An ideal two-stage rocket has identical specific impulse and structural coefficient
for its two stages. For an optimized rocket, the two stages have identical payload
ratio as well. The payload is 2 tons and the initial mass of the rocket is 200 tons.
The mass of the second stage of the rocket (including the final payload mass) is
____ tons.
Options:
(A) 100
(B) 10
(C) 20
(D) 50
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2025 Q.32)**
**1. Intuitive Physical Setup & Given Parameters**:
An ideal two-stage rocket:
- Identical specific impulse ($I_{\text{sp}1} = I_{\text{sp}2}$) and structural coefficient ($\epsilon_1 = \epsilon_2$).
- Identical stage payload ratios ($\lambda_1 = \lambda_2 = \lambda$).
- Final payload mass: $m_{\text{payload}} = 2\text{ tons}$.
- Total initial launch mass of the rocket: $m_{01} = 200\text{ tons}$.
We need to determine the total mass of the second stage $m_{02}$ (including the final payload mass) in tons.
**2. Step-by-Step Rocket Staging Derivation**:
1. **Definition of Payload Ratio ($\lambda_i$)**:
For stage $i$, the payload ratio is defined as:
$\lambda_i = \frac{m_{\text{payload}, i}}{m_{\text{initial}, i} - m_{\text{payload}, i}} \quad \text{or} \quad \lambda_i' = \frac{m_{\text{payload}, i}}{m_{\text{initial}, i}}$
Using the standard staging definition $\lambda_i = \frac{m_{0, i+1}}{m_{0i}}$:
- For Stage 1: Payload is the entire second stage (mass $m_{02}$).
$\lambda_1 = \frac{m_{02}}{m_{01}}$
- For Stage 2: Payload is the final satellite payload ($m_{\text{payload}}$).
$\lambda_2 = \frac{m_{\text{payload}}}{m_{02}}$
2. **Condition of Identical Payload Ratios**:
$\lambda_1 = \lambda_2 \implies \frac{m_{02}}{m_{01}} = \frac{m_{\text{payload}}}{m_{02}}$
3. **Solve for Second Stage Mass $m_{02}$**:
$m_{02}^2 = m_{01} \times m_{\text{payload}}$
$m_{02} = \sqrt{m_{01} \times m_{\text{payload}}}$
4. **Substitute Given Values**:
- $m_{01} = 200\text{ tons}$
- $m_{\text{payload}} = 2\text{ tons}$
$m_{02} = \sqrt{200 \times 2} = \sqrt{400} = 20\text{ tons}$
**3. Comprehensive Options & Distractor Analysis**:
- **(A) 100**: ❌ **INCORRECT**. Arithmetic mean $(200+2)/2 \approx 100$, invalid for optimal staging.
- **(B) 10**: ❌ **INCORRECT**.
- **(C) 20**: ✅ **CORRECT**. Geometric mean of initial mass and final payload: $\sqrt{200 \times 2} = 20\text{ tons}$.
- **(D) 50**: ❌ **INCORRECT**.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: In optimal multi-stage rockets with identical stage properties, stage mass distribution forms a geometric progression: $m_{02} = \sqrt{m_{01} \times m_{\text{pl}}} = \sqrt{200 \times 2} = 20\text{ tons}$.
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