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Consider a rocket being launched vertically.

Let $T(t)$ denote the thrust from the engine and $M(t)$ be the total mass of the rocket at time $t$.

At $t=0$, $T(0)=M(0)g$ (so that the normal force due to the launch pad needs not to be considered).

The acceleration $a(t)$ of the rocket at time $t$ can be obtained (along with other variables like the ejection speed of the fuel that are less important to my question) from Newton's second law of motion:

$$T(t)-M(t)g=\frac{dp}{dt}=\frac{d(M(t)v(t))}{dt}$$

$$=M(t)\frac{dv}{dt}+v(t)\frac{dM}{dt}=M(t)\frac{dv}{dt}=M(t)a(t)\tag{1}$$

So it seems to me that in general, we do not need to consider the $\frac{dM}{dt}$ term? But shouldn't $\frac{dM(t)}{dt}$ be non-zero if the total mass of the rocket is decreasing over time.

Or is it that the change in mass over time is accounted for by $M=M(t)$ alone already?

And when do we need we to consider the $\frac{dm}{dt}$ term in $N2$?

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Your second equation in $(1)$ isn't valid when the mass is changing, see here.

When you have a variable mass (or rather the mass of the body of concern is changing), you need to think carefully about the system on which you are applying the second law. Here are two ways to go about this:

At time $t - \delta t$, the rocket mass is $M(t) + \delta m$ and at time $t$ it is $M(t)$. Apply the second law to the system that is only the mass that will remain at time $t$, i.e. the mass $M(t)$. This mass isn't changing during this time interval. We can write $$T(t) - M(t)g = M(t)\frac{dv}{dt}.\tag{1}$$ This equation is instantaneously valid during the entire motion of the rocket.

Now consider the same situation, but this time choose the system to be the entire rocket mass $M(t) + \delta m$ at time $t - \delta t$, including the mass $\delta m$ that will have been ejected by time $t$. This mass again does not vary during the time interval $\delta t$. The only external force applied on this system is the weight. Suppose the mass $\delta m$ is ejected from the rocket at a speed of $v_e$ relative to the rocket, and $M(t)$ picks up a velocity increment $\delta v$. The second law now states $$-(M(t) + \delta m)g = \frac{1}{\delta t}\left[M(t)(v+\delta v)+\delta m(v -v_e) - (M(t)+\delta m)v \right].$$ As $\delta t\to 0$, we get $$-M(t)g=M(t) \frac{dv}{dt} +\frac{dM(t)}{dt}v_e.\tag{2}$$ Here $dM/dt$ the rate of change of mass, not the rate at which mass is ejected, i.e. $\delta m$ was positive but $dM/dt$ is negative.

Comparing $(1)$ and $(2)$, you see that $$T(t) = -\frac{dM(t)}{dt}v_e$$ so the ejection speed isn't unimportant after all. When $v_e$ is constant, neglecting the weight term and integrating $(2)$ yields the famous Tsiolkovsky rocket equation.

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  • $\begingroup$ Thanks. So under what circumstances is $\mathbf{F_{EXT}}=\mathbf{v(t)}\frac{dM}{dt}+M(t)\frac{d\mathbf{v}}{dt}$ applicable? (It seems that this is invalid for all non-closed system?) $\endgroup$ Commented Jun 22, 2020 at 9:16
  • $\begingroup$ I'm not aware of any circumstances under which that would hold. Note that because of the velocity in the first term, $\vec{F}_{ext}$ changes when you switch to a different inertial reference frame, which is a red flag. $\endgroup$
    – Puk
    Commented Jun 22, 2020 at 9:21
  • $\begingroup$ I'm confused, why wont you take account of the change in the mass (partial)m in equation(1), despite it being a change in the mass of the rocket, is it that it's negligible or is it that (partial)m isn't a part of the system anymore?? $\endgroup$ Commented Jul 31, 2021 at 15:37
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    $\begingroup$ @KhaledOqab Because for $(1)$ we selected the system to be the mass that will remain at time $t$, so $\delta m$ is not part of the system during this time interval. If you choose the system to be the mass that you start with at time $t - \delta t$ (including $\delta m$), you will end up with $(2)$. The difference is that in $(1)$, the mass $M(t)$ is acted upon by the ejected mass $\delta m$ in addition to gravity, while in $(2)$ the mass $M(t) + \delta m$ is acted upon solely by gravity. $\endgroup$
    – Puk
    Commented Aug 4, 2021 at 5:19
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You can't cancel the $\frac{dM}{dt}$ term in this case. And no, $M(t)$ does not have the information of the change in mass. It just knows how much mass there is in the system given a time $t$. You need to consider the mass derivative term when indeed the mass of the system changes during the process you're studying.

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The derivation of $F=ma$ comes from $F=\frac{\mathrm{d}p}{\mathrm{d}t}$. So if mass of system changes we can say $$F\neq m\frac{\mathrm{d}v}{\mathrm{d}t}$$ because it's equal to $m\frac{\mathrm{d}v}{\mathrm{d}t}+v\frac{\mathrm{d}m}{\mathrm{d}t}$ according to $F=\frac{\mathrm{d}p}{\mathrm{d}t}$.

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