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Path: utzoo!utgpu!radio.astro!helios.physics!griffin
From: griffin@helios.physics.utoronto.ca (Prof. A. Griffin)
Subject: Re: Exhaust velocity
Message-ID: <1989Oct25.020104.10782@helios.physics.utoronto.ca>
Reply-To: griffin@helios.physics.utoronto.ca (Christopher Neufeld)
Organization: University of Toronto Physics/Astronomy/CITA
References: <538.252A3A3B@mamab.FIDONET.ORG> <34577@srcsip.UUCP> <126311@sun.Eng.Sun.COM> <1307@accuvax.nwu.edu> <15596@netnews.upenn.edu> <1321@accuvax.nwu.edu> <1989Oct18.174154.23242@utzoo.uucp> <2639@ganymede.inmos.co.uk>
Date: Wed, 25 Oct 89 02:01:04 GMT

Disclaimer: I am NOT professor Griffin. If you use "F", please check the
attribution against the signature.

In article <2639@ganymede.inmos.co.uk> 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.
>
   All right, here's a high-school physics description. We assume that the
reason the gas is moving so quickly out the back of the rocket is that the
gas is hot, and that it has thermalized according to a Maxwell-Boltzmann
distribution. Fancy talk which means that the average velocity (careful,
not the root mean square velocity) is sqrt( 8 k T /(pi m) ) where m is the
molecular weight of the particle which has been thermalized, and k is the
Boltzmann constant. So, if you're going to throw one kilogram of gas heated
to T degrees out the back of your rocket, you would do best to choose one
with a low molecular weight, because of the m^-1/2 dependence.
   By the way, about that equation. I've been known to slip a decimal on
occasion, as history will show, but the m^-1/2 dependence is as described
in a book, and we all know that textbooks are never wrong :-)

>-- 
>Conor O'Neill, Software Group, INMOS Ltd., UK.
>UK: conor@inmos.co.uk		US: conor@inmos.com
>"It's state-of-the-art" "But it doesn't work!" "That is the state-of-the-art".
-- 
 Christopher Neufeld....Just a graduate student  |  "Scotty..now _would_
 cneufeld@pro-generic.pnet01.crash               | be a good time!"
 griffin@helios.physics.utoronto.ca              |   - Pavel Chekov
 "Don't edit reality for the sake of simplicity" |