Frequently I see the expression $$F = \frac{dp}{dt} = \frac{d}{dt}(mv) = \frac{dm}{dt}v + ma,$$ which can be applied to variable mass systems.

But I'm wondering if this derivation is correct, because the second law of Newton is formulated for point masses.

Furthermore if i change the inertial frame of reference, only $v$ on the right side of the formula $F = \frac{dm}{dt}v+ma$ will change, meaning that $F$ would be dependent of the frame of reference, which (according to me) can't be true.

I realize there exists a formula for varying mass systems, that looks quite familiar to this one, but isn't exactly the same, because the $v$ on the right side is there the relative velocity of mass expulsed/accreted. The derivation of that formula is also rather different from this one.

So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

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    $\begingroup$ You may find this article useful. $\endgroup$
    – valerio
    Mar 14, 2019 at 16:37

4 Answers 4


So my question is: is this formula, that I frequently encounter in syllabi and books, correct? Where lies my mistake.

The equation and the subsequent expression of the derivative $$\mathbf F_{ext} = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v) = \frac{dm}{dt}\mathbf v + m\mathbf a,~~~(1)$$

are based on the erroneous idea that the equation

$$ \mathbf F_{ext} = \frac{d\mathbf p}{dt}~~~(2) $$

is valid for systems that lose or gain mass (where $\mathbf F_{ext}$ is external force on the system and $\mathbf p$ is momentum of the system.

The system in this context is often a rocket without the expelled gases far away. Obviously such system has variable mass. More generally, we can consider any mass system in a specified well-delimited control volume as the subject for which we seek the equation of motion.

The above idea keeps around (probably) in part because of some special relativity texts saying $\mathbf F_{ext} = d\mathbf p/dt$ is more general than $\mathbf F_{ext} = m\mathbf a$ because the former is valid equation for relativistic particles.

But in non-relativistic mechanics this equation is valid only when the system does not lose or gain parts.

In the subtler case where the system of interest (inside the control volume) does acquire or lose material parts (like a rocket), the equation (2) is no longer valid. An easy way to see this is that this equation is not Galilei-invariant: when written in different frame, the external force does not change, but the right-hand side does.

However, a different and correct equation for $\mathbf p$ (a variable-mass system's momentum) may be derived from Newton's laws when applied to each particle the system is made of 1:

$$ \mathbf F_{ext} + \mathbf F_{parts~outside} = \frac{d\mathbf p}{dt} + \frac{d\mathbf p_{lost}}{dt}~~~(4). $$

1) This can be done because each particle obeys Newton's law $\mathbf F= m\mathbf a$, as it does not lose or gain parts. One way to derive this equation goes like this.
Let us use convention where $\mathbf F_{a}$ means force due to body $a$ on something, $\mathbf F_{-b}$ means force acting on body $b$ due to something, and $\mathbf F_{a,-b}$ means force due to body $a$ acting on the body $b$.
It is easy to see that at time $t$ $$ \sum_{i\in V(t)} \mathbf{F}_{ext,-i} + \sum_{i \in V(t)} \mathbf F_{parts~outside,-i} = \sum_{i\in V(t)} m_i \mathbf a_i.~~~(a) $$ We would like to express this without summing over index $i$, using only quantities referring to the system inside and outside as a whole.
Let $\mathbf p$ be momentum inside the control volume $V$. This changes in time for two reasons:
1. particles that stay inside change their momentum
2. some particle leave or come in the control volume

So we can write $$ \frac{d\mathbf p}{dt} = \sum_{i\in V} m_i \mathbf a_i - \frac{d\mathbf p_{lost}}{dt}~~~(b) $$
where $d\mathbf p_{lost}$ is momentum lost from the control volume due to particles leaving, per time $dt$).
Comparing $(a),(b)$ we see that the sum equation can be expressed more over $i$ can be removed resulting equation of motion for the system (in the control volume) can be written more concisely as $$ \mathbf F_{ext} + \mathbf F_{parts~outside} = \frac{d\mathbf p}{dt} + \frac{d\mathbf p_{lost}}{dt} $$ which is the equation (4).

In case of mass leaving the system (a rocket), we can write this in an easier-to-remember way $$ \mathbf F_{ext} + \mathbf F_{parts~outside} - \frac{d\mathbf p_{lost}}{dt} = \frac{d\mathbf p}{dt}~~~(5). $$

When we apply this to a rocket, we can see that momentum of the rocket changes due to 1) external force, 2) force of the exhaust gases acting back on the rocket, but decreased by momentum lost from the rocket per unit time (due to exhaust gas leaving the system).

Although more general, this is somewhat foreign to the engineering viewpoint on rockets, because for the purpose of travel, rather than in momentum of the rocket, we are interested in its velocity.

In the simplest case where the lost particles all leave in direction same or opposite to the body velocity $\mathbf v$ (idealized rocket), this can be further simplified, as is common in textbooks. Let the boundary of the control volume be far from the rocket, so that velocity of particles crossing the boundary (relatively to the rocket) is constant $\mathbf{c}$, and $\mathbf{F}_{parts}$ acting back on the system in the control volume is negligible (the exhaust gas is rarified). Then the lost momentum per unit time is

$$ \frac{d\mathbf{p}_{lost}}{dt} = - \frac{dm}{dt}(\mathbf v+\mathbf c)~~~(6) $$ and the equation of motion simplifies into

$$ \mathbf{F}_{ext} + \frac{dm}{dt} \mathbf c = m\frac{d\mathbf{v}}{dt}.~~~(7) $$ It is important to realize also that $\mathbf v$ is not velocity of center of mass of the system, but is defined as $\mathbf p/m$ where $\mathbf p$ and $m$ are momentum and mass inside the control volume. The necessity of this distinction is best seen from this example: let the body have constant velocity $\mathbf v$, but let the control volume shrink so that less and less of the body is inside. Center of mass of the control volume has different velocity from $\mathbf v$, in fact it accelerates due to moving boundary of the control volume. However, velocity of the material particles does not change at all.

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    $\begingroup$ +1. It might be worth mentioning that using $\mathbf F = \frac{d \mathbf p}{dt}$ would mean that force becomes a frame-dependent quantity in a variable mass system. Most people who work with variable mass systems see this as anathema and use $\mathbf F = m\mathbf a$ instead. $\endgroup$ Oct 23, 2014 at 20:02
  • $\begingroup$ Indeed, only the meaning of $\mathbf a$ may be a little obscure in some cases, like when mass is removed from the body while both external and parts' force vanishes - the center of mass may accelerate, but clearly $\mathbf a$ has to be zero. This is because $\mathbf a$ is not acceleration of the center of mass anymore (this is valid only for constant mass cases), but is defined as $\mathbf p/m$. $\endgroup$ Oct 23, 2014 at 20:16
  • $\begingroup$ Correction: $\mathbf a$ is defined as $\frac{d}{dt}(\mathbf p/m)$. $\endgroup$ Nov 14, 2018 at 13:02
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    $\begingroup$ @claws, I don't think so, because in the paper author argues for validity of the equation $\dot{\mathbf F}=\dot{\mathbf p}$ where $\mathbf{F}$ is a nonstandard, Galilei variant "force" that includes not only external force, but also weird term $\dot{m}\mathbf u$ that depends on the chosen inertial frame of reference. He does not argue for the equation $\mathbf F_{ext} = \dot{\mathbf p}$. The paper is misguided in its direction, the term $\dot{m}\mathbf u$ is not a physical force. $\endgroup$ Jan 10, 2020 at 20:55
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    $\begingroup$ @JamesWirth you're right that part wasn't very clear. I changed those sections, please read them again. However, note that the derivation I had in mind actually does not work with momentum of the whole super-system. The actual argument is application of Newton's laws to each individual particle, and then summing the equations for the relevant particles that form the system (with variable-mass). $\endgroup$ May 17, 2020 at 13:27

You'll probably find the wikipedia article useful: http://en.wikipedia.org/wiki/Variable-mass_system

It is formulated so that $F+v_{rel} \frac{dm}{dt} = ma$, where $v_{rel}$ is the relative velocity of the mass being ejected to the center of mass of the body. This takes care of your question about reference frames, because $v$ will be the same in all frames. The term gets moved to the left side of the equation because $-v$ describes the velocity of the center of mass relative to the ejected matter.


You can only apply Newton's second law to closed systems. But, since you are applying second law to a open system, you are getting contradictory results. The correct procedure for solving variable mass system, is by calculating the change in momentum and then equating it to

    Force = (change in momentum)/small time interval in which change occurred.

Here is a article on it that you can find useful, apart from the wikipedia article. See, this website. http://www.thestudentroom.co.uk/wiki/Revision:Motion_With_Variable_Mass

  • $\begingroup$ Pratik, that website no longer exists. Do you know if it's available in a different location? $\endgroup$
    – Matt
    Jun 11 at 0:48
  1. Whether a mass can be considered a point or not depends on the scale at which it is studied. The earth can be considered to be a point mass when we are studying its motion around the sun but not so when we are studying its own rotation. Newton's laws are applied to systems of many particles.

  2. Newton's second law says that the rate of change of momentum of a system is proportional to the applied force. We choose units in such a manner that the constant of proportionality is 1. With this definition, the equation $md\vec{v}/dt + \vec{v} dm/dt = \vec{F}$ makes sense with $\vec{v}$ being the velocity of the body in the same frame of reference in which other vectors are measured.


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