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In one of my lectures our physics professor gave a derivation of the ideal rocket equation as follows:

Let $v_G > 0$ be the velocity at which the gas is emitted from the rocket. Let $m$ and $v$ be the mass and the velocity of the rocket at some point in time. The mass and velocity of the rocket change with $dm$ and $dv$ when some gas is emitted, so the new mass and velocity are $m - dm$ and $v + dv$. Let $v_T = v + dv + v_G$ be the velocity of the gas relative to the outside observer after it is emitted. The momentum of the whole system before and after the gas is emitted is equal:

$$mv = (m - dm)(v + dv) + dm \cdot v_T \Leftrightarrow mv = mv + mdv - vdm - dmdv + vdm + dmdv + v_Gdm \Leftrightarrow 0 = mdv + v_Gdm$$

By integrating this one can derive the ideal rocket equation.

Now as far as I can tell, my professor has done two mistakes here, which coincidentally seem to compensate each other, so that she reaches the correct solution nevertheless.

Firstly, the velocity of the emitted gas relative to the outside observer should be $v_T = v + dv - v_G$, since the gas is emitted in the opposite direction of the acceleration of the rocket.

Secondly, she implicitly says that $dm$ is positive, which is wrong since the mass of the rocket $m$ is decreasing over time. Thus, $dm < 0$. So the mass and velocity of the rocket after the gas is emitted should be $m + dm$ and $v + dv$. (See edit at the bottom)

With these two changes, the derivation looks like this:

$$mv = (m + dm)(v + dv) + (-dm) \cdot v_T \Leftrightarrow mv = mv + mdv + vdm + dmdv - vdm - dmdv + v_Gdm \Leftrightarrow 0 = mdv + v_Gdm$$

So we reach the same equation at the end.

Are the points I made valid and is it a mere coincidence that her mistakes compensate each other to reach a correct final solution? Or did I wrongly identify what she did as mistakes and there is a good explanation as to why she did those two things differently?

EDIT: Later in the derivation she integrates like this:

$\int_{m_0}^{m_1}-v_G\frac{dm}{m} = -v_G \int_{m_0}^{m_1}\frac{1}{m} dm = -v_G [ln(m)]_{m_0}^{m_1}$

So she is using $dm$ as the change of $m$, which means it has to be negative. Just to point this out; if $dm$ was actually positive, it should be labelled $d(-m)$ and the integration should go like this:

$v_G \int_{m_0}^{m_1}\frac{1}{m} d(-m) = -v_G \int_{m_0}^{m_1}\frac{1}{-m} d(-m) = -v_G [ln(m)]_{m_0}^{m_1}$

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    $\begingroup$ If you add a negative or substract a positive $dm$ makes no difference. Your changes do not improve the equation, your teacher did it the right way. You should apologize to her and never question her authority again. $\endgroup$
    – Yukterez
    Commented Mar 15, 2016 at 15:51
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    $\begingroup$ I'll ignore the irrelevant second half of your comment. Do you have anything to say regarding the other mistake? Also check out my edit. $\endgroup$
    – Marcus Thyssen
    Commented Mar 15, 2016 at 15:58
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    $\begingroup$ If your professor explicitly defined $v_G>0$, she probably took as reference the gas, and assumed the rocket going in the negative direction, $v, dv <0$. In this case the gas gains mass $dm>0$ and its velocity to the outside observer is $v+dv+v_G$. Otoh if you take as reference the rocket and assume it's going in the positive direction, $v, dv>0$, then the rocket looses $dm<0$, while the gas is expelled in the negative direction and its velocity relative to the observer is $v+dv -v_G$. Likewise, integration may refer to the mass of the rocket or to the mass of the gas ejected. $\endgroup$
    – udrv
    Commented Mar 15, 2016 at 16:42
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    $\begingroup$ There are also several explicit derivations of the rocket equation on this site, you can find them all by clicking on the rocket-science tag. $\endgroup$
    – Kyle Kanos
    Commented Mar 15, 2016 at 16:47
  • $\begingroup$ I don't understand your use of negative signs at the beginning of your last integral. What happened to the negative sign in from of the $v_G$ at the first? And if you integrate $d(-m)/(-m) =ds/s$ [$s=-m$], shouldn't you get$\ ln (-m)$? $\endgroup$
    – Bill N
    Commented Mar 15, 2016 at 16:56

1 Answer 1

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Regarding your edit, you say:

So she is using dm as the change of m, which means it has to be negative.

When doing integration, the limits actually tell you how the quantity (e.g., $m$) is changing, not the differential. Consider a simple calculation of $\Delta m$: $$\Delta m = \int_{m_0}^{m_1} \mathrm{d}m = m \bigg|_{m_0}^{m_1} = m_1-m_0$$

If $m_1<m_0$, the $\Delta m$ will be negative. Notice, we didn't have to make d$m$ negative for this to work.

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  • $\begingroup$ Why is it labelled $dm$ then? If I build the derivative of $m$ I'll get $\dot m = \frac{dm}{dt} = \lim_{t_2 \rightarrow t_1} \frac{m_2 - m_1}{t_2 - t_1}$ with $t_2 > t_1$. If $\dot m$ is positive, then $m_2 > m_1$ which says that $m$ is increasing over time, which is false. So this $dm$ and the $dm$ you're talking about are two different things? $\endgroup$
    – Marcus Thyssen
    Commented Mar 15, 2016 at 17:33
  • $\begingroup$ Notice that in the rocket problem, if $m$ is the mass of the rocket (including unburned fuel), $\dot{m}$ itself is negative. that means $m_2 < m_1$ in the limit. While the notation lends itself to treating $\frac{dm}{dt} $ as a ratio, you can't assign specific values to $dm$ and $dt$ separately. Also, while $dm$ itself can be conceptually positive or negative, we let the limits of integration handle that. The derivative is a limit. The notation allows us to manipulate the symbols, but we can't assign signs to the pieces. We must allow the limits on the integral to do that. $\endgroup$
    – Bill N
    Commented Mar 15, 2016 at 17:45
  • $\begingroup$ Look at Taylor's handling of the issue and his treatment of positive and negative signs versus positive and negative concepts. Classical Mechanics by John R. Taylor, page 85-86. $\endgroup$
    – Bill N
    Commented Mar 15, 2016 at 17:47
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    $\begingroup$ We face the same issue in regular language. Someone might say "I went on a diet and lost 1 kg." But that's not the right way to say it in physics, because it presupposes the sign of the quantity before we report the quantity, so in physics we say "my weight changed by -1 kg." See how that lets the quantity itself report its own sign? We don't force signs onto our quantities with our language, our symbols refer to changes and we just let the signs be however they come out. Of course, one must be careful, because sometimes we revert to common language, and it can be confusing. $\endgroup$
    – Ken G
    Commented Apr 8, 2017 at 14:01
  • $\begingroup$ This is a pretty bad answer. He's right in saying that by definition, $dm = m(t+dt)-m(t)$, so she should have written $(m+dm)(v+dv)$ for the rocket's momentum. $\endgroup$
    – Jahan Claes
    Commented Aug 25, 2017 at 15:59

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