# Rocket/Thrust/Gas/Free Expansion of Gas

We know, the rockets in space use Newton's 3rd law to increase their velocity and hence move. What I don't understand is how it is possible in space aka vacuum-state without air? From what I know, Joule's "Free Expansion of Gas" says that free-expansion compresses the gas and is therefore "affected" by vacuum so it can't make the rocket move as the gas will have zero press/force. Could someone please explain me how rockets do really work and the above-mentioned statement?

Actually, please have a look at this site: http://cluesforum.info/viewtopic.php?f=23&t=1632

Not: The site appears to include some conspiracy theory thingummies, but made me wonder anyway.

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How does free expansion compress a gas? –  shortstheory Dec 30 '13 at 10:44
Before I think of drafting an answer, I must say I am quite close to losing my sanity after reading that forum you linked. There are so many flaws with Boethius's argument and most stem from just having absolutely NO understanding of physics. Please, do not believe that rubbish. –  shortstheory Dec 30 '13 at 10:51
But NASA's site doesn't say that, it says: F = m dot * Ve + (pe - p0) * Ae Here, the "m dot" term is important. It means mass expelled per unit time and multiplying it with velocity you get exactly the same units as force. So, there is no mistake with NASA's explanation here, I could go on and on, but that wouldn't be answering your question and I'm not too sure whether an answer like this would fit with Physics SE guidelines. –  shortstheory Dec 30 '13 at 10:53
Geez, I must say that it was pretty entertaining too; that guy is so passionately wrong. With regards to your question I don't think the free expansion term is even relevant here. Think about it, if the rocket was expelling stones at high velocity rather than fuel in space, wouldn't the rocket move the same way? –  shortstheory Dec 30 '13 at 10:58
Recommend migrating to seriously_funny.stackexchange.com –  Carl Witthoft Dec 30 '13 at 13:22

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.

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 $Ps^2$ pushing on it.

Now remove one wall. 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.

On the side, students who memorize contextless phrases and key words ("free expansion," "time dilation," "entropy is always increasing," ...) will almost certainly apply them incorrectly. 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.

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"Physics is not about magic combinations of words that one can invoke like some sort of incantation." +1 –  Ján Lalinský Jan 7 '14 at 18:32

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. 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.

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:

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. 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 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:

The reason this matters is that the atoms can't escape to the right or up or down because the walls of the combution 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:

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. 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.

So burning the fuel and allowing it to escape to the left means the rocket must have accelerated to the right. 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. In fact we can be more precise about the force. If the rocket produces $N_s$ particles of exhaust gas per second then the momentum change of the rocket per second is $-N_s mv$. Momentum change is force times time, so the force on the rocket is simply:

$$F = N_s mv$$

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.

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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

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