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Here's the problem I have, specifically relating to a rocket in deep space:

A rocket burns fuel at a constant rate. Assuming its mass remains constant, what happens to its acceleration? I seem to get two different answers depending on how I solve it (using conservation of energy or conservation of momentum):

Conservation of Energy

$\frac12mv^2=E_{exhaust}$, where $E_{exhaust}$ varies with time.

$v=\sqrt{\frac2mt}$, meaning that velocity varies with $\sqrt t$, meaning that acceleration decreases over time.

This also means that depending on your frame of reference's velocity relative to the rocket, the ship's acceleration must change...which makes about 0 sense,

Conservation of Momentum

$P_{exhaust}=mv$

Meaning that as more fuel is consumed, the velocity of the ship increases too (as $P_{exhaust}$ increases linearly with time, $v$ must also increase linearly with time).

I am assuming that the mass of the spaceship is constant. While not quite true, the effect would be negligible and the paradox I'm running into would remain true anyways.

Can anyone tell me what I'm doing wrong? The acceleration cannot possibly change depending on where you observe it from, but the acceleration also cannot remain constant as that violates conservation of energy.

Thanks in advance!

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  • $\begingroup$ "A rocket burns fuel at a constant rate. Assuming its mass remains constant, ..." -- is it getting mass from somewhere else? otherwise, the only scenario consistent with this is that the fuel burn rate is exactly zero. $\endgroup$
    – Stan Liou
    Apr 25 '14 at 0:37
  • $\begingroup$ @StanLiou You can assume that the mass of fuel is negligible compared to the mass of the rocket itself, and then the question makes sense. $\endgroup$
    – garyp
    Apr 25 '14 at 0:42
  • $\begingroup$ Yes, sorry I should've been more clear about that. It's burning fuel, and its mass is decreasing, but negligibly so. Even if it's not negligible, the issue of acceleration being different depending on the frame of reference observed from would still apply (and confuse me!) $\endgroup$
    – Mavvie
    Apr 25 '14 at 0:47
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    $\begingroup$ Possible duplicate of Where does the extra kinetic energy of the rocket come from? $\endgroup$
    – knzhou
    Sep 27 '18 at 9:38
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Your conception of Conservation of Energy is wrong. First of all, what do you mean by $E_{exhaust}$ ? The kinetic energy of the exhaust? If so, you have the KE of the exhaust, and the KE of the rocket ... and where did all of this KE come from?

To begin applying conservation of energy, you'd have to say something like: the sum of the KE of the rocket plus the KE of the exhaust is equal to the chemical energy released during the burning of the fuel".

Your conservation of momentum argument is more or less correct, but you would have to explain why the exhaust momentum increases linearly. Won't exhaust be expelled at different velocities as the speed of the rocket increases? A more complete analysis is needed.

But your original question has a simple answer. If you assume, as you do, that the mass of the rocket+fuel does not change significantly, and also that the fuel is burned at a constant rate, then the thrust (force) will be constant, and you can apply Newton's second law.

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  • $\begingroup$ Thank you so much! I left out lots of math because I figured it was a concept error, and am glad that it was! My complete solution involves: $$E_p=E_{k(rocket)}+E_{k(exhaust)}$$ $$2E_p=m_{rocket}v_{rocket}^2+m_{exhaust}v_{exhaust}^2$$ Since the exhaust's velocity relative to the space ship is constant, $v_{rocket}-v_{exhaust}=constant$, so after substitution and ignoring other constants I'm left with $Ft^2=Gt^2$ where F and G are some combination of constants. So...it works! Thanks again, working through a complete analysis has been very helpful. $\endgroup$
    – Mavvie
    Apr 25 '14 at 1:17
  • $\begingroup$ I'm glad it worked out for you, but you've left out enough steps that I can't see it. I am a little concerned about the KE of the exhaust, as the exhaust velocity changes as the rocket accelerates. In calculating the total KE of the exhaust you'd have to take into account the fact that all of those exhaust molecules are traveling at different speeds. (No need to justify your solution to me, but you might want to justify it to yourself.) $\endgroup$
    – garyp
    Apr 25 '14 at 12:40
  • $\begingroup$ Thanks, I hadn't thought of that. Would you be able to show me a full solution? I'd really appreciate it. I keep missing things, and have been unable to convince even myself of how the Ek of the rocket can go up with the square of time...where does that square energy come from? $\endgroup$
    – Mavvie
    Apr 26 '14 at 20:47

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