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While studying the dynamics of a meteorite entering the atmosfphere, the book "Impact Cratering - A geological Process" by H.J. Melosh considers the forces of drag, lift and gravity and compares their sum to $m\vec{a}$. This is okay if the mass of the meteorite is constant, but then he writes that, due to the process of ablation, th mass changes following this equation: $$\frac{dm}{dt}=-\frac{C_H\rho Av}{2\xi}(v^2-v_{cr}^2)$$ where $v=|\vec{v}|$ and the other terms are constants depending on the shape and material of the meteorite and the density of the atmosphere.

My question is: if the mass changes during the motion, shouldn't we add a term $$\vec{v}\frac{dm}{dt}$$ to Newton's second law? And, if not, how do I know in which cases I must not consider that term even if mass is changing? (Why is ablation different from other ways the mass of an object changes?)

Rhino
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  • Try calculating the magnitude of the missing term to the other terms in the equation. Is it large or small compared to them? If small, then one can ignore it. – Anders Sandberg Sep 12 '19 at 16:04
  • I used this equations to study the motion of a meteorite (using the computer, with the Runge-Kutta method using coefficients that are plausible) and the results are completely different if I add that term to the differential equation, so I think it's not negligible – Rhino Sep 12 '19 at 17:23
  • @Rhino: The drag force is v or v^2 right? So this term appears to be a combination of a quadratic $v^2 v_{cr}^2$ piece and a $v^4$ piece (in the differential equation). So at small $v$, this can be neglected as linear drag dominates, but at large $v$ this term would dominate. The specific transition region would depend on the constants. – levitopher Sep 13 '19 at 01:08

2 Answers2

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The equation $\vec{F}=\frac{d\vec{p}}{dt}$ cannot just blindly be applied to systems in which mass is entering or leaving the system. You have to consider how mass enters or leaves a system. For example, if the mass is being ejected in the opposite direction of $\vec{p}$ (a rocket) this will increase $\vec{p}$, while if the mass is being isotropically ejected this will decrease $\vec{p}$ (since the mass of the body decreases, but the speed does not). The equation $\vec{F}=m\frac{d\vec{v}}{dt}+\frac{dm}{dt}\vec{v}$ is simply not a true equation, and should not be applied. I would go so far as to call it a dangerous equation, because there are specific cases in which it is true and this often misleads students (and even occasional textbook authors!) to think it is a general rule.

The general rule for a variable mass system is

$$ \vec{F}+\vec{v}_{rel}\frac{dm}{dt}=m\frac{d\vec{v}}{dt} $$ where $\vec{v}_{rel}$ is the relative velocity between the mass that is lost and the object you care about. Notice that if $\vec{v}_{rel}$ is opposite $\vec{v}$, it tends to increase $\vec{v}$, as in the case of a rocket. If $\vec{v}_{rel}=-\vec{v}$, we get the "dangerous" equation, which only holds in this case. Finally, in the case of meteor ablation, we have $\vec{v}_{rel}=0$, because the pieces falling off the meteor have the same initial speed as the meteor. Thus, we can ignore the $\frac{dm}{dt}$ term entirely.

Jahan Claes
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I think you are right, but we should think about it like this:

$$\frac{d\vec{p}}{dt}=\vec{F}\rightarrow \frac{d(m\vec{v})}{dt}=\frac{dm}{dt}\vec{v}+m\frac{d\vec{v}}{dt}=\vec{F}$$

Assuming those forces do not depend on the mass, you should actually be subtracting that from the acceleration:

$$\frac{d\vec{v}}{dt}=\frac{F}{m(t)}-\frac{1}{m}\frac{dm}{dt}\vec{v}$$

But just to be clear, that $\vec{F}$ (which is net force, of course), is probably velocity-dependent as well (since the drag force is). So we are now solving some complicated differential equation, which is a polynomial in $\vec{v}$.

EDIT: To try and clarify and summarize what's wrong with this (my) answer. The relationship between Newton's laws and momentum relies on interactions transferring momentum. This makes sense in the case of objects either collecting mass (via interactions) or ejecting mass (via interactions). However, in the case of a meteorite, the prevailing model uses interactions that do not transfer momentum, at least not in the direction of motion. The model of a meteorite traveling through the atmosphere is to treat the solid rock as a collection of smaller rocks, traveling together but not interacting. The interaction of the atmosphere pulls rocks away from the clump (possibly transferring momentum in the longitudinal direction), but the rest of the meteorite "doesn't know anything about it". As long as this happens roughly evenly across the body of the asteroid, it's path remains approximately 1-dimensional.

In this way, $\vec{F}=\frac{d\vec{p}}{dt}$ can be applied consistently, as long as one carefully specifies how momentum is being transferred.

EDIT EDIT: Another answer with more detail on this consistency, and an article from EJP.

levitopher
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    I don't think this makes sense. You can't just apply $\dot{\vec p}=\vec F$ to a case with changing mass. In general, the instantaneous acceleration should depend only on the instantaneous mass and instantaneous force, and not the derivative of the mass. – Jahan Claes Sep 12 '19 at 19:41
  • It's not like the mass is vanishing; it's just becoming "not a part of the system". – Jahan Claes Sep 12 '19 at 19:42
  • @JahanClaes: I didn't say anything about mass vanishing, and this is how you you traditionally deal with a time-dependent mass in Newtonian physics. Check out any of the traditional textbooks; i.e. Fowles and Cassiday eq (7.7.6). Like I said, it does assume the forces are external and do not depend on the mass. – levitopher Sep 12 '19 at 19:51
  • (of course, if the mass is becoming "not part of the system", it might as well be vanishing, but still doesn't change the discussion) – levitopher Sep 12 '19 at 19:53
  • That equation applies to a system whose changing mass corresponds to adding or subtracting particles that are stationary. There's a whole discussion in the preceding paragraph about the relative velocity of the particles and the object under consideration, and equation 7.7.6 only applies when the particles that cause the gain/loss of mass are stationary. – Jahan Claes Sep 12 '19 at 19:58
  • That is NOT the case for ablation, in which a meteorite is losing parts of itself. If you want to apply the arguments in section 7.6 to this meteorite, you have to accept that the particles that cause the mass loss start out with the same velocity as the meteorite (because they are part of the meteorite) and thus have relative velocity zero. – Jahan Claes Sep 12 '19 at 20:00
  • ^^ I'm sorry that is just not true. $dp/dt=F_{net}$ is a statement of Newton's second law. It doesn't care how an object looses mass, what the object is made of, what the system is - it's the equation of motion for the center of mass of an object in NR motion. $p=mv$ is the definition of momomentum. Then I used the chain rule. The discussion in that section of F+C might have started with the specific case of rocket motion, but the derivation is based on first principles. In fact, the final equation of rocket motion DOES have an $\dot{m}$ in it, in the differential equation. – levitopher Sep 13 '19 at 00:49
  • (It's something along the lines of $(v+v_e)\dot{m}=m\dot{v}$, I don't have the book right in front of me). Shall we move to chat? – levitopher Sep 13 '19 at 00:51
  • Again, rocket motion is ALSO not the same as ablation. In a rocket, the speed of the exhaust is different from the speed of the rocket, so there is a relative speed $v_e$ that is nonzero. This is different from ablation, in which the relative speed of the particles leaving the body is zero.

    Surely you agree that it DOES matter how an object loses mass, since the rocket equation has a $v_e$ in it, which characterizes HOW the exhaust is being expelled.

     – Jahan Claes Sep 13 '19 at 12:27
  • Let us continue this discussion in chat. – levitopher Sep 13 '19 at 13:12
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    @JahanClaes Is correct here – BioPhysicist Sep 13 '19 at 21:02
  • You can see I've agreed with him and explained how to make it consistent. Seems helpful to me, but if not I'll simply delete the answer. – levitopher Sep 14 '19 at 01:49
  • @levitopher Not to pile on, but I don't think I agree with your edit either. $\vec{F}=\dot{\vec p}$ just can't be applied, even if you know there is an interaction that causes the mass to be ejected. See my answer: It only works in ONE very specific case. It doesn't work for the rocket, it doesn't work for a meteor that's ejecting mass with strong interactions, it simply doesn't work. It's not a true equation. – Jahan Claes Sep 14 '19 at 01:54
  • So that means you are just fine with "Newton's laws work in some cases with some kinds of mass loss, but not others"? That's not even close to good enough for me. Like I said, I'm getting heavy flame so I'll probably just remove it, but we should not be happy with this. – levitopher Sep 14 '19 at 02:06
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    I've read that article you link to before. They just redefine what the equation and forces means depending on the context so that the time derivative can always be used. So they are saying the same thing Jahan is saying, essentially. I think it's misleading to say $F=\dot p$ for variable mass systems because if you don't know about these nuances you will incorrectly just say $F=m\dot v+\dot mv$. Saying $F\neq\dot p$ isn't saying Newton's laws aren't valid for variable mass systems. – BioPhysicist Sep 14 '19 at 05:31
  • @Aaron: When you say "$F\neq \dot{p}$ for variable mass systems", it's just like saying "$F\neq ma$ if you include friction"....because it does work if you include friction correctly, and if you include all the masses in the case in question, it does in fact work. It has to work, because if you broke the asteroid into a mole of particles, and considered the interactions of everything, it had better describe nature correctly. It's a question of finding a consistent model when moving from the particle picture (where is has to work) to the extended object picture. – levitopher Sep 14 '19 at 16:28
  • A system of multiple particles is not necessarily a variable mass system. I'm not talking about multiple particles where we keep track of everything. A variable mass system is one where we don't keep track of everything and instead just say the mass of the system changes. In that case if $m$ is the mass of the system and $v$ is it's velocity, then you cannot simply say $F=m\dot v+\dot mv$. – BioPhysicist Sep 14 '19 at 16:46
  • @Aaron: Exactly! No one should EVER say "Newton's Laws work always, you don't even have to keep track of the masses!" We just have to construct a model in which you do keep track of them all, which is what I've attempted to do in my (modified, since I didn't understand the physical situation at first) answer. – levitopher Sep 14 '19 at 17:09