I've been attempting to simulate a rocket launch, using:

ΣF= ma + md/dt*v

If ΣF is simplified to only include F_m, being the rocket's thrust, the equation, if solved for a, is:

a = (F_m - md/dt*v)/m

Simulating with


gives a value much higher than using Tsiolkovskys, but if it's dropped, the two values are approximately the same. Why would you drop using variable mass?

Note: by variable mass I mean m' itself in:


but not entirely, as m is given as

inital mass - m' * time (in seconds)


  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Nov 2, 2022 at 1:03

2 Answers 2


$$\sum F= ma + md/dt*v$$

This is erroneous, and everything after this is in turn erroneous. You should be using $$\sum F= ma = m \frac{dv}{dt}$$

Accounting for variable mass is non-trivial, but as a starter you should start with the correct form of Newton's second law.

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    $\begingroup$ If the mass is not assumed to be constant, isn't this also an incorrect statement of Newton II? I am used to seeing $\sum F = \frac{dp}{dt}$. $\endgroup$ Nov 2, 2022 at 9:24
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    $\begingroup$ @preferred_anon That expression works fine for an object with a fixed mass where $dp/dt = ma$. You do not want to use that $F_\text{net} = dp/dt$ for a variable mass object as this yields wildly different results based on the choice of the inertial frame. Acceleration should be the same in all inertial frames in Newtonian mechanics. Use $F_\text{net} = ma$. Compensating for variable mass (and the center of mass moving within the rocket body) is non-trivial, but $F_\text{net} = ma$ is a good start, much better than $F_\text{net} = dp/dt$. $\endgroup$ Nov 2, 2022 at 9:39
  • $\begingroup$ Just think about it: when mass decreases, does it transfer its momentum to the remaining mass to keep the total momentum the same? Or does its momentum also disappear from the system? (the answer is obviously the second one) $\endgroup$ Nov 2, 2022 at 18:24
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    $\begingroup$ @user253751 Here's a silly example: Consider a long rod sliding horizontally on a frictionless track. Once the rod crosses a certain line, we'll arbitrarily say the part of the rod that has crossed the line as outside the system boundary. The rod is moving, so $v$ is non-zero, and the mass is decreasing thanks to the silly system boundary. Yet the acceleration of any point on the rod that remains within that system boundary is zero despite $\dot m v$ being non-zero. $\endgroup$ Nov 2, 2022 at 19:12
  • $\begingroup$ @DavidHammen indeed, and it's the same with the rocket: when the mass "disappears" so does its associated momentum, so this has no impact on the velocity of the remaining mass. $\endgroup$ Nov 2, 2022 at 19:36

Isaac Newton stated that the forces equal the change in momentum:

$\Sigma F=\frac{d(mv)}{dt}$

When the rocket has a flow of $\dot{m}$ the result is

$\Sigma F=-\dot{m}v_e+m\frac{dv}{dt}$

where $v_e$is the exhaust velocity. Then $fv_e$ is identified as the thrust and we would write:

$ \Sigma F+\dot{m}v_e=m\frac{dv}{dt}$

The derivation of the differentiation has been done on Stackexchange before and in any good university textbook.

The last two equations take into account the variable mass since $m$ is the time changing mass. As a commenter pointed out, the thrust is often lumped in with the external forces.

One last note on thrust, the full amount is

$Thrust = \dot{m}v_e + (p_e - p_0)A_e$

where $A_e$ is exit area, $p_e$ and $p_0$ are the exit and local atmospheric pressures.

  • $\begingroup$ Shouldn't we account for the change in mass as well? $\endgroup$
    – Rsf
    Nov 2, 2022 at 9:22
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    $\begingroup$ @Rsf See my reply to what essentially is same comment by preferred_anon to my answer. Accounting for change in center of mass with respect to the vehicle structure is non-trivial. Ignoring this change results in an error in the tens of meters for a launch vehicle that discards 85 to 90% of its mass in 8 to 10 minutes and makes a 90° turn while doing so. Using $F_\text{net} = dp/dt$ results in much larger errors. $\endgroup$ Nov 2, 2022 at 9:48
  • $\begingroup$ Another way to look at the $\dot m\,v_e$ term is that it is just another external force, making the resulting equation $\sum F = m \frac{dv}{dt} = ma$. $\endgroup$ Nov 2, 2022 at 11:42

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