Does $F = dp / dt$ apply to a rocket ejecting mass? In Sean Carroll's book "The Biggest Ideas in the Universe, space, time, and motion", he makes the following claim:

$$\overset\rightarrow{F} = \frac{d\overset\rightarrow{p}}{dt}$$
Not only is this pleasingly compact, it's more general than $\overset\rightarrow{F} = m\overset\rightarrow{a}$, as this form remains valid even when the mass of the object is changing (for example, as a rocket gradually loses mass by ejecting exhaust).

Does this make any sense? I think it's just false.
Suppose we have a rocket going east at 100 m/s. It weighs 1000 kg, of which 500 kg is fuel. We eject the 500 kg of fuel back west. We do it at the rate of 1 kg/s, at a very small velocity -1 m/s relative to the rocket.
According to my calculations, the rocket is experiencing a constant 1 N of force while its mass is decreasing from 1000 kg to 500 kg during the 500 s burn. Which means its acceleration is growing. But it's a very small acceleration. The velocity increases from 100 m/s to 100.69 m/s.
Anyway: the momentum of the rocket decreases, from 100,000 kg m/s to 50,345 kg m/s, because of the lost mass.
If $F = \frac{dp}{dt}$ were correct in this variable mass scenario, it would imply that the average force on the rocket during the 500 s burn was not +1 N, but rather -99.3 N (i.e. backwards, west).
This just makes no sense to me, but to clarify: is Sean Carroll's statement about rockets just a mistake, or is there something more to it?
Edit:
Since people are requesting formulas in the comments, here is all the math in detail:
Initial mass $m_0 = 1000 \text { kg}$.
Initial velocity $v_0 = 100 \text { m/s}$
Burn time $T = 500 \text { s}$.
Mass rate $\frac{dm}{dt} = -1 \text{ kg/s}$
Exhaust velocity relative to rocket $v_e = -1 \text { m/s}$
Thrust $F = \frac{dm}{dt}v_e = (-1) \cdot (-1) = 1 \text { N}$
Mass $m_t = m_0 + \frac{dm}{dt}t$
Final mass $m_T = m_0 + \frac{dm}{dt}T = 1000 - 1\cdot 500 = 500\text{ kg}$
Acceleration $a_t = \frac{F}{m_t}$
Final velocity $v_T = v_0 + \int_0^T a_t dt = v_0 + \int_0^T \frac{F}{m_0 + (dm/dt)t} dt = 100 + \int_0^{500} \frac{1}{1000 - t} dt = 100 + \int_{500}^{1000}\frac{1}{u} du = 100 + (\ln 1000 - \ln 500) = 100 + \ln 2 \approx 100.69 \text { m/s}$
Initial momentum: $p_0 = m_0v_0 = 1000 \cdot 100 = 100000 \text{ kg m/s}$
Final momentum: $p_T = m_Tv_T \approx 500 \cdot 100.69 = 50345 \text{ kg m/s}$
$\Delta p = p_T - p_0 = 50345 - 100000 = -49655 \text { kg m/s}$
$\Delta T = T - 0 = T = 500 \text{ s}$
$\frac{\Delta p}{\Delta T} = \frac{-49655}{500} \approx -99.3\text{ N}$
$F = 1\text{ N} \ne \frac{\Delta p}{\Delta T} \approx -99.3\text{ N}$
 A: The most general form of the momentum equation reads
$\dfrac{d \mathbf{p}}{dt} = \mathbf{R}^{e}$,
being $\mathbf{p}$ the momentum of the system, $\mathbf{R}^{e}$ the resultant of the external forces on the system.
In your example on the rocket, we can write along the trajectory of the rocket
$\dfrac{d}{dt} (m_{rocket} v) = R^e$,
being $R^e$ the sum of the gravitation, the air drag and the rocket thrust. Usually the rocket thrust is proportional to the mass flow of ejected, $\dot m_{fuel}$ and the mass of the rocket is provided by the sum of the empty mass and the fuel stored $m_{rocket}(t) = m_{empty} + m_{fuel}(t)$.
I won't go into the details of the dynamical equation here (if not requested later), but I'll leave two hints/references:

*

*the dynamical equation can be derived from integral balance of the open system rocket+contained fuel;

*you can take a look at the Tsiolkovski equation for rockets

EDIT: derivation
The balance equations of the system, rocket (solid, $s$) and fluid $f$, can be derived from the mass and momentum balance equation, of the solid and fluid contained in the volume delimited by the external surface of the rocket and the area of its nozzle. Let's call the contained volume $V_t$ and its boundary $S_t = \partial V_t$. The system is not closed, so the dynamical equations read:

*

*mass equation:
$0 = \displaystyle \dfrac{d m^{tot}_{V_t}}{dt} + \oint_{S_t} \rho ( \mathbf{u} - \mathbf{u}_s ) \cdot \mathbf{\hat{n}} $,
being $\mathbf{u}$ the velocity of the solid and the fluid on the boundary and $\mathbf{u}_s$ the velocity of the surface of boundary $S_t$. They are equal on every part of $S_t$, except for the nozzle of the rocket where $\mathbf{u}_s$ is the velocity of the rocket, and $\mathbf{u}$ is the velocity of the ejected fluid, $\mathbf{u}_{outflow}$. The boundary integral is equal to the mass flow ejected from the nozzle, $\dot{m}^f_{outflow}$, while the total mass inside the volume can be written as the sum of the empty rocket and the fluid therein, $m^{tot} = m^s + m^f_{inside}$. Since the mass of the empty fluid $m^s$ is constant, we can write
$\dfrac{d}{dt} (m^s + m^f_{inside}) + \dot{m}^f_{outflow} = 0$$\qquad \rightarrow \qquad$$\dot{m}^f_{inside} + \dot{m}^f_{outflow} = 0$$\qquad \rightarrow \qquad$$ \dot{m}^f_{outflow} = -\dot{m}^f_{inside} = - \dot{m}^{tot}$


*momentum equation:
$\displaystyle  \dfrac{d \mathbf{Q}^{tot}}{dt} + \oint_{S_t} \rho \mathbf{u} ( \mathbf{u} - \mathbf{u}_s ) \cdot \mathbf{\hat{n}} = \mathbf{R}^{ext} = \mathbf{0}\text{ if we ignore weight and drag, and neglect pressure stress on the nozzle air.}$
Assuming that the velocity of the fluid and the solid inside the rocket is the same, $\mathbf{v}$, and assuming that the velocity is uniform on the nozzle surface, we can write:
$\dfrac{d}{dt} (m^{tot} \mathbf{v}) + \mathbf{u}_{outflow} \dot{m}^f_{outflow} = \mathbf{0} $.
Explicitly writing the time derivative, we get
$m^{tot} \dfrac{d \mathbf{v}}{dt} = - \dot{m}^{tot} ( \mathbf{v} - \mathbf{u}_{outflow})$
Now, we can project the vector equation along the trajectory to get a scalar equation. And we can further assume that the nozzle is designed to have a constant relative velocity $v_e = v - u_{outflow}$
$m^{tot} \dfrac{d v}{dt} = - \dot{m}^{tot} v_e$,
that can be easily integrated to get the relation between the velocity of the rocket and its mass
$\dfrac{v_1 - v_0}{v_e} = \ln \left(\dfrac{m^{tot}_0}{m^{tot}_1}\right)$
As the mass of the rocket decreases $m^{tot}_1 < m^{tot}_0$, the velocity of the rocket increases $v_1 > v_0$.
Comments. If you want, you can split the equations writing them for the solid and the fluid independently, whose sums give the equation for the overall system:

*

*mass equation:

*

*solid $\dfrac{d m^s }{dt} = 0$;

*fluid $\dfrac{d m^f_{inside} }{dt} + \dot{m}^f_{outflow} = 0$;



*momentum:

*

*solid $\dfrac{d}{dt} (m^s v) = F_{sf}$, being $F_{sf}$ the force acting on the solid due to the fluid

*fluid $\dfrac{d}{dt} (m^f f) + \dot{m}^f_{outflow} u_{outflow} = F_{fs}$, being $F_{fs}$ the force acting on the fluid due to the solid, $F_{fs} = - F_{sf}$, for the principle of action/reaction.



A: Newton's second law describes the dynamics of a point particle. You can write it either as
\begin{align}
\vec{F} = m\vec{a}
\end{align}
or the way Newton wrote it as
\begin{align}
\vec{F} = \frac{d\vec{p}}{dt}.
\end{align}
These are completely equivalent since $\vec{p} = m \vec{v}$, and particle masses don't change with time.
Considering the motion of a system of particles, you can use Newton's 2nd law and the principle of conservation of momentum (or, equivalently, Newton's 3rd law) to show that
\begin{align}
\vec{F}_{\text{ext}} = \frac{d\vec{P}}{dt}
\end{align}
where $\vec{P}$ is the total momentum of the system. A common mistake is to try to apply this to a "system of particles with variable mass," which is a non-sensical concept: a system is a definite collection of particles, the mass of which cannot change.
The correct way to treat this problem is described in An Introduction to Mechanics by Kleppner and Kolenkow and with a slightly different exposition in Physics for Mathematicians: Mechanics I by Spivak. The idea is to apply momentum conservation over a small time interval to a system consisting of the rocket and a small amount of fuel ejected over the time interval.
Let the $m(t)$ be the mass of the rocket plus its enclosed fuel at time $t$, let $\vec{v}(t)$ be the velocity of the rocket, and let $\vec{U}(t)$ be the velocity at which fuel is ejected with respect to the rocket. Note that we work in an inertial frame instantaneously coincident with the non-inertial frame of the rocket. The mass of the fuel expelled during the small interval from $t$ to $t+\Delta t$ is
\begin{align}
m(t) - m(t + \Delta t)
\end{align}
so the momentum of the fuel expelled by the rocket is
\begin{align}
\Delta \vec{p}_e = \left[m(t) - m(t + \Delta t)\right]\vec{U}(t)
\end{align}
Taking the limit $\Delta t\rightarrow 0$, the rate of change of the momentum of the expelled fuel is
\begin{align}
\frac{d\vec{p}_e}{dt} = - \frac{dm}{dt} \vec{U}(t)
\end{align}
In the absence of external forces, $d\vec{P}/dt = 0$ for the system consisting of the rocket (momentum $\vec{p}_r$) and its expelled fuel (momentum $\vec{p}_e$), so
\begin{align}
\frac{d\vec{p}_e}{dt} + \frac{d\vec{p}_r}{dt} &= 0\\
\end{align}
Since
\begin{align}
\frac{d\vec{p}_r}{dt} = m(t)\frac{d\vec{v}}{dt}
\end{align}
we get
\begin{align}
\boxed{\frac{dm}{dt} \vec{U}(t) = m(t) \frac{d\vec{v}}{dt}}
\end{align}
Note that this is not the same equation we would get by erroneously taking
\begin{align}
\frac{d}{dt}\left[m(t)\vec{v}(t)\right] = 0\\
\rightarrow \frac{dm}{dt}\vec{v}(t) + m(t) \frac{d\vec{v}}{dt} = 0
\end{align}
Unfortunately, to add to the confusion, one sometimes encounters examples where $\vec{U}(t)=-\vec{v}(t)$, in which case the wrong equation and the right equation accidentally coincide.
A: It's always good to start with pictures. At time $t$, we have an intact rocket that hasn't yet emitted the parcel of mass $\Delta m$ (so total mass is $m+\Delta m$). At time $t+\Delta t$, the mass $\Delta m$ is ejected with velocity $V_e$ and causes an increase in the rocket's velocity, $\Delta V$ (and now of mass $m$).

(source)
Since conservation of momentum applies to the two states (the system is closed as matter is not lost), then we must have that the momentum at $t$ be equal to the momentum at $t+\Delta t$. This means that,
$$p(t)\triangleq (m+\Delta m)V=p(t+\Delta t)\triangleq m(V+\Delta V)+\Delta m V_e$$
Or, after some rearranging,
$$ p(t+\Delta t)-p(t)\equiv m(V+\Delta V)+\Delta m V_e-(m+\Delta m)V=0.\tag{1}$$
From here, it is rather trivial to determine that we end up with something akin to $\mathrm{d}p/\mathrm{d}t$:
$$\frac{p(t+\Delta t)-p(t)}{\Delta t}=m\frac{\Delta V}{\Delta t}+(V-V_e)\frac{\Delta m}{\Delta t}$$
However, saying that this "something" is equal to the total force (via $F=\mathrm{d}p/\mathrm{d}t$) is not really correct as there is the presence of the $V_e$ term that is not included in applying the product rule (unless $V_e=0$, which I think means you're not really accelerating anyway). While this is loose with terminology, note that the book in question is a popularization of physics and probably doesn't claim to be rigorous (not having read it, I cannot say for certain).
More correctly for this scenario, one should be writing the formula for a variable mass system,
$$\mathbf F_\text{ext}+\mathbf{v}_\text{rel}\frac{\mathrm dm}{\mathrm dt}=m\frac{\mathrm d\mathbf{v}}{\mathrm dt}\tag{2}$$
which aligns with Equation (1) when $\mathbf{F}_\text{ext}=0$, but still wouldn't be the relation expressed by OP.

As an aside, some have questioned OP's calculations. Note that they have derived the Tsiolkovsky rocket equation,
$$\Delta v=v_e\log(m_0/m_f)$$
where $m_0$ is the initial mass and $m_f$ the final mass. Simply plugging in their numbers ($v_e=1$, $m_0=1000$ and $m_f=500$), one does indeed arrive at $\Delta v=0.69$.
However, as I discuss in my answer to Why are rockets so big, we can re-arrange this to determine the exhaust mass fraction required to get a specific increase in velocity,
$$M_f\triangleq1-\frac{m_f}{m_0}=1-\mathrm{e}^{-\Delta v/v_e} $$
Using OP's numbers, $v_e$ is too small a value to provide a meaningful velocity increase as the entirety of the rocket must be fuel (i.e., $M_f=1$). If we chose say $v_e=4.5$ m/s and a desired velocity increase of 10 m/s, then $M_f\sim0.9$.
A: It's an awkward statement I agree. The formula indeed works for a varying mass, but it has to be handled with care, due to the fact that the rocket is losing matter. Here's how I do it in class.
Let's study the problem only between instants $t$ and $t+dt$. The system is the rocket and the fuel still inside it at instant $t$. Let $m(t)$ be the mass of the system.
Linear momemtum at $t$
At instant $t$, the system's mass is $m(t)$, and both the rocket and its fuel are going at velocity $v(t)$, so linear momentum is:
$$p(t)=m(t)v(t)$$
Linear momentum at $t+dt$
Let $dm$ be the mass lost by the rocket during $dt$ (=mass of the ejected fuel), $dv$ the change in velocity of the rocket and $u$ the ejection velocity of the fuel. Then:

*

*mass of the rocket: $m(t)-dm$

*velocity of the rocket: $v(t)+dv$

*mass of the ejected fuel: $dm$

*velocity of the ejected fuel: $v(t)+dv-u$
So the linear momentum of the system is:
$$p(t+dt)
=(m(t)-dm)(v(t)+dv)+dm(v(t)+dv-u)
=m(t)v(t)+m(t)dv-dm\,u$$
Variation of linear momentum
The overall variation between $t$ and $t+dt$ is:
$$p(t+dt)-p(t)=m(t)dv-dm\,u$$
Taylor's expansion to order 1 also yields:
$$p(t+dt)-p(t)=\frac{dp}{dt}(t)\,dt
\quad\Rightarrow\quad
\frac{dp}{dt}(t)=m(t)\frac{dv}{dt}(t)-\frac{dm}{dt}\,u$$
Conclusion
Now apply Newton's second law (assuming no drag):
$$\frac{dp}{dt}=-mg$$
Combining the two:
$$m(t)\frac{dv}{dt}(t)=-m(t)g+\frac{dm}{dt}(t)\,u$$
While $dp/dt=F$ is indeed correct for a system losing mass, the way to use it is a bit tricky. Note that $dm/dt$ is a commonly used quantity (mass flow rate).
A: 
Does $F = dp / dt$  apply to a rocket ejecting mass?

Yes. It does. Product rule in calculus says that : $$ (u\cdot v)'=u'\cdot v+u\cdot v' \tag 1$$
So according to (1) :
$$ \begin{align}
F &= \frac{dp}{dt}\\
 &=\frac{d(mv)}{dt}\\
 &=m\frac{dv}{dt}+v\frac{dm}{dt}\\
 &=m(t)~a+v(t)\frac{dm}{dt}
\end{align}
\tag 2
$$
Thus, extended form of Newton second law applies equally well to systems which acquires or loses mass.
When system loses mass (for example in case of rocket burns fuel), then $\frac{dm}{dt} \lt 0$, so (2) becomes:
$$
F = m(t)~a-v(t)\left|\frac{dm}{dt}\right|
\tag 3
$$
Meaning that same rocket thrust will produce greater and greater acceleration to the rocket, because constant engine force must push forward less and less mass.
Keep in mind that Newton second law $F=ma$ is only valid when $m=const$,- system neither loses nor acquires mass, otherwise one needs to use (2) form of equation. So the final answer is that your quote is 100% right,-
$$F=\dot p \tag 4$$ is most universal second Newton law interpretation.
