A single-stage sounding rocket has an initial liftoff mass $m_0 = 1200\text{ kg}$ and a burnout mass $m_f = 400\text{ kg}$. The solid motor produces an effective exhaust velocity $C = 2500\text{ m/s}$ in gravity-free, drag-free flight. Applying Tsiolkovsky's rocket equation, what is the ideal burnout velocity $\Delta V$ (in m/s) attained by the rocket? (Round off to the nearest integer)

Answer Key:

2747

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Governing Principle:** In the absence of atmospheric aerodynamic drag and gravitational potential loss, the maximum theoretical velocity gain attained by a rocket propellant expulsion is governed by Konstantin Tsiolkovsky's rocket equation. 2. **Stage 2: Mathematical Formulation:** * $\Delta V = C \ln\left(\frac{m_0}{m_f}\right) = I_{sp} g_0 \ln\left(\frac{m_0}{m_f}\right)$. 3. **Stage 3: Mass Ratio Evaluation:** * Mass ratio $R = \frac{m_0}{m_f} = \frac{1200}{400} = 3.0$. * $\ln(3.0) \approx 1.09861229$. 4. **Stage 4: Numerical Calculation:** * $\Delta V = 2500 \times 1.09861229 = 2746.53\text{ m/s}$. * Rounding to the nearest integer yields $2747\text{ m/s}$. 5. **Stage 5: Official Reference:** George P. Sutton and Oscar Biblarz, *Rocket Propulsion Elements*, Chapter 4 (Flight Performance).

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