Rocket drive and conservation of momentum

Question

I am currently reading through some lecture notes of Physics 1 and in a chapter about the dynamics of the mass point, there is an example covering the rocket drive.

Let $v$ be the velocity of the rocket relative to some external observer, $u$ be the speed of output of the gas relative to the rocket and $M(t)$ be the mass of the rocket at time $t$. The total momentum of this system after a short instance is $$p(t+\mathrm dt) = (M-\mathrm dm)(v + \mathrm dv) + \mathrm dm(v-u)\\ = Mv + M \, \mathrm dv - v \, \mathrm dm - \mathrm dm \, \mathrm dv + v \, \mathrm dm - u \, \mathrm dm\\ \approx Mv + M \, \mathrm dv - u \, \mathrm dm, $$ where $\mathrm dm$ is the mass of the ejected gas, and due to conservation of momentum we have $$p(t+\mathrm dt) - p(t) \approx Mv + M \, \mathrm dv - u \, \mathrm dm - M v = M \, \mathrm dv - u \, \mathrm dm = 0,$$ i.e. $M \, \mathrm dv = u \, \mathrm dm$ or $M \frac{\mathrm dv}{\mathrm dt} = u \frac{\mathrm dm}{\mathrm dt}.$ By integrating, we arrive at $$\int_{t_0}^t \frac{\mathrm dv}{\mathrm dt'} \, \mathrm dt' = \int_{t_0}^t \frac{u}{M(t')} \frac{\mathrm dm}{\mathrm dt'} \, \mathrm dt'$$ and in order to solve the integral on the right hand side, the writer uses $\mathrm dm = - \mathrm d M$.

Question: Why is there a negative sign and not just $\mathrm dm = \mathrm dM$? How does it matter? Up to now, I always understood the notion $\mathrm dm$ to be a very tiny change of mass which would always be positive. Is that wrong? If so, how else do I have to interpret that notion?

Thank you very much in advance for any help.

Answer 1

$M$ is reducing. Thus, $\mathrm dM$ has a negative value.

In contrast, in the above equations, you can see an $M-\mathrm dm$ term. Here, we can see that $M$ will reduce only if $\mathrm dm$ has a positive value.

In other words, when time goes forwards, the mass that got thrown out ($m$) is increasing, thus $\mathrm dm$ is positive. In contrast, $M$ decreases, thus $\mathrm dM$ is negative. But $|\mathrm dm|=|\mathrm dM|$ by conservation of mass, so $dm=-dM$.

The $\mathrm d$ symbol is always used for increase while integrating.

One can also look at it as this: if $m$ increases by $\mathrm dm$, them $M$ decreases by $\mathrm dm$. But, $\mathrm dM=\text{change in M}=(M-\mathrm dm)-M=-\mathrm dm$