Filip Larsen said:
No, you read it wrong.
I wrote the kinetic energy out like I did in terms of ##m_f## and the total (constant) mass ## m = m_f + m_r## to clearly show that it varies quadraticly, and not linearly, in ##m_f## and with the first part (i.e. ## \frac{1}{2} m_f v_e^2 ##) separate because that is the equivalent energy released by the the fuel in the continuous case
Thanks!
I think I see what's happening here. There are
different reference frames to consider and they are not the same nor is the kinetic energy required to be equivalent.
First, in the rocket's frame ##R##, an observer sees the mass ejecta at ##v_e## as specified by the rocket engine and combustion chemistry or just by whatever is ejecting the mass. That is always ##m_f## moving at ##v_e## and its kinetic energy in that frame is always ## K_r = \frac{1}{2}m_f \large v_e ^2 ## and linearly proportional to ##m_f##. The rocket velocity is by definition zero so that is the total kinetic energy. Also, one could look in the reference frame of the recoiling fuel mass where the total energy is ## K_f = \frac{1}{2}m_r \large v_e ^2 ## since ##v_f## is zero in that frame.
Next, we look at the velocities in the instantaneous center of mass reference frame. That is where you calculated;
$$K_{cm} = \frac{1}{2}m_f \large v_e ^2 \frac{m-m_f}{m}$$ or
$$K_{cm} = \frac{1}{2}\large v_e ^2 \frac{m_rm_f}{m_r + m_f}$$ I agree that is the total kinetic energy in the center of mass reference frame.
for ##m_r \gg m_f## we are essentially in the rocket frame ##R##
$$K_{cm} ≈ \frac{1}{2}\large m_f \large v_e ^2$$
and for ##m_r = m_f##
$$K_{cm} = \frac{1}{2}\frac{1}{2}\large m_f \large v_e ^2$$
The point is that in general ##K_r ≠ K_f ≠ K_{cm}## but all are valid.It is in the rocket frame however where the fuel is burned or just ejected that we measure the chemical or stored energy released. A real rocket is all about constant thrust or power which is mass flow times velocity or ## \large m_f' ⋅ v_e## where the energy per unit mass is ≈constant and ##v_e## is fixed by reaction chemistry.
The original OP question was simply;
Why does ejecting fuel over time cause rocket to go faster than ejecting all at once?
I think as long as we fix the relative velocity of the fuel mass wrt the rocket, as rockets do, we are keeping with the spirit of the question and we both saw that the rocket does better with a slow burn than throwing off a single lump.