Consider a launch vehicle of mass 10 tons being launched vertically. The vehicle
has 8 tons of propellant, which burns completely at a constant rate over 50 s. If the
engine specific impulse is 250 s, and the acceleration due to gravity at sea level is
g0, the acceleration experienced by the vehicle at lift-off is ________.
Options:
(A) g0
(B) 2g0
(C) 3g0
(D) 4g0
Official Answer Key:
C
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.41)**
**1. Intuitive Physical Setup & Given Parameters**:
- Total initial launch vehicle mass: $m_0 = 10\text{ tons} = 10,000\text{ kg}$
- Propellant mass: $m_p = 8\text{ tons} = 8,000\text{ kg}$
- Burn time: $t_b = 50\text{ s}$ (constant mass burn rate $\dot{m}$)
- Specific impulse: $I_{\text{sp}} = 250\text{ s}$
- Sea-level gravitational acceleration: $g_0$
We need to find the net vertical acceleration experienced by the vehicle at **lift-off** ($t = 0$).
**2. Step-by-Step Rocket Flight Mechanics Derivation**:
1. **Compute Propellant Mass Flow Rate $\dot{m}$**:
Since the propellant burns uniformly over 50 s:
$\dot{m} = \frac{m_p}{t_b} = \frac{8\text{ tons}}{50\text{ s}} = \frac{8000\text{ kg}}{50\text{ s}} = 160\text{ kg/s}$
2. **Compute Engine Thrust $T$**:
Thrust produced by the rocket engine is:
$T = \dot{m} g_0 I_{\text{sp}} = 160\text{ kg/s} \times g_0 \times 250\text{ s} = 40,000 g_0\text{ N} = 40\text{ tons} \times g_0$
3. **Formulate Equation of Motion at Lift-Off ($t = 0$)**:
The vertical forces acting on the rocket at launch are upward thrust $T$ and downward gravitational weight $W = m_0 g_0$:
$m_0 a = T - m_0 g_0$
4. **Calculate Lift-Off Acceleration $a$**:
$a = \frac{T}{m_0} - g_0 = \frac{40,000 g_0}{10,000} - g_0 = 4g_0 - g_0 = 3g_0$
**3. Comprehensive Options & Distractor Analysis**:
- **(A) g0**: ❌ **INCORRECT**. Corresponds to thrust-to-weight ratio $T/W = 2$.
- **(B) 2g0**: ❌ **INCORRECT**. Corresponds to $T/W = 3$.
- **(C) 3g0**: ✅ **CORRECT**. $T/m_0 = 4g_0$, so net upward acceleration is $4g_0 - g_0 = 3g_0$.
- **(D) 4g0**: ❌ **INCORRECT**. This is the gross thrust acceleration $T/m_0$, failing to subtract the downward gravitational pull $-g_0$.
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **C**
- **Exam Takeaway**: Never forget to subtract gravity from vertical launch thrust: $a_{\text{liftoff}} = \left(\frac{T}{m_0}\right) - g_0 = \left(\frac{m_p I_{\text{sp}}}{m_0 t_b} - 1\right)g_0 = \left(\frac{8 \times 250}{10 \times 50} - 1\right)g_0 = (4 - 1)g_0 = 3g_0$.
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