Infinitesimals meaning in this context

Question

I was solving this rocket propulsion's classic mechanics exercise: M is the instantaneous rocket's mass, and v its velocity. The exhaust gases are ejected with speed 𝑢 relative to the speed 𝑣 of the rocket (there's the weight and friction forces acting on the rocket, but that's not relevant for now). So with $\frac{d\vec{p}}{dt} = \sum{}\vec{F}_{ext}$ in mind, I went for $$d\vec{p} = M(t+dt)\vec{v}(t+dt)+[(M(t)-M(t+dt))(\vec{v}(t+dt)+\vec{u})]-M(t)\vec{v}(t) \qquad \textbf{(1)}$$ what lead me to $$d\vec{p} = M(t)\vec{v}(dt)-M(dt)\vec{u} \qquad \textbf{(2)}$$ I don't have sure if the expression "t+dt" in (1) is acceptable or makes sense. I think it does, I'm adding an infinitesimal quantity of time since I'm taking an infinitesimal change of the momentum. But my main doubt is about the "M(dt)" and the "$\vec{v}$(dt)" in (2). Does M(dt) stands for dM? M(dt) is an infinitesimal change in rocket's mass, but that change could be different depending on the point we're taking into consideration, i.e., if M(t) has different slopes through time, changing different starting points will also change this dM, right? Am I interpreting this correctly? I have some uncertainties when dealing with infinitesimals, any help is welcome.

Answer 1

The notation $M(dt)$ or $v(dt)$ is not acceptable. Typical notations are like:

At the time $t$:

• mass of rocket $M(t)$
• velocity of rocket $\vec{v}(t)$

Then at the time $t+dt$

• mass of rocket $M(t+dt) = M(t) + dM$
• velocity of rocket $\vec{v}(t+dt) = \vec{v}(t) + d\vec{v}$
• mass ejected $ dm = -dM = M(t) - M(t+dt) $ ( Note that $d M$ is negative.)
• eject velocity w.r.t ground $\vec{v}(t) + \vec{u}(t) + d\vec{v}$

Therfore, the momentum change from $t$ to $t+dt$

$$ d\vec{p} = \vec{p}(t+dt) - \vec{p}(t) = \{ (M(t)+dM) (\vec{v}(t) + d\vec{v}) - (\vec{u}(t) + \vec{v}(t) + d\vec{v}) dM\} -\{ M(t) \vec{v}(t) \} $$ $$ d\vec{p} = M(t) d\vec{v} - \vec{u} dM $$

Where $dM = M(t+dt) - M(t) \lt 0$ and $d\vec{v} = \vec{v}(t+dt) - \vec{v}(t)$. Also note that $\vec{u}$ is in the opposite direction of $\vec{v}$. In the scalar expression, giving $v$ positive, $u$ should be negative.

Answer 2

$d M$ would be equal to $M(t+dt) - M(t)$, as an infinitesimal $M$ is the change in $M$ after an infinitesimal time $dt$.