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Assumptions:

  • A rocket engine requires a constant mass flow rate to maintain a given force
  • F = ma

Dilemma:

Let's say I take a 1 Newton rocket engine and attach it to a rocket with a mass including the starting fuel and engine itself of 1 kilogram. During a burn, both will experience an acceleration of 1 m/s^2.

For the first 10 seconds, the average velocity will be 5 m/s. The distance covered will be 50 m, so the work done will be 50 J.

For the second 10 seconds, the average velocity will be 15 m/s. The distance covered will be 150 m, so the work done will be 150 J.

For the third 10 seconds, the average velocity will be 25 m/s. The distance covered will be 250 m, so the work done will be 250 J.

How does the work done increase if the mass flow rate of the fuel, and therefore the input energy flow rate, is constant?

Caveats:

In a real rocket, the mass is lost over time, so the mass flow rate decreases or the acceleration increases over time. For a rocket with enough mass and a high-enough energy fuel to burn for more than a few hundred seconds, this effect is negligible relative to the linear work increase over time above.

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  • $\begingroup$ The answer that was here disappeared. I would only add that the work done on the exhaust is negative in the moving frame. And it becomes larger in magnitude as we go faster, as does the positive work on the rocket. But wait, is there any work done at all. If the momentums are conserved (?). I’ll check the linked question i guess $\endgroup$
    – Al Brown
    Jul 30 '21 at 15:51
  • $\begingroup$ The concept of work is fine, you just have to consider the exhaust too. See my answer in particular at the link posted above: physics.stackexchange.com/questions/428952/… $\endgroup$
    – Dale
    Jul 30 '21 at 15:56
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    $\begingroup$ While the referenced question is the same, none of the answers actually answer it. They answer why kinetic energy isnt going up. The one answer that considers work claims more and more work IS done which cant be true for a constant burn rate. $\endgroup$
    – Al Brown
    Jul 30 '21 at 16:10
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Work done is $W=\int_0^x F\,dx$, and in your case $F$ is constant. If you want to calculate this in smaller intervals you should choose space intervals and not time intervals (10s in your case). In each space interval $\Delta x$ the work done is $\Delta W=F\Delta x$ (same for all space intervals).

Clearly, as the rocket accelerates it takes less and less time to cover one such interval. In other words: over time intervals of equal length the work done must increase.

The derivation of the kinetic energy is as always: $$ E=\int_0^xF\,dx=\int_0^xma\,dx=\int_0^xm\frac{dv}{dt}\,dx=\int_0^vmv\,dv=\frac{1}{2}mv^2. $$

A more realistic rocket that looses mass due to fuel consumption is discussed here where also a reference is given.

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  • $\begingroup$ But why not a time interval? We should be able to correctly analyze that. It seems the work done should be equal for two of them but also seems not. Also, Im not sure you showed that increase in energy for two equal space intervals is the same. But if you did then that cant be right because for two different space intervals the engine burns for different times so energy should be different. $\endgroup$
    – Al Brown
    Jul 30 '21 at 16:40
  • $\begingroup$ By definition of $W$ as the integral over $x$ the natural discretization is in space. I am going to add to my answer the following: it takes less and less time to cover a space interval, in other words, over equal time intervals the work done must increase. Let me know if that's convincing. $\endgroup$
    – Kurt G.
    Jul 30 '21 at 16:58
  • $\begingroup$ But that makes no sense, because a constant mass flow rate provides constant energy per time interval $\endgroup$ Jul 30 '21 at 23:44
  • $\begingroup$ @TheEnvironmentalist It also makes no sense because we have a constant burning of fuel to make that mass flow rate, so if energy is being conserved... I think the answer is in the exhaust. It is going backwards at a lower backward speed as seen from a still frame, less energy. Thats all I can figure. And the rocket is losing mass which reduces kinetic energy linearly not quadratic. So the exhaust going slower relative to a still frame and the rocket losing mass combine to overcome the energy gains from increasing v^2 of the rocket? $\endgroup$
    – Al Brown
    Jul 31 '21 at 0:38
  • $\begingroup$ Constant mass flow rate does not mean constant energy per time. It means constant change of momentum per time. The force that propells the rocket is constant. The fact that we generate that mass flow by burning a constant amount of fuel per time is secondary. We must not mix up the mechanics with the thermodynamics. $\endgroup$
    – Kurt G.
    Jul 31 '21 at 4:58
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Very simple to explain with words. The amount of energy may remain constant over the intervals you have chosen, but the work is the sum of all of the energies across all of the intervals considered.

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