1. Take a look at these two setups for the rocket equation. Are these two scenarios the same?

No. The top setup is the same setup I used in the derivation, as do many other videos and explanations of the rocket equation. However, I’ve seen the bottom setup as well, most notably on Wikipedia (as recently as Feb 2023). At first glance, both setups might seem to describe the same physical scenario: a mass of $\Delta m$ is ejected at velocity $v_e$ , increasing the velocity of the rocket by $\Delta v$ .

👉🏻 Here’s what the same conservation of momentum equation would look like in the bottom setup:

$\begin{aligned} (m+\Delta m)v &= m(v+\Delta v) + (v-v_{e})\Delta m \\ mv + v\Delta m&= mv + m\Delta v + v\Delta m - v_{e}\Delta m \end{aligned}$

Note that the bottom setup has the benefit of not having the pesky $\Delta m \Delta v$ term! You don’t need to explain why you can ignore this term (as I do in FAQ 2 below).

🚨 Additionally, notice that the starting masses are different! If we follow the same mathematical steps to Eq. 3, here’s how the bottom setup’s equation would look:

$\sum^{v_f}_{v_i} \frac{\Delta v}{v{e}} = \sum^{m_f}_{m_i+\Delta m} \frac{-\Delta m }{m}$

The bottom setup essentially starts with an extra amount of $\Delta m$ .

Now, the rocket equation turns out to be the same regardless of which setup you choose.

The $\Delta m \Delta v$ term can be ignored in our derivation of the rocket equation (see FAQ 2 below), so its premature disappearance here doesn’t matter.
In taking the $\lim_{\Delta m \to 0}$ , this extra mass goes to 0.

⚠️ But while both setups lead to the same rocket equation, the reasoning in the bottom setup is simply incorrect! Sweating the small stuff is important in physics; it’s things like a summation index that’s off by 1 or an accidental negative sign that can completely change a solution. If we’re not careful with analyzing our physical setup, we’ll get wrong answers in the future.

For example, the extra mass is most definitely not 0 for finite $\Delta m$ , and so the bottom approach would lead you astray in a problem that requires you to sum up finite amounts.

2. Why can we just remove $\Delta m\Delta v$ because it’s “small”?

Here’s the less hand-wavy explanation.

Eq. 1 can be rewritten as a function of $\Delta t$ instead of $\Delta v$ and $\Delta m$ :

$m(t) \frac{\Delta v(t)}{\Delta t}\Delta t - \frac{\Delta m(t)}{\Delta t}\Delta t\frac{\Delta v(t)}{\Delta t}\Delta t - v_e \frac{\Delta m(t)}{\Delta t}\Delta t = 0$

which can then be rewritten as

(S1) $m(t) v'(t)\Delta t - m'(t)v'(t)(\Delta t)^2 - v_e m'(t)\Delta t = 0$

What we’ve done here is rewrite $\Delta v = \frac{\Delta v(t)}{\Delta t} \Delta t \equiv v'(t) \Delta t$ , and likewise for $m$ . Instead of directly using the change in rocket velocity or amount of mass ejected, we’re writing it as the rate of velocity change and the rate of mass ejected, then multiplying this rate by the total time elapsed. These two expressions represent the same quantity, but using rate allows us to work with a single variable $\Delta t$ .

Let’s now take the Taylor expansion for each of these terms around $\Delta t = 0$

As a reminder, the formula for this expansion is $f(a) \approx f(0) + f'(0)a + \dots \text{(higher order terms)}$

$\begin{aligned} m(t)v'(t) \Delta t &\approx 0 + m(t)v'(t) \Delta t \\ m'(t)v'(t)(\Delta t)^2 &\approx 0 + 0 \\ m'(t) \Delta t&\approx 0 + m'(t) \Delta t\end{aligned}$

As shown, all the zero-order terms go to 0, while for the middle expression the first-order term is also 0. Eq. S1 becomes

$m(t) v'(t)\Delta t - v_e m'(t)\Delta t = 0$

From here on out, we’re going to follow the same mathematical steps as we did in the original derivation. Rearranging terms and adding in a negative sign gives us

$\frac{v'(t)\Delta t}{v_e} = \frac{-m'(t)\Delta t}{m(t)}$

Next, we’re going to sum both sides over the entire time interval, then take the limit as $\Delta t \to 0$ . This gets us to the integrals, at which point we rewrite the rate $v'(t)$ as the ratio $\frac{dv(t)}{dt}$ , then cancel out the $dt$ s. We have now arrived at the same Eq. 4 in the original derivation, and can return to that path.

$\begin{aligned} &\text{ }\lim_{\Delta t \to 0} \sum^{t_f}<em>{t = 0}\frac{v'(t)\Delta t}{v_e} &&= \lim</em>_{\Delta t \to 0} \sum^{t_f}_{t = 0}\frac{-m'(t)\Delta t}{m(t)} \\ &\Rightarrow \frac{1}{v_e}\int^{t_f}_0v'(t)dt &&= -\int^{t_f}_0\frac{1}{m(t)}m'(t)dt \\ &\Rightarrow \frac{1}{v_e}\int^{t_f}_0\frac{dv(t)}{dt}dt &&= -\int^{t_f}_0\frac{1}{m(t)}\frac{dm(t)}{dt}dt \\ &\Rightarrow \frac{1}{v_e}\int^{v_f}{v_i} dv && = -\int^{m_f}_{m_i} \frac{dm}{m}\end{aligned}$

Note that we’ve summed/integrated from 0 to $t_f$ instead of 0 to $t$ to avoid confusion with $t$ as the variable of integration, but it is the same range we’re summing/integrating over as we did in the original derivation.

Evidently, the $\Delta m \Delta v$ term disappears in the first-order (aka linear) approximation for small $\Delta t$ . That’s not a problem for our rocket equation because a rocket is essentially firing continuously, so the $\Delta t$ is sufficiently small for this approximation to work extremely well.

💡 And if $\Delta t$ weren’t small enough to take the first-order approximation, then we should also be concerned with taking $\lim_{\Delta t \to 0}$ to arrive at the integral, a far more crucial part of the derivation.

⚠️ But that doesn’t mean that the $\Delta m \Delta v$ term is 0! This term is real, and we haven’t proven that this term goes to 0 as $\lim_{\Delta t \to 0}$ , just that it is approximately 0 for small $\Delta t$ .

While I’m pretty sure it does, I couldn’t prove it. If any of you folks out there who are much better at math than I am can, shoot me a message!

Rocket equation FAQ

1. Take a look at these two setups for the rocket equation. Are these two scenarios the same?

2. Why can we just remove because it’s “small”?

2. Why can we just remove $\Delta m\Delta v$ because it’s “small”?