A monopropellant thruster produces a constant thrust $F = 220\text{ N}$ while consuming propellant at a mass flow rate $\dot{m} = 0.10\text{ kg/s}$. Taking standard sea-level acceleration due to gravity $g_0 = 9.80665\text{ m/s}^2$, the specific impulse $I_{sp}$ (in $\text{seconds}$) of the thruster is closest to:

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### 1. Strategy & Theoretical Principles By definition, specific impulse in seconds is: $$I_{sp} = \frac{F}{\dot{m} g_0} = \frac{c}{g_0}$$ where $c = F / \dot{m}$ is the effective exhaust velocity in $\text{m/s}$. ### 2. Analytical Formulation & Calculation Given: - $F = 220\text{ N}$ - $\dot{m} = 0.10\text{ kg/s}$ - $g_0 = 9.80665\text{ m/s}^2$ 1. Effective exhaust velocity: $$c = \frac{220\text{ N}}{0.10\text{ kg/s}} = 2{,}200\text{ m/s}$$ 2. Specific impulse: $$I_{sp} = \frac{2{,}200}{9.80665} \approx 224.337\text{ seconds} \approx 224.3\text{ s}$$ ### 3. Distractor Forensics - Candidate reporting effective exhaust velocity $c = 2200\text{ m/s}$ confuses units of velocity with seconds. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Always divide effective exhaust velocity $c$ by $g_0 = 9.81\text{ m/s}^2$ to express $I_{sp}$ in seconds. - **Standard Reference:** Sutton, G.P. & Biblarz, O., *Rocket Propulsion Elements*, Chapter 2.

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