I am confused about the equation describing the reactive force of a rocket. I read from this source this equation:
where M is mass of the rocket, u is the velocity of exhaust gas in an inertial frame of reference, and v is the speed of the rocket. And after taking the limit as $\Delta t$ goes to zero it becomes:
My question is how is $u$ still $u$ after taking the limit. Does that mean that part of gas with mass $dm$ accelerated instantaneously to that speed?
Also when calculating the thrust of the engine $F=q C$ ($q$ is the rate of change of the mass and C is the exhaust velocity of the gas relative to the rocket) we take the speed at the end of the nozzle. But isn't the moment when that little part of gas separated from the rocket (it is not "connected" with rocket anymore it is free to move outside) the moment when it entered the combustion chamber? Shouldn't that be velocity u?
UPDATE:
I think I made some progress with the equations. It would be nice if someone could confirm that or point to the mistake. If we look at the rocket plus fuel inside it as one body of mass $M$ (not like in the previous case) its center of mass is moving with the velocity $\vec{v_{cm}}$ at some instance of time. We still don't care how fuel inside the rocket is moving, we only need speed of center of mass of rocket and fuel inside together. After some interval of time $\Delta t$. The rocket with the fuel inside of it has mass $M-\Delta M$ and velocity $\vec{v_{cm}}+\Delta \vec{v_{cm}}$. There is also part of exhaust with mass $\Delta M$ and velocity $\vec{v_{ex}}$. After using law of conservation of momentum and taking the limit as $\Delta t \rightarrow 0$ it can be concluded that: \begin{equation} M \vec{a_{cm}}=-\dot M \vec u, \end{equation} where $\vec u $ is velocity of exhaust relative to rocket plus fuel. Now, $a_{cm}$ can be expressed as: \begin{equation} a_{cm}=\frac{1}{M}(m_r \vec{a_r}+m_f \vec{a_f}), \end{equation} where $m_r$ is mass of rocket and $m_f$ is mass of fuel inside of rocket. Acceleration $\vec{a_f}$ is equal to $\vec{a_r}+\vec{a_{fr}}$, where $\vec{a_{fr}}$ is acceleration of fuel inside of rocket relative to rocket. If we use this in equation above, we will get: \begin{equation} \vec{a_{cm}}=\vec{a_r}+\frac{m_f}{M} \vec{a_{fr}}, or \end{equation}
\begin{equation} M \vec{a_r}=-\dot M \vec u-\frac{m_f}{M} \vec{a_{fr}}. \end{equation} So will $\vec{a_{fr}}$ be zero and if it will then under which conditions (this system is not inertial)?
NOTE: Velocity and acceleration of fuel are taken to be the effective (like treating all fuel as one body).
