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The mathematical representation of the net external force on a system is $\vec F_{net} = \frac {d\vec P}{dt}$, which is the rate of change of linear momentum of the system. If we substitute $\vec P = m\vec v$ into the formula for force and differentiate, we get:$$\vec F_{net} = m\frac{d\vec v}{dt} + \frac {dm}{dt}\vec v.$$ What do the terms $m$, $\frac{d\vec v}{dt}$, $\frac {dm}{dt}$, $\vec v$ mean exactly? Could you use rocket propulsion as an example to explain the meaning of these terms? I have an idea of what each of these could be in this case, please check whether they are right or wrong. Here, the system I am choosing is the total mass of the rocket and the ejected gas and the term "rocket" refers to the rocket and the gas present in the rocket.

$m$ is the total mass of the rocket at any given instant.

$\frac{d\vec v}{dt}$ is the rocket's acceleration at a given instant in time.

$\frac {dm}{dt}$ is the rate at which the gas is ejected out of the rocket to propel it.

$\vec v$ is the rocket's velocity at any given instant(I am especially not sure of this one, but the same goes for the others.). Thank you for answering.

A message to the community regarding the question suggested: The question suggested does not answer my query as it does not convincingly explain what each of the terms in the formula is, and only seeks to know the difference between the formula in the rocket equation and the one I seek to use. And most importantly, it fails to convince me why I can not use my equation to solve for solving variable mass problems. Thank you for understanding.

EDIT: I tried to solve other problems involving the concept of variable mass like the falling chain problem and got a bit of clarity as to what each of the terms could mean, I am convinced that the 1st two terms are correct(hopefully) but I still have a doubt in the variable mass part.

Aditya Bansal
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The rocket equation derivation does not utilize $F_\text{ext}=\dot{p}=\dot{m}v+m\dot{v}$. Instead, you start with the conservation of momentum and arrive at a similar expression (see this answer of mine): $$\frac{\mathrm{d}p}{\mathrm{d}t}=m\frac{\mathrm{d}v}{\mathrm{d}t}+\left(v-v_e\right)\frac{\mathrm{d}m}{\mathrm{d}t}$$ Note that there is an extra term here that your expression does not have, which is the momentum of the ejected mass from the rocket (hence it being negative). You could write $V_e=v-v_e$ to "simplify" the equation to have the same form as your expression, but it's still a different velocity in the second term than the velocity of the rocket.

Otherwise, the interpretation of your remaining terms are correct: $m(t)$ is the mass at time $t$ and $\dot{v}(t)$ the instantaneous acceleration.

Kyle Kanos
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  • But why can it not utilise the 1st expression? Do we still not use Newton's second law for the derivation? – Aditya Bansal Apr 01 '23 at 19:58
  • You can't because you do not account for the ejected mass that way, which means you're not describing the physical scenario. – Kyle Kanos Apr 02 '23 at 01:25
  • I get what you mean, but why is that so? What does the $v$ term in the second law of motion mean exactly, and how is it different from the rocket equation? Also, what exactly is $\frac {dm}{dt}$? I suspect that it is -ve for my equation and +ve for your equation, given that the 2nd law even works. – Aditya Bansal Apr 02 '23 at 03:50
  • It's so because my formula starts from an analysis of momentum conservation whereas yours does not. The $v$ in the 2nd law means the velocity of the object; in the rocket equation is also means the same thing except we now also have to consider the velocity of the ejected mass in the conservation equations, hence the $v_e$ term. $\mathrm{d}m/\mathrm{d}t$ is the rate at which the fuel is burned (or, equally, the rate of the mass of the rocket lost in time). Your equation clearly has it positive while mine contains both positive and negative values. – Kyle Kanos Apr 02 '23 at 11:16
  • I get what you mean, I think. Though I used the velocity vector, which can take any sign. The thing I do not understand is why can I use this equation for other variable mass problems and not for the rocket problem. – Aditya Bansal Apr 02 '23 at 20:29
  • Well that's part of the problem with using $\mathrm{d}p/\mathrm{d}t$: it isn't actually valid for variable mass systems, you have to consider the momentum gain/loss due to mass loss/gain and that does not occur in the formula you are trying to use. – Kyle Kanos Apr 03 '23 at 10:46
  • See also this question or this question which both discuss the use of $\mathrm{d}p/\mathrm{d}t$ with regards to variable mass systems. – Kyle Kanos Apr 03 '23 at 10:52