Rocket Motion treatment by Kleppner and Kolenkow

Question

I was going through the "Rocket Motion" section of An Introduction to Mechanics by Kleppner and Kolenkow, where something caught my attention. To begin with, I am going to put a part of derivation of rocket motion(in my words) as is done in the book:

Consider a rocket moving in space, with a mass $M + \Delta m$, coasting at a velocity of $\textbf{v}$. At time $t$, its engine starts and the mass $\Delta m$ is ejected in time $\Delta t$ at a velocity $\textbf{u}$ (with respect to the rest of the rocket) as shown in figure. During this time, the velocity of the rocket changed by $\Delta\textbf{v}$. Now, the system is taken to be the whole mass $M + \Delta m$ as indicated by the dotted lines. Then the initial, final momentum of the system would be:

$$ \textbf{P}(t) = (M + \Delta m)\textbf{v} $$$$ \textbf{P}(t + \Delta t) = M(\textbf{v} + \Delta\textbf{v}) + \Delta m(\textbf{v} + \Delta\textbf{v} + \textbf{u}) \tag{1} $$ Then, $$ \Delta\textbf{P} = \textbf{P}(t + \Delta t) - \textbf{P}(t) = M\Delta\textbf{v} + \Delta m(\Delta\textbf{v} + \textbf{u}). $$ And, if there is external force,$\textbf{F}_{ext}$ on the whole system, we get: $$ \textbf{F}_{ext} = \frac{\Delta \textbf{P}}{\Delta t}. $$ After this, they take the limit as $\Delta t \to 0$ and ignores the term $\Delta m\Delta\textbf{v}$ as "it is a non-first order term". Finally, we get the correct equaion as: $$ \textbf{F}_{ext} = M\frac{\text{d}\textbf{v}}{\text{d}t} - \textbf{u}\frac{\text{d}M}{\text{d}t} \tag{2} $$ Last step follows as: $\frac{\text{d}m}{\text{d}t} = -\frac{\text{d}M}{\text{d}t}$.

My doubt starts from the $eq(1)$.

• How can the final velocity of mass $\Delta m$, which is actually ejected, also be increased by $\Delta \textbf{v}$? As, the ejection of the mass $\Delta m$ with velocity $\textbf{u}$ itself causes the rocket to change the velocity by $\Delta \textbf{v}$. How can a mass accelerate itself (to say)?

But it doesn't stop here, the term containing this increase i.e. $\Delta m \Delta\textbf{v}$ is later ignored. So, it doesn't affect our final result.

• Then, is there any significance of this term in any way?

What I tried

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I thought that the problem might be in how the initial setup was defined and I started with another. I assumed the initial mass of the whole rocket to be $M$, and then ejected the mass $\Delta m$ from it and got these equations (assuming timing and velocities to be the same):

$$ \textbf{P}(t) = M\textbf{v} $$

$$ \textbf{P}(t + \Delta t) = (M-\Delta m)(\textbf{v} + \Delta\textbf{v}) + \Delta m(\textbf{v} + \textbf{u}) $$

Again, $$ \Delta\textbf{P} = \textbf{P}(t + \Delta t) - \textbf{P}(t) = M\Delta\textbf{v} + \Delta m(\textbf{u}-\Delta\textbf{v}). \tag{3} $$

Lo and behold! I again got that term but this time with a minus sign (right hand side of $eq(3)$). Can somebody explain its significance (if any) and is my above interpretation even correct (in saying that it means that the mass is accelerating itself).

Answer 1

That term has no significance. You are free to include or omit it as you like.

But if you want to understand where it comes from, simply break the one single time step into more sub-time-steps. For example, in the case of two-sub-steps, the convenient way to see this is that you have $$ \begin{align} \tag1\vec p&=(M+m_1+m_2)\vec v\\ \tag2&=(M+m_1)(\vec v+\vec\delta)+m_2(\vec v+\vec u)\\ \tag3&=M(\vec v+\vec\delta+\vec\Delta)+m_1(\vec v+\vec\delta+\vec u)+m_2(\vec v+\vec u) \end {align} $$ Now, if you consider the mean velocity of the ejected mass $m_1+m_2$, you will see that it is $\vec v+\vec u+\frac{m_1}{m_1+m_2}\vec\delta$; this last term is clearly a term that encapsulates that fact that the ejected mass took some time to eject and accelerate to its leaving velocity, and during that time the yet-to-eject mass is also accelerated forwards, leading to a slightly altered velocity from naïve expectation.

Again, it totally does not matter, because it is a second order term that vanishes when you take the limit.

Answer 2

You need to remember that the entire derivation accounts for the possibility that there is also an external force $\mathbf{F}_{\text{ext}}$, that's acting on the entire system that contains both the ship and the ejected mass, as you yourself actually noted. It is for that reason that we must take the general case where this same force accelerates the system during the time $\Delta t$, to a new velocity, given as $\mathbf{v}+\mathbf{\Delta{v}}$.

In your attempt to rederive this, you have actually written an equation that describes a situation corresponding to the mass $\Delta m$ being ejected before the time interval $\Delta t$ elapses. Therefore it is not subjected to any net change of the system's velocity during the interval $\Delta t$, and hence the momentum change imparted to it comes out different. But as you noted correctly, this turns out to make no difference once we take the limit where $\frac{\Delta \mathbf{P}}{\Delta t} \rightarrow \frac{\mathrm{d}\mathbf{P}}{\mathrm{d}t}$.

Answer 3

I think I have found the mistake I was doing here. In this scenario, consider this line

mass $\Delta m$ is ejected in time $\Delta t$ at a velocity $\textbf{u}$ (with respect to the rest of the rocket).

Here, "with respect to the rest of the rocket" contains it all. Since we consider the rest of the rocket to have its velocity changed by amount $\Delta\textbf{v}$, making the final velocity to be: $\textbf{v} + \Delta\textbf{v}$. So, if the final velocity of the ejected mass with respect to the rocket is $\textbf{u}$, then with respect to a stationary observer in an inertial frame, the ejected mass would have the exact velocity of $\textbf{v} + \Delta\textbf{v} + \textbf{u}$, as used by the authors.

And this makes sense. I mean, the ejection of mass(fuel) with a constant velocity $(\textbf{u}$), with respect to the rocket, is indeed what one would expect for a particular process happening in the combustion tank in every cycle, at least, when we ignore slight variations in reality. And, the velocity with respect to external observer would vary depending on the velocity of the rocket and change in velocity.

As of how this doesn't mean that the mass accelerates itself? For an external observer, the velocity of ejection would be $\textbf{v} + (\Delta\textbf{v} + \textbf{u})$. So, apart from the initial velocity $(\textbf{v}$) of the mass with the rocket, we observe that the mass is ejected with velocity $\Delta\textbf{v} + \textbf{u}$. Which represents just some direction(in which the mass actually is ejected) expressed as the sum of the two velocities for the sake of expressing velocity with respect to the rocket and making calculations more intuitive and easier.

To say, from exterior, the mass $\Delta m$ is ejected with velocity $\textbf{b} (= \Delta\textbf{v} + \textbf{u})$ and makes the velocity of rocket be changed by $\Delta \textbf{v}$.

But, at last all of this is for nothing since we just ignore the term $\Delta m\Delta \textbf{v}$, as it is second order (which would not suffice when $\Delta t \to 0$), as mentioned by others in their answers.