A test like it could be done. But it hasn't been, or hasn't been released to the public.
Well? Go ahead and get started, if you are convinced it would (ahem) work.
I'd like to review some answers to Boethius' explanation, which are appearing at the
Stock Stack Exchange site, which — to me, at least — is indicative of the inadequate explanations for why it supposedly should. The posted question references our own CluesForum thread:
http://physics.stackexchange.com/questi ... ion-of-gas
Chris White (my bolds) wrote:If someone ever says "free expansion does no work" all they mean is that it does no work on the vacuum, which is pretty obvious in retrospect. This is because 19th century experimenters and 21st century high schools find it easiest to talk about gas properties in terms of pistons pushing on containers of gas. If the piston is replaced by nothingness, well clearly no work will be extracted from the system.
Credible beginning? The onus of work is on the vacuum instead of the rocket? I am not so sure ...
This doesn't mean the gas doesn't do anything. Think of it this way: First, you have a closed container, sitting in vacuum and containing a gas with some nonzero pressure P inside. The force on the walls is the same in all directions, no matter the shape of the container, but for simplicity you can picture it as a cube with side length s. Each wall will have a force Ps2 pushing on it.
The following is a priceless explanation, worthy of a NASA charlatan.
Now remove one wall.
With
what? Oh, well. Now it's easy. Dispense with Ps2. We have physics working one way the first moment, and another way the next, due to the conditions being changed with imagination.
There will no longer be any force acting on it (your "free expansion" principle), but until the gas is fully evacuated there will be a force on the opposite wall. So your container has a net force in the opposite direction from the gas expulsion lasting for some time. Momentum is conserved; rockets work.
— Sincerely (ostensibly), Mister 'Chris White'
Aaaaah. Brilliant.
Step 1. Act as if all six walls are there up until the moment you need the rocket to move.
Step 2. Add fuel from somewhere.
Step 3. Remove one wall from the design
but keep all the old physics of the previous design for "some time"!
You've got rocket power! Holy shit.
Unfortunately, for all the prefacing 'Chris White' does, the questions remain: how a wall appears and disappears; or what the reinforced, super strong, but extremely agile, vacuum-proof membrane is made of; whether it takes sufficient time at all for the "combusting" gas to evacuate once the membrane is open larger than a single molecule's width; what those time periods look like; etc. As you can read yourself, he concluded curtly, "momentum is conserved," and, "rockets work." before making this quotable NASA apology:
"One always needs to understand context: What has no work done? Whose perspective says time is dilating? Physics is not about magic combinations of words that one can invoke like some sort of incantation."
One could modify his/her/its/their statement to read something like this:
"Statements are not about combinations of words that one can say."
~or, roughly~
"Shut up, don't think, and let me tell you how it is without needing to actually explain my perspective."
However, a slightly more illustrated (if still inadequate) explanation follows, thanks to 'John Rennie':
'John Rennie' (my bolds) wrote:When you're considering the properties of gases there are often two ways to look at the problem. The first is to use the continuum approximation leading to the usual laws like Boyle's law, Charles' law etc.
... which is probably closer to what we are seeking, but which nobody has produced the formula for yet!
The second is to treat the gas as many tiny particles (i.e. the gas atoms/molecules) and use Newtonian mechanics. In this case I think the second way is to understand what's going on.
In other words, treat us like children who cannot possibly understand the high concepts (that they fail to provide, anyway). Hmm.
The rocket motor burns a mixture of fuel and oxygen to produce a very hot gas. By very hot we mean that the gas molecules have very high random velocities
[LOL! Orly? -hp]
- aO8hX.gif (3.39 KiB) Viewed 12689 times
This diagram is supposed to show a representative sample of the atom/molecules in the flame. They are all moving in random directions, so the total momentum of all the atoms is going to be
close to zero.
... so we're being told to imagine the closed box system given to us by 'Chris White'. But surely, we aren't going to be given the old "switcheroo" explanation again, are we?
This means burning the fuel has not changed its momentum - this may seem a funny thing to say, but bear with me.
If the fuel were burning in a vaccum
[sic] the random directions of the atom velocities would mean the ball of atoms expands in a roughly spherical way and the total momentum stays zero. But the fuel is not burning in a vacuum,
it's burning inside a combustion chamber:
- HdcTg.gif (3.99 KiB) Viewed 12689 times
AAAAAHH!!! They did it again!
Imagine a cup full of water sitting on a table. How does the water get in your mouth without your touching it? It's actually not sitting on the table; it's being suspended in the air, at an angle, and the water is pouring into your mouth right now!
I ... see.
So, again, with illustrations, we're meant to imagine a totally fictional boxed bit of non-vacuum,
built strong enough to resist an infinite vacuum, in which the expansion occurs. Then, with no regard as to the construction of this mechanism, we are meant to — once more, like 'Chris White' — picture the structure morphing into something else while preserving the imagined activities of the original!
The reason this matters is that the atoms can't escape to the right or up or down because the walls of the combution
[sic] chamber are in the way. So they will bounce around
until some random collision (with the walls or other atoms) gives them a velocity pointing to the left:
- fgP09.gif (4.28 KiB) Viewed 12689 times
Now, this is novel. The nozzle (
formerly imagined as a closed system) won't direct atoms due to its design as a nozzle; it will simply be caused by random collisions between the chamber and by
less than half of the explosive force going "right" — which is made evident by all the atoms going to the "left". No mention of collimating parabolas or further dissection of a continuum approximation. Just a box.
Presumably, what he/she/it/they means to say is that when the chamber is exposed to vacuum (still waiting for a single explanation as to how
that is done) then restores pressurization (an even more physically remarkable feat, if one is to presume the chamber must be constantly in a state of depressurization and repressurization — all from some extremely strong, extremely compressed pressure slowly being doled out from inside the main body of the rocket) the molecules all rush to the vacuum due to free expansion.
So very quickly all the atoms are going to end up with their velocities pointing in roughly the same direction, because at that point they can escape from the combustion chamber and go flying off into space.
Right. So ... this debunks Boethius' explanation ... how, again?
Now let's calculate the momentum of all those atoms. If there are N atoms and the mass of each atom is m and their average velocity is v then the total momentum is now Nmv (we'll take velocity to the left to be positive). The momentum of the fuel before burning was zero, and after burning it's Nmv, so the momentum has changed by Nmv. Conservation of momentum means the rocket must have changed its momentum by −Nmv so that the total momentum change adds up to zero.
Ha ha ha! Right. Because the atoms unattached to the rocket are escaping into the vacuum, that means the rocket must have changed its momentum. Interesting move — swapping the words "spent fuel" for the word "rocket". Still doesn't work, but ...
So burning the fuel and allowing it to escape to the left means the rocket must have accelerated to the right.
Oh sure, it must have! That's what we see in a childhood drawing of flames rushing out the back end of something. So, right. It totally flew in a vacuum. Makes sense.
In other words the rocket engine has produced a force on the rocket, and we've calculated this without needing to think of pressures or other macroscopic quantities.
Ooo. Ouch. The word "calculated" is so stretched here, it's going to need a deep tissue massage after their post.
In fact we can be more precise about the force. If the rocket produces Ns particles of exhaust gas per second then the momentum change of the rocket per second is −Nsmv. Momentum change is force times time, so the force on the rocket is simply:
F=Nsmv
This force is produced simply because atoms moving to the right bounce off the end of the combustion chamber, and hence push the rocket to the right, but atoms moving to the left don't.
- 'John [hobby programmer and photochemist] Rennie'
- NFoAF.png (4.04 KiB) Viewed 12689 times
Real professionals!
But even on that site of willing believers, you have people asking questions of the so-called explanations given.
Interestingly, the flow will be choked at some point along your hypothetical shaft. I've always struggled with exactly what choked flow means in the molecular model of gases. Nonetheless, you could use your model to justify several aspects of a De Laval nozzle. For instance, the outward facing cone is at least a little parabolic-ish, so we can tell ourselves that rocket engine designs collimate the particles. But that's not the exact same justification as the continuum model. Exactly how you jump between those models is a funny question. – Alan Rominger Jan 7 '14 at 19:38
@AlanSE: the discussion above is highly simplified, but at the end of the day, and regardless of how the rocket engine is designed, the thrust generated is the rate of change of momentum of the exhaust gases. – John Rennie Jan 8 '14 at 7:12
An amateur physicist and his student walk into a bar.
The students says to his teacher,
"Uh, excuse me, sir, but on the last test, we were working on rockets and we learned that the momentum was not adequate to propel the rocket at necessary velocities due to the lack of atmosphere. Yet you gave me a bad mark for catching the trick question. How, exactly, did you switch models in the problem from an atmosphere to a vacuum?"
The other replies,
"Regardless of how I did it, it is done! It is up to you to figure out how the switch occurred, not to change it back just because you don't like it."
The student replies,
"Well, sir, would you get me a beer then?"
The other says,
"But didn't you say you were treating? I thought I taught you well."
The amateur physicist student concludes,
"You did, sir. If you can't make a rocket fly where you want it to go, you can at least give it a little motivation."
Alternative ending:
"... with just a bit of hot air." 
