# A question about $I_{s}=\frac{F}{\dot{w}}=\frac{c}{g}$ with an example

I'm new to rocket propulsion elements and I'm reading Rocket Propulsion Elements by George P. Sutton, 4th Edition.

In the second chapter, it introduced me the following expression,

$$I_{s}=\frac{F}{\dot{w}}=\frac{c}{g}$$

$$I_{s}: \text{Specific impulse}$$

$$F: \text{Thrust}$$

$$\dot{w}: \text{Weight flow rate}$$

$$c: \text{Effective exhaust velocity}$$

$$g: \text{Gravitational constant}$$

Now, I created an example myself. I considered that I have an amateur rocket engine that contains $$1kg$$ of propellant and I ignite this amateur rocket engine on the ground of the Earth (so, $$g$$ is basically $$~9.8m/s^2$$). So the initial weight of the propellant is $$w_{i}=m_{i}g=1\times 9.8=9.8 kg.m/s^2$$. I consider that the burn time of the propellant is $$0.3$$ seconds. So, the average weight flow rate is,

$$\frac{\Delta{w}}{\Delta{t}}=\frac{w_{f}-w_{i}}{t_{f}-t_{i}}=\frac{0-9.8}{0.3}=32,66kg.m/s^3$$

I give $$F=50N$$, so I can find the $$I_{s}$$ of my amateur rocket engine.

$$I_{s}=\frac{F}{\Delta{w}/ \Delta{t}}=\frac{50}{32,66}=0,153s$$

Now, I want to find out the effective velocity of my amateur rocket engine,

$$I_{s}=\frac{c}{g}\Rightarrow c=I_{s}g=0,153\times 9.8=1,5 m/s$$

Are my calculation correct and logic correct?

• Missed this, but I'm glad you got a copy of the 4th edition. Commented Feb 17, 2020 at 2:16
• Yeah, when I saw the book on my university's library i was very happy! I have never thought that I would find the book at the library :P Commented Feb 18, 2020 at 16:47