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.


$M$ 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, $\mathrm dM=\text{change in M}=(M-\mathrm dm)-M=-\mathrm dm$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.