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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

md/dt*v

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:

md/dt*v

but not entirely, as m is given as

inital mass - m' * time (in seconds)

Thanks.

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  • $\begingroup$ 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
    Commented Nov 2, 2022 at 1:03

2 Answers 2

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

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 $\sum F = \frac{dp}{dt}$. $\endgroup$
    – preferred_anon
    Commented 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 $F_\text{net} = dp/dt$. $\endgroup$
    – David Hammen
    Commented Nov 2, 2022 at 9:39
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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. $\endgroup$
    – David Hammen
    Commented Nov 2, 2022 at 19:12
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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.

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  • $\begingroup$ Shouldn't we account for the change in mass as well? $\endgroup$
    – Rsf
    Commented 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. $\endgroup$
    – David Hammen
    Commented 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 $\sum F = m \frac{dv}{dt} = ma$. $\endgroup$
    – David Hammen
    Commented Nov 2, 2022 at 11:42

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