Let $m$ be the mass of a rocket in free space (including fuel) at the time $t$. Now suppose that the rocket ejects a mass $\Delta m$ during the interval $\Delta t$ with velocity $v_e$ relative to the rocket.

I was told that the correct ansatz is as follows:

$$ mv = (m - \Delta m) \cdot (v + \Delta v) + \Delta m \cdot (v + \Delta v - v_e). $$

If you work this out, it reduces to:

$$ m \Delta v - \Delta m v_e = 0. $$

However some books start with the equation

$$ mv = (m - \Delta m) \cdot (v + \Delta v) + \Delta m \cdot (v - v_e) $$

which reduces to

$$ m \Delta v - \Delta m v_e - \Delta m \Delta v = 0. $$

Then you need to "neglect" the higher order term $\Delta m \Delta v$ to get the correct result.

Is there any good conceptual reason why the first ansatz is the more correct one?

However if you use $\tilde m = m(t+ \Delta t)$ in the derivation, you get the correct result without the need of neglecting higher order terms in the line of the second derivation:

$$ (\tilde m + \Delta m) \cdot v = \tilde m (v + \Delta v) + \Delta m (v - v_e) $$

so why is the following equation the conceptually wrong ansatz:

$$ (\tilde m + \Delta m) \cdot v = \tilde m (v + \Delta v) + \Delta m (v + \Delta v - v_e)? $$

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    $\begingroup$ Too short to be an answer: Neither one is correct with regard to non-infinitesimal quantities of mass. The reality is somewhere in between. Both are the same with regard to infinitesimal quantities of mass. $\endgroup$ – David Hammen May 2 '18 at 13:51
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    $\begingroup$ The first method is clearer in terms of when and when you are conserving momentum. Conserve momentum at the snapshot in time $t$: $m(t)v(t)$. Now conserve momentum at the snapshot in time $t + \Delta t$: $m(t+ \Delta t)v(t+\Delta t) + (m(t) - m(t + \Delta t))v_{\text{mass element relative to ground at} t + \Delta t}$. By velocity addition, we have $v_{\text{element with respect to ground}}(t + \Delta t) = v(t+\Delta t) + v_e(t + \Delta t)$. Notice how since $v_e$ is a variable, I'm leaving it as $v_e$. Why you have a bunch of $-v_e$'s I'm not sure. I don't like how it's taught that way $\endgroup$ – DWade64 May 2 '18 at 14:12
  • $\begingroup$ As it teaches you to put minus signs everywhere for what you think the direction is, when in reality the math doesn't care and will always take care of the signs for you. $v_e$ is as much of a variable as $-v_e$ $\endgroup$ – DWade64 May 2 '18 at 14:17
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    $\begingroup$ You answered your own question. When discussing quantities of 1st order ($v\Delta m$ and $m \Delta v$) you can neglect quantities of 2nd order ($\Delta m \Delta v$). $\endgroup$ – sammy gerbil May 4 '18 at 10:31