Hi DH,
Thank you for the very well-worked and informative post. I think it is clear that conservation of momentum is (1) more fundamental than conservation of energy and (2) more general than f=ma. I really see no reason to bring energy considerations into the rocket equations, particularly for teaching purposes.
However, for the purpose of this thread and to demonstrate that energy is conserved I thought it would be useful to expand on your post and include KE terms.
At the beginning of the interval, the rocket has KE
KE_r(t) = 1/2 (m_r(t))(\vec v_r(t))^2
At the end of the interval the KE of the rocket is
KE_r(t+\Delta t) = 1/2 (m_r(t)-\dot m_f(t)\,\Delta t)(\vec v_r(t)+\Delta \vec v_r(t))^2
The KE of the exhaust is
KE_e(t+\Delta t) = 1/2 (\dot m_f(t)\,\Delta t)(\vec v_r(t)+\vec v_e(t))^2
So the change in KE for the rocket and exhaust system is
\Delta KE(t+\Delta t)) = (KE_e(t+\Delta t)+KE_r(t+\Delta t))-KE_r(t)
By solving the conservation of momentum expression we obtain
\Delta \vec v_r(t) = - \, \frac {\Delta t \,\vec v_e(t)^2 \,\dot m_f(t)\,}{m_r(t)-\Delta t \,\dot m_f(t)\,}
Substituting this into the KE expression and simplifying we obtain
\Delta KE(t+\Delta t) = \frac {\Delta t \, m_r(t) \,\vec v_e(t)^2 \,\dot m_f(t)\,}{2\, m_r(t) - 2 \,\Delta t \,\dot m_f(t)\,}
Dividing by \Delta t and taking the limit \Delta t \to 0,
\frac {dKE(t)}{dt} = \frac {1}{2}\,\vec v_e(t)^2 \,\dot m_f(t)
So, in the end the expression for the change in KE of the whole system (rocket and exhaust) is not a function of the rocket's velocity. It is, in fact, exactly equal to the KE of the exhaust in the rocket's comoving frame, which should have been expected from the beginning although it surprised me. This change in KE is then equal to the useable chemical energy in the fuel (chemical energy minus engine ineffeciencies such as waste heat), so energy is conserved.
Note, however, that although energy is conserved, energy conservation cannot be used by itself to solve this problem. Without conservation of momentum there is no way to determine \Delta \vec v_r(t). In other words, we know that some of the chemical energy goes into kinetic energy of the rocket and rest goes into kinetic energy of the exhaust, but without conservation of momentum there is no way to determine what fraction goes to each. So, since the problem can be solved by conservation of momentum alone, and since the problem cannot be solved by conservation of energy alone, the problem should be approached strictly with a conservation of momentum approach. Energy conservation can be demonstrated, but not used.
