GATE AE 2025 • Q53 NAT • 2 Marks Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Space Dynamics & Orbital Mechanics

Subtopic: Kepler's Laws & Two-Body Problem

Question Statement:

F and G denote two points on a spacecraft's orbit around a planet, as indicated in the figure. O is the center of the planet, P is the periapsis, and the angles are as indicated in the figure. If OF = 8000 km, OG = 10000 km, $\theta$F = 0 deg, and $\theta$G = 60 deg, the eccentricity of the spacecraft's orbit is ____ (rounded off to two decimal places).

Official Answer Key:

0.665

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2025 Q.53)** **1. Intuitive Physical Setup & Given Parameters**: Spacecraft orbit around a planet with center $O$ and periapsis $P$: - True anomaly at point $F$: $\theta_F = 0^\circ$ (so $F$ is at periapsis $P$, $r_F = r_p = 8000\text{ km}$). - True anomaly at point $G$: $\theta_G = 60^\circ$ ($r_G = 10000\text{ km}$). We need to determine the eccentricity $e$ of the orbit (rounded off to two decimal places). **2. Step-by-Step Orbital Mechanics Derivation**: 1. **Keplerian Orbit Trajectory Equation**: r(\theta) = \frac{p}{1 + e\cos\theta} where $p = a(1 - e^2)$ is the semi-latus rectum. 2. **Apply at Point $F$ ($\theta = 0^\circ$)**: r_F = \frac{p}{1 + e\cos 0^\circ} = \frac{p}{1 + e} = 8000\text{ km} \implies p = 8000(1 + e) 3. **Apply at Point $G$ ($\theta = 60^\circ$)**: r_G = \frac{p}{1 + e\cos 60^\circ} = \frac{p}{1 + 0.5e} = 10000\text{ km} \implies p = 10000(1 + 0.5e) 4. **Equate the Expressions for $p$**: 8000(1 + e) = 10000(1 + 0.5e) Divide by 2000: 4(1 + e) = 5(1 + 0.5e) 4 + 4e = 5 + 2.5e 4e - 2.5e = 5 - 4 1.5e = 1 \implies e = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 **3. Comprehensive Options & Distractor Analysis**: - **Evaluation Criteria**: Numerical Answer Type (NAT). - **Exact Analytical Result**: $e = 2/3 \approx 0.667$. - **Rounding Requirement**: Two decimal places $\implies \mathbf{0.67}$ (nominal **0.665**). **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **0.665** (Acceptable Range: **0.65 to 0.68**) - **Exam Takeaway**: Solve orbit equation at two known true anomalies: $\frac{r_G}{r_F} = \frac{1 + e}{1 + e\cos\theta_G}$. For $10/8$ and $\cos 60^\circ = 0.5$, $e = \frac{10/8 - 1}{1 - 0.5(10/8)} = \frac{0.25}{0.375} = \frac{2}{3} \approx 0.67$.
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