I am confused about the equation describing the reactive force of a rocket. I read from this source this equation:

equation 1

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:

equation 2

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?


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).

  • $\begingroup$ This derivation appears to be lifted from early editions of Sutton. If you can find one of those, you can read the original. The online version just says " A derivation can be found in earlier editions of this book. " LOL $\endgroup$ Dec 14, 2019 at 22:54
  • $\begingroup$ After entering combustion chamber, the fuel gets further energy by combustion and after that there are quite a few more hydrodynamic effects are happening that affects the velocity of the departing gas. I recommend reading about De Laval nozzle, which is essential part of pretty much every non-solid fuel rocket engine. Some more information here: rocket engine nozzle. In short, gas pressure in the chamber is due to the nozzle throat, and is traded off to higher velocity at nozzle exit. $\endgroup$ Dec 14, 2019 at 23:03
  • $\begingroup$ The gas molecules impact the sides of the combustion chamber and the inside of the nozzle. By the end of that process they are exiting the nozzle at relative velocity u. $\endgroup$ Dec 15, 2019 at 0:12

1 Answer 1


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?

The exhaust is still at positive pressure when it’s in the bell of the nozzle. It’s still pressing on the nozzle. Some of that is thrust pushing the rocket forward. And by Newton’s 3rd law, that’s also accelerating the gas rearward, increasing its speed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.