My book derived the formula for the acceleration of a rocket at any instant in the following way:

$v_r=$ velocity of gas released from the nozzle relative to the rocket

$dt=$ infintesimal time interval

$dm=$ mass of the gas released from the nozzle in time interval $dt$

$dP=$ momentum of the gas released in time interval $dt$

$F=$ thrust force acting directly opposite to the direction of the release of gas

$M=$ mass of the rocket after time interval $dt$

We know from Newton's 2nd law,


$$\implies F=\frac{dm}{dt}v_r$$

$$\implies a=\frac{1}{M}\frac{dm}{dt}v_r$$

In this derivation, the velocity of the released gas, $v_r$, has been calculated from the perspective of the rocket, which is constantly accelerating, making it a non-inertial frame of reference. Newton's laws don't hold true in non-inertial frames of reference, but we used newton's 2nd law in this derivation. So, how is this derivation correct?

PS: A similar derivation can be found in Fundamentals of Physics by Halliday, Walker & Resnick

  • $\begingroup$ Actually $a$ cannot be acceleration of the rocket in rocket's frame as this acceleration is trivially zero and $dP$ cannot be momentum, but a change in momentum, because force is given by change in momentum in a time interval and not by momentum itself. $\endgroup$
    – Umaxo
    Commented Jun 30, 2021 at 11:55
  • $\begingroup$ Halliday et.al.'s first rocket equation can be written as $M\,a=-\dot{M}\,\nu_{rel}$ (9-86). They derive it assuming "we are at rest relative to an inertial reference frame". In that frame we have $F=\dot{P}=\dot{M}\,\nu+M\,\dot{\nu}=\dot{M}\,\nu+M\,a=-R\,\nu+R\,\nu_{rel}$ where $R$ is their fuel consumption rate $R=-\dot{M}$ and $\nu$ is the velocity of the rocket. The only non constant quantity in this equation for $F$ is $\nu$ which is increasing. Therefore $F$ becomes negative when $\nu$ exceeds $\nu_{rel}\,.$ I think this is an interesting example of an apparent force. ... $\endgroup$
    – Kurt G.
    Commented Jul 2, 2021 at 8:39
  • $\begingroup$ The force measured by an accelerometer in the rocket is $M\,a=R\,\nu_{rel}\,.$ This is a real force. The equation is also interesting in the sense that it is tempting to conclude from $F_{real}=M\,a$ that this force holds in the rest frame (because $a$ does). But this is incorrect. $\endgroup$
    – Kurt G.
    Commented Jul 2, 2021 at 8:44

5 Answers 5


Newton's second law

$$F = \frac{dp}{dt}$$

only holds in inertial frames.

While $v_r$ is defined as the velocity of the released gas relative to the rocket, it is also equal to the change in velocity $\Delta v$ of the fuel as observed from any inertial frames. This is a consequence of Galilean relativity, which holds at low velocities. Thus the formula $$F = \frac{dm}{dt}v_r$$ is written in an inertial frame of reference.


This derivation actually uses $v_{rel}$ because the author found more convenient to write the equations this way. But this does not mean that the system is being described from the rocket reference frame.


The derivation that you show is not calculated from the perspective of the rocket. It is calculated from the perspective of a stationary outside observer. The calculation simply uses the relative velocity of the gas to calculate what force the rocket will experience.

Newton's laws are valid in non-inertial reference frames as well (with the usual non-relativistic limitations), as long as you treat the reference frame acceleration as a gravity-like, external force on all masses.

In your specific example that would not be very practical, because the purpose of the calculation is to determine the size of that external force. But you can use Newton's laws to determine what will happen to objects inside the rocket while it is being accelerated. For example, if you want to calculate the trajectory of a ball that is being tossed by an astronaut in the rocket, from the perspective of that astronaut.


So, what you need to understand is that the calculations are all being done from any arbitrary inertial frame of reference. We are not calculating $a$ or acceleration of the rocket from the reference frame of the rocket (a non-inertial reference frame) itself. It is just easier to express the velocities of the rocket and the released gas using the relative velocity of the released gas with respect to the rocket.

Moreover, see Halliday's derivation. It is much better.


This derivation makes the most sense in momentarily comoving reference frame.

Say you want to derive the equation at some arbitrary time $t$. At this particular time, there exists an inertial reference frame where the velocity of the rocket is exactly zero. In this reference frame, the rocket was moving backwards and decelerating at times $t'<t$, and will be moving forward at increasing speed at times $t'>t$.

Since this frame is inertial, Newton's second law

$$F = \frac{dP}{dt}$$

does hold. Here, the change in momentum is

$$dP = v_r dm$$

where $v_r$ the velocity of the released gas. But since the velocity of the rocket is zero, $v_r$ can also be interpreted as the velocity of gas relative to the rocket.


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