Velocity change over time calculation (Rockets) - Integration vs Momentum-approach

[Tonnz]

So i'm trying to figure out the change in velocity of a rocket after a t-seconds burn. I came up with one working approach, and then figured out there may be another way. Both return almost the same results. ALMOST

m - some mass,

v(t) velocity change after t-seconds burn,

Ve - exhaust velocity,

mDot - fuel-mass used per second,

m0 - total mass at t=0

F - force,

a - acceleration

The naive approach would be by a*t=v, where a is assumed to be constant. So this doesn't work obviously.

I'd really like to use LATEX or something similar here, but i don't think this forum allows it.

I started with F=m(t)*a(t). Note that nor m(t) or a(t) are constant. I rearranged it to be:

a(t) = F/m(t)

I replace m(t), knowing that m(t)=m0-mDot*t. mDot is the mass-loss per second, which is equal to = F/Ve. Ve is the exaust-velocity.

So i end up with: a(t) = F / (m0-t*F/Ve)

Now all i need to do to figure out how much the velocity has increased in t seconds is to calculate the integral (lower bound 0, upper t) of this function with regards to t.

So: ∆v(t) = ∫a(t) dt (upper bound is t, lower bound is 0),

which turns out to be (i wont bother typing the hole calculation out, will only lead to a mess without more tools to write proper equations):

∆v(t) = Ve*ln( (m0*Ve) / (m0*Ve - F*t) ), where ln is the natural logarithm.

This is all fine and works, BUT:

I thought "Why don't you just calculate the momentum the engine developes in t seconds and apply it to the craft?" Sure, why not? The "force" is just momentum per second for an engine.

So, gain in momentum: F*t, change in velocity: F*t/(m0-t*F/Ve) . Looks much easier.

I tried it out, difference is neglitible. Plottet both functions to compare, resulst are almost the same (but get bigger for large values of t). But for low values... difference is neglitible.

(first approach returns larger values for positive values of t than the second approach)

What do you guys think? I think the problem with the second approach is that it doesn't take into account that momentum is transferred gradually, which actually is the case, but how is this a problem? I'm almost certain it is because momentum is, like velocity, relative. Which would mean that from an inertial frame where the observer is in the middle and rests, the exhaust gasses blown out earlier have a higher momentum than those which where blown out later. It would be the other way round for the rocket, resulting in a higher change in velocity compared to the second approach (which is actually the case).

Any thoughts or ideas, or found an error? Anything will be welcome!

Edited June 25, 2015 by Tonnz

[Iskierka]

The problem with the momentum approach is that that momentum is shared between the craft, fuel that is yet to be burnt, and fuel which you have already burnt and thrown out the back to gain that momentum. Calculating the distribution of this momentum sharing, you'll end up at an equation that should be directly equivalent to the integration approach, since the immediate suggestion of how to do such would be to convert the momentum equation into an acceleration equation, which then ends up at the same first equation.

[Kryten]

What's wrong with just using Tsiolkovsky's rocket equation?

[Tonnz]

The problem with the momentum approach is that that momentum is shared between the craft, fuel that is yet to be burnt, and fuel which you have already burnt and thrown out the back to gain that momentum. Calculating the distribution of this momentum sharing, you'll end up at an equation that should be directly equivalent to the integration approach, since the immediate suggestion of how to do such would be to convert the momentum equation into an acceleration equation, which then ends up at the same first equation.

That makes alot sense, thanks Especially the part about "Calculating the distribution of this momentum sharing" convinced me , this would be equivalent to my thoughts about the relative momentum.

What's wrong with just using Tsiolkovsky's rocket equation?

Now that i think of it... the equation i came up with is exactly the same as the rocket equation, just with time-dependency.

Replacing the "m1" in the rocket equation with m0-t*F/Ve, and extending* the resulting fraction (inside the ln(...)) with the velocity (m/v) gives exactly what i came up with. If i undo the last step, the resulting equation would also require less floating point operations to compute.

:facepalm:

It could've been so much easier. Without the nasty integration stuff. Thank you for your careful observation anyways

*not sure if it is the correct term, i mean multiplying both numerator and denomination of the fraction with something.

Edited June 26, 2015 by Tonnz