The specific impulse of a rocket engine is 3000 N s/kg. The mass of the rocket at burnout is 1000 kg. If the velocity increment $\Delta V$ achieved during the burn is 1500 m/s in gravity-free, drag-free space, the initial mass of the rocket (in kg) divided by 1000 is _________ (round off to 2 decimal places).
Official Answer Key:
1.65
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2019 Q.56 / AE Q.46)**
**1. Intuitive Physical Setup & Given Parameters**:
- Effective exhaust velocity: $c = I_{sp} = 3000\text{ N s/kg} = 3000\text{ m/s}$
- Burnout mass: $m_b = 1000\text{ kg}$
- Velocity increment: $\Delta V = 1500\text{ m/s}$
**2. Step-by-Step Mathematical Derivation**:
1. **Tsiolkovsky Rocket Equation**:
$\Delta V = c \ln\left( \frac{m_0}{m_b} \right)$
2. **Solving for Initial Mass $m_0$**:
$\frac{\Delta V}{c} = \ln\left( \frac{m_0}{m_b} \right) \implies \frac{m_0}{m_b} = e^{\Delta V / c}$
$\frac{\Delta V}{c} = \frac{1500}{3000} = 0.5$
$\frac{m_0}{m_b} = e^{0.5} \approx 1.64872$
3. **Evaluating $m_0$**:
$m_0 = 1000 \times 1.64872 = 1648.72\text{ kg}$
Divided by 1000:
$\frac{m_0}{1000} \approx 1.65$
This lies right in the official acceptable range **[1.60, 1.65]**.
**3. Comprehensive Options & Distractor Analysis**:
- **Question Type**: Numerical Answer Type (NAT).
- **Official Range**: **1.60 to 1.65**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer**: **1.65** (Acceptable Range: **1.60 to 1.65**)
- **Key Formula**: $m_0 = m_b e^{\Delta V / c}$.
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