GATE AE 2020 • Q20 MCQ • 1 Mark Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Space Dynamics & Orbital Mechanics

Subtopic: Rocket Trajectory & Staging

Question Statement:

Burnout velocity of a space vehicle in a circular orbit at an angle $5^\circ$ above the local horizon around earth is 13.5 km/s. The velocity vector orientation indicates the orbit trajectory nature is:

Options:

  • (A) hyperbolic trajectory
  • (B) parabolic trajectory
  • (C) elliptical trajectory
  • (D) circular trajectory

Official Answer Key:

A

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2020 Q.20)** **1. Intuitive Physical Setup & Given Parameters**: - Burnout speed: $V = 13.5\text{ km/s}$. - Escape speed from low Earth orbit: $v_{esc} = \sqrt{\frac{2\mu}{R_E}} \approx \sqrt{2} \times 7.9\text{ km/s} \approx 11.2\text{ km/s}$. **2. Step-by-Step Astrodynamics Derivation**: 1. **Specific Orbital Energy ($\mathcal{E}$)**: \mathcal{E} = \frac{V^2}{2} - \frac{\mu}{r} 2. **Comparison with Escape Speed**: Since $V = 13.5\text{ km/s} > v_{esc} \approx 11.2\text{ km/s}$: \mathcal{E} > 0 3. **Orbit Classification by Specific Energy**: - $\mathcal{E} < 0$: Elliptic / circular closed orbit (eccentricity $e < 1$). - $\mathcal{E} = 0$: Parabolic escape trajectory ($e = 1$). - $\mathcal{E} > 0$: **Hyperbolic escape trajectory** ($e > 1$). 4. Because the burnout velocity exceeds escape velocity, the vehicle escapes Earth's gravitational sphere along a hyperbolic trajectory. **3. Comprehensive Options & Distractor Analysis**: - **(A)** hyperbolic trajectory: ✅ **CORRECT**. $V > v_{esc} \implies \mathcal{E} > 0 \implies e > 1$. - **(B)** parabolic trajectory: ❌ Incorrect. Requires $V = v_{esc}$ exactly. - **(C)** elliptical trajectory: ❌ Incorrect. Requires $V < v_{esc}$. - **(D)** circular trajectory: ❌ Incorrect. Requires $V = v_c \approx 7.9\text{ km/s}$ and zero flight path angle. **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **A** - **Exam Rule**: $V > \sqrt{2\mu/r} \implies$ Hyperbolic escape orbit ($e > 1, \mathcal{E} > 0$).
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