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[QUOTE]Originally posted by Timo: [QB] Identity Crisis already got the collision thing exactly right with his momentum analysis. In kinetic gas theory, one usually assumes the collisions will be elastic, so that energy is preserved in addition to momentum. In reality, this assumption actually holds pretty well, but we don't need energy conservation. We only need momentum conservation. Momentum is always conserved *for the entire system in isolation*. Here our system is the ship and the exhaust (in vacuum, so there's nothing to thrust against and no air resistance). Initially, they are at rest wrt the camera, so momentum p(system)=p(ship)+p(gas) is zero. Regardless of what the gas does within the ship (it could stop for tea at 10-fwd for five minutes if it wanted), eventually it exits the vessel. When it does, it carries momentum p(gas)=m(gas)v(gas). The other half of our system, the ship, therefore MUST carry the momentum p(ship)=-m(gas)v(gas), no matter what. The sum of momenta, p(system)=p(ship)+p(gas), MUST remain zero. And don't even think about ranting about the gas hitting the walls at odd angles. You CAN'T lose momentum in collisions. You can lose mechanical energy into heat and vibrations and whatnot if the collision is inelastic, but you can't lose momentum. You can *redistribute* it from the gas to the ship and back for all week if you want, you can send some of it aboard a shuttlecraft to Risa and then bring it back, but eventually the gas exits and carries p(gas), which must exactly match p(ship), no matter what route the gas took. All the physical, chemical or theological interactions between the gas and the ship are irrelevant - the gas is not even obligated to hit the engine nozzle walls at all, even though that's how most rocket engines tend to work. The absence or presence of an easily recognizable coupling between the gas and the ship will not alter the fact that momentum for the system MUST be conserved, or God Himself comes down and whacks you silly. So in the end, the ship IS moving at momentum p(ship)=m(gas)v(gas) opposite the direction of v(gas). Of course, when you write out p(ship)=m(ship)v(ship) and equate that with -m(gas)(v(gas), you find out that v(ship)=-v(gas)m(gas)/m(ship) is pretty darn small, because m(gas)/m(ship) is bound to be low. Unless your v(gas) is very high, which it certainly is in a Trek impulse engine. (In fact, it's probably so high that one has to evoke Einstein to correct the momenta, but that's another discussion altogether.) Timo Saloniemi [/QB][/QUOTE]
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