# Rocket equation derivation, question about signs

My question is about signs in the derivation of the rocket equation. Considering a rocket of mass $m$, its momentum at time $t$ is $p(t)=mv$. At time $t+dt$ it ejects a quantity $dm$ so its momentum at $t+dt$ is $p(t+dt)=(m+dm)(v+dv)$. The students I tutor asked why it was written m+dm as opposed to m-dm. I was able to convince them that if you considered the mass of the rocket at time $t$ to be $m+dm'$ it worked out to be equivalent, but I think I must have forgotten something fundamental as I was not able to provide a satisfactory explanation for the former case.

If you the rate of change of mass is $\dot{m} = \mathrm{d}m/\mathrm{d}t$ and we're talking about passing into the future direction ($\mathrm{d}t > 0$), then $\mathrm{d}m$ is necessarily negative. Thus, $m\mapsto m+\mathrm{d}m$ represents a decrease of the rocket mass.
The opposite convention would require that $\dot{m} = -\mathrm{d}m/\mathrm{d}t$ works against the conventional meaning of the derivative in calculus. You could treat that consistently, but it's an extra hassle to intentionally flip certain signs but not others.
Really, the ambiguity is in the "ejects a mass $\mathrm{d}m$". If one understands this as a shorthand for the magnitude, or perhaps something like "ejects a mass, changing the rocket's mass by $\mathrm{d}m$", there's no problem.
• @k-selectride Right, you can remind them of the the definition of derivative of $f$ as the limit of $[f(t+\Delta t)-f(t)]/\Delta t$. This corresponds to $\dot{f} = \mathrm{d}f/\mathrm{d}t$ in infinitesimal form. Since we know $m$ is decreasing with time, $\mathrm{d}m$ must be negative if $t$ is going forward, etc. – Stan Liou Jul 13 '15 at 23:55