simply stating some equation and saying "see, dummy?" would care to
post one.

-Johnny Non-physicist