Physics Solution: Propulsion Efficiency
1. Definitions
Let $v$ be the instantaneous speed of the rocket. Let $u$ be the speed of ejected gases relative to the rocket. Mass flow rate $\dot{m} = \frac{dm}{dt}$.
Useful Power ($P_{thrust}$): The power generated by the thrust force propelling the rocket.
$$\text{Thrust } F = \dot{m} u$$ $$P_{thrust} = F \cdot v = \dot{m} u v$$Wasted Power ($P_{waste}$): The kinetic energy carried away by the ejected gases (which is lost to the system).
Velocity of gases relative to ground: $v_{gas} = v – u$ (assuming ejection opposite to motion). Note: Since we are dealing with energies, the term is $(v-u)^2$.
$$P_{waste} = \frac{1}{2} \dot{m} v_{gas}^2 = \frac{1}{2} \dot{m} (v – u)^2$$ $$P_{total} = P_{thrust} + P_{waste}$$ $$P_{total} = \dot{m} u v + \frac{1}{2} \dot{m} (v – u)^2$$ Wait, $(v-u)^2 = v^2 + u^2 – 2uv$. $$P_{total} = \dot{m} u v + \frac{1}{2} \dot{m} v^2 + \frac{1}{2} \dot{m} u^2 – \dot{m} u v = \frac{1}{2} \dot{m} (v^2 + u^2)$$