# Second law of Newton for variable mass systems

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.

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

The procedure $$\mathbf F_{ext} = \frac{d\mathbf p}{dt} = \frac{d}{dt}(m\mathbf v) = \frac{dm}{dt}\mathbf v + m\mathbf a,$$

is based on the erroneous idea that the equation

$$\mathbf F_{ext} = \frac{d\mathbf p}{dt}$$

is valid for systems with variable mass ($$\mathbf F_{ext}$$ is external force on the system). It keeps around probably also because of some 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 in relativity.

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

In the more rare case where the system of interest (a control volume = imaginary , possibly moving region in space) does acquire or lose material parts (like a rocket), the equation $$\mathbf F_{ext} = \frac{d\mathbf p}{dt}$$ is no longer valid. Another equation may be derived from this one applied to the whole supersystem (system + incoming/leaving parts). This can be done because the supersystem does not lose or gain particles. The new equation is

$$\mathbf F_{ext} + \mathbf F_{parts} = \sum_k m_k \mathbf a_k$$ where $$\mathbf F_{parts}$$ is force on the system due to parts no longer inside the control volume and summation is to be done over all particles in the control volume.

It can be written also in this way:

$$\mathbf F_{ext} + \mathbf F_{parts} = \frac{d\mathbf p}{dt} + \frac{d\mathbf p_{lost}}{dt}$$ where $$d\mathbf p_{lost}$$ is momentum lost from the control volume per time $$dt$$.

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. 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 $$\mathbf{F}_{parts}$$ is negligible. Then the lost momentum per unit time is

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

$$\mathbf{F}_{ext} + \frac{dm}{dt} \mathbf c = m\frac{d\mathbf{v}}{dt}.$$ 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.

• +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. – David Hammen Oct 23 '14 at 20:02
• 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 $\mathbf p/m$. – Ján Lalinský Oct 23 '14 at 20:16
• Correction: $\mathbf a$ is defined as $\frac{d}{dt}(\mathbf p/m)$. – Ján Lalinský Nov 14 '18 at 13:02

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

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.