COM BYU 25

Solution 25: Rocket Efficiency

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, $(v-u)^2 = v^2 + u^2 – 2uv$. $$P_{total} = \dot{m} u v + \frac{1}{2} \dot{m} v^2 + \frac{1}{2} \dot{m} u^2 – \dot{m} u v = \frac{1}{2} \dot{m} (v^2 + u^2)$$

2. Calculation of Efficiency

$$\eta = \frac{P_{thrust}}{P_{total}} = \frac{\dot{m} u v}{\frac{1}{2} \dot{m} (v^2 + u^2)}$$ $$\eta = \frac{2 u v}{u^2 + v^2}$$

Answer:

$$\eta = \frac{2vu}{u^2 + v^2}$$