The first law of motion states that :

"Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it."

Now consider a body with initial mass $m_{0}$ that constantly changes with respect to time. If the net force acting on it is zero according to Newton's second law of motion the body has a constant momentum. Which means that its velocity is changing* beside the fact that the net force acting on the body is zero, which implies that the first law of motion does not hold true. Is there a fault in my logic or was the first law formulated for a body of constant mass?

*That's because : $ p(t)=m(t)*u(t)=constant$ and the mass of the body is changing.

Important note to give some context : I came up with this question when I was trying to solve a textbook exercise where there was a satellite (consider that the satellite has a cylindrical shape)moving through a stationary cloud of interplanetary dust of constant density and all the dust that collides with the satellite sticks to its surface changing its mass.

And the exercise was asking to find the velocity of the satellite in respect to time. Due to the fact that the cloud of dust is stationary and there are no external forces exerted on the system the total momentum: $$p_{T}=m_{s}(t)u_{s}(t)+p_{cloud}<=>p_{T}=m_{s}(t)u_{s}(t)$$ must be conserved and hence velocity must change while the net force acting on the satellite is zero because:


  • $\begingroup$ Why does the mass change in time? $\endgroup$
    – Qmechanic
    Dec 29, 2023 at 17:09

2 Answers 2


The thing is, you're never going to find an object whose mass changes as a function of time in the way that you describe. One way that a mass could change over time is by throwing off or picking up massive matter, as happens to a rocket when it jettisons its fuel at high speed. But this isn't really what you're asking about, because we have to consider the force that the rocket exhaust exerts on the rocket, giving it thrust.

One way to think about your problem that might help is via Lagrangian mechanics. In this framework, a mechanical system is defined by its Lagrangian, a function $L(x, \dot{x}, t)$ of position, velocity, and time, such that the action $S = \int dt\, L(x(t),\dot{x}(t),t)$ is minimized for the physical trajectory $x(t)$. The Lagrangian for a single massive particle experiencing some force is $$ L = \frac{1}{2} m \dot{x}^2 - U(x),$$ where $U(x)$ is a potential such that the force is $F(x) = - U'(x)$. The calculus of variations tells us that the path that minimizes the action is the one for which the Euler-Lagrange equations are satisfied: \begin{align} \frac{d}{dt} \frac{\partial L}{\partial\dot{x}} &= \frac{\partial L}{\partial x}\\ \frac{d}{dt} (m \dot{x}) &= - U'(x). \end{align} If the mass is constant, this is just Newton's second law $ m \ddot{x} = F$. If the mass is not, we have instead $\dot{m} \dot{x} + m \ddot{x} = F$. For zero force, this is the situation that you describe, where the momentum $m(t) \dot{x}(t)$ is conserved, so changes to the mass can be viewed as an effective force, as $m \ddot{x} = - \dot{m} \dot{x}$. But this is all a little silly, because again, we don't know of any systems that are defined by a Lagrangian with a time-varying mass.

  • $\begingroup$ So according to Newton's law of motions we can conclude that there does not exist a body that that has a mass a that changes constantly while at the same time its net force is equal to zero? $\endgroup$ Dec 29, 2023 at 9:54
  • $\begingroup$ It's not really from Newton's law of motion. It's just that we don't usually have to describe systems like the one that you propose, so it's not clear how to generalize Newton's laws to describe them. The Lagrangian formalism gives us a cleaner way to talk about it, by turning constant $m$ into $m(t)$. $\endgroup$ Dec 29, 2023 at 10:49
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    $\begingroup$ Of course mass of the body can change without any net force - consider a body which ejects mass in such a way that it carries equal amount of momentum in opposite directions. $\endgroup$ Dec 29, 2023 at 13:57

Now consider a body with initial mass $m_0$ that constantly changes with respect to time. If the net force acting on it is zero according to Newton's second law of motion the body has a constant momentum.

No, in general with such bodies, momentum is changing. This is because you can't apply the original formulation of Newton's 2nd law $\mathbf F_{net} = d\mathbf p / dt$, valid for systems of constant mass, to bodies of variable mass. The mass variation makes the right-hand side

$$ \frac{d\mathbf p}{dt} = \frac{dm}{dt}\mathbf v + m\frac{d\mathbf v}{dt} $$

depend on the frame of reference, but the force $\mathbf F_{net}$ does not depend on that. So the equation does not hold in all frames for variable mass systems. People think it does because the same equation (for different $\mathbf p$) holds in special relativity, but that is a wrong idea transplant because in special relativity, that equation holds only for constant mass systems too (to be precise, I mean constant invariant mass systems).

Instead, it is the equation $$ \mathbf F_{net}=m\mathbf a, $$ where $\mathbf F_{net}$ takes into account also forces due to ejected mass, that applies to bodies of both constant and variable mass.

If the mass is expelled in equal amounts in opposite directions so that net propelling force due to this loss of mass is zero, and if sum of external forces is zero, then net force is zero. According to this equation, since $m$ is not zero, acceleration must be zero, and thus such body has constant velocity. It does not have constant momentum, since mass changes in time (unless we observe the body in a frame where the velocity is zero - then also momentum is zero and remains so).


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