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Let's say a rocket is moving along $\hat{i}$ and is ejecting mass at a rate of $r$ with relative velocity $-k \hat{i}$ under constant external force $F$

The rocket equation, $$m\frac{dv}{dt} = F + u\frac{dm}{dt}$$ gives me $$(m_0-rt)\frac{dv}{dt} = F + (-k)(-r)$$

However, using $F = dP/dt$ I get $$F = m_0\frac{dv}{dt} + vr$$ which gives me a different $v(t)$

I want to know why using $$F = \frac{dP}{dt} = m\frac{dv}{dt} + v\frac{dm}{dt}$$ is failing me.

I suspect that $F = dP/dt$ might be only valid for closed systems, but why?

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    $\begingroup$ What is the question? $\endgroup$
    – JMac
    Commented May 29, 2017 at 16:38
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    $\begingroup$ Personally I can't follow much of what you've done here, some more description would likely help you. Also, you should add your actual question to the question. $\endgroup$
    – JMac
    Commented May 29, 2017 at 16:44
  • $\begingroup$ Newton's laws are valid only for constant mass systems. That's simply an assumption made when transitioning from point-mass systems to systems made of multiple particles. Otherwise, the equations of motion for extended systems would not resemble the point-mass equations of motion. For more info, see this SE answer $\endgroup$
    – garyp
    Commented May 29, 2017 at 17:07
  • $\begingroup$ @garyp Oh, I see. Using $F = dP/dt$ for an open system leads to an equation that is not Galilean invariant! So my suspicion that F = dP/dt being applicable to only closed systems is correct in that case. Could you post that as an answer so I can accept it? $\endgroup$
    – xasthor
    Commented May 29, 2017 at 17:32

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