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This is from pages 32-33 of Physics for Mathematicians. A rocket + fuel system of mass $m(t)$ is moving along at velocity $v(t)$, with its fuel being ejected at the velocity (relative to the rocket) $q(t)$.

In a short time interval $[t, t + h]$, the amount of fuel ejected is $m(t) - m(t + h)$, and therefore the momentum of expelled fuel will be close to $[m(t) - m(t+h)]q(t)$. Thus, the momentum of the fuel in the other direction, pushing the rocket forward, will be the negative of this.

How do we deduce that the momentum of the fuel remaining in the rocket is the negative of the momentum of the fuel flying out of the rocket? This could be justified if we assumed that there are no external forces on the fuel, and that the fuel is initially at rest in the rocket, but I'm not sure about either of those assumptions.

...so the force on the rocket must be the derivative, $m'(t)q(t)$...

Why? All I can recognize is that $m'(t)q(t)$ is, given the expressions mentioned previously, the derivative of the momentum of the expelled fuel at time $t$. This is the force on the expelled fuel, which by what was said before is the negative of the force on the fuel pushing into the rocket, so we're saying

Force on fuel pushing into rocket = - Force on rocket

Which seems kind of like Newton's third law, but who says that the force on the left hand side is being applied by the rocket?

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  • $\begingroup$ The sentence starting "Thus, the momentum ..." sounds like gibberish to me. I don't understand what the author is saying. $m'(t)q(t)$ is not the derivative of the momentum of the expelled fuel. And I don't understand what you mean by the sentence "Force on fuel pushing into rocket ...". I think you need to clarify you concerns, and add the details of your reasoning. $\endgroup$ – garyp Jun 18 '16 at 13:30
  • $\begingroup$ @garyp By "fuel pushing into rocket", the author means that when the fuel ignites, some of it flies out the end of the rocket, and some of it pushes into the rocket, propelling it forward. $\endgroup$ – Jack M Jun 18 '16 at 15:51
  • $\begingroup$ Thanks. The phrasing is still bad. No fuel is ejected. Exhaust, the products of combustion, are ejected. Some of the exhaust goes straight out the back. The rest of it is directed toward the front, and encounters the nozzle, where much of it is redirected backwards. No fuel remains in the rocket to propel it forward. $\endgroup$ – garyp Jun 18 '16 at 17:59
  • $\begingroup$ You don't need your assumptions to justify the statement. Imagine that the combustion is a series of explosions of short duration $\mathrm{d}t$. Whatever momentum goes straight back has to be balanced by something, either the rocket itself and its contents (remaining fuel), or whatever exhaust manages to leak forward (assumed to be negligible). $\endgroup$ – garyp Jun 18 '16 at 18:00
  • $\begingroup$ As your question "Why? All I can ..." , I don't think that question can be answered without seeing what the author's argument is. $\endgroup$ – garyp Jun 18 '16 at 18:05

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