GATE AE 2026 • Q36 MCQ • 2 Marks Flight Mechanics & Space Dynamics

Flight Mechanics & Space Dynamics

Space Dynamics & Orbital Mechanics

Subtopic: Kepler's Laws & Two-Body Problem

Question Statement:

An earth satellite has the instantaneous position vector r⃗ and velocity vector v⃗ as given below. Here p̂ and q̂ denote the unit vectors along the x and y axes of the perifocal frame, respectively. Assume that the value of gravitational parameter is 398600 km3/s^2. Which one of the following trajectories does the satellite follow? r⃗ = (8000p̂ + 9000q̂) km and v⃗ = (-6p̂ + 6q̂) km/s

Options:

  • (A) Circle
  • (B) Hyperbola
  • (C) Parabola
  • (D) Straight line

Official Answer Key:

B

Step-by-Step Solution & Explanation

### **Detailed Step-by-Step Solution (GATE AE 2026 Q.36)** **1. Intuitive Physical Setup & Given Parameters**: An Earth satellite has instantaneous position vector $\vec{r}$ and velocity vector $\vec{v}$ in the perifocal frame: - $\vec{r} = (8000\hat{p} + 9000\hat{q})\text{ km}$ - $\vec{v} = (-6\hat{p} + 6\hat{q})\text{ km/s}$ - Earth gravitational parameter: $\mu = 398600\text{ km}^3/\text{s}^2$ We need to determine the trajectory type (Circle, Hyperbola, Parabola, or Straight line). **2. Step-by-Step Orbital Mechanics Derivation**: 1. **Specific Mechanical Energy Relation**: The orbit shape (conic section) is uniquely determined by the satellite's specific mechanical energy $\mathcal{E}$: \mathcal{E} = \frac{v^2}{2} - \frac{\mu}{r} - If $\mathcal{E} < 0$: Elliptic / Circular orbit - If $\mathcal{E} = 0$: Parabolic escape trajectory - If $\mathcal{E} > 0$: Hyperbolic escape trajectory 2. **Compute Magnitude of Position Vector $r$**: r^2 = 8000^2 + 9000^2 = 64 \times 10^6 + 81 \times 10^6 = 145 \times 10^6\text{ km}^2 r = \sqrt{145 \times 10^6} \approx 12041.59\text{ km} 3. **Compute Magnitude of Velocity Vector $v$**: v^2 = (-6)^2 + 6^2 = 36 + 36 = 72\text{ km}^2/\text{s}^2 v = \sqrt{72} \approx 8.485\text{ km/s} 4. **Evaluate Specific Energy $\mathcal{E}$**: \text{Kinetic energy term: } \frac{v^2}{2} = \frac{72}{2} = 36\text{ km}^2/\text{s}^2 \text{Potential energy term: } \frac{\mu}{r} = \frac{398600}{12041.59} \approx 33.102\text{ km}^2/\text{s}^2 \mathcal{E} = 36 - 33.102 = +2.898\text{ km}^2/\text{s}^2 > 0 5. Since $\mathcal{E} > 0$ and the angular momentum $\vec{h} = \vec{r} \times \vec{v} \neq 0$ ($h_z = 8000(6) - 9000(-6) = 48000 + 54000 = 102000\text{ km}^2/\text{s} \neq 0$), the orbit is non-rectilinear with eccentricity $e = \sqrt{1 + \frac{2\mathcal{E}h^2}{\mu^2}} > 1$, which represents a **Hyperbola**. **3. Comprehensive Options & Distractor Analysis**: - **(A) Circle**: ❌ **INCORRECT**. Requires $e = 0$ and $v = \sqrt{\mu/r} \approx 5.75\text{ km/s}$, whereas actual speed is $8.49\text{ km/s}$. - **(B) Hyperbola**: ✅ **CORRECT**. $\mathcal{E} > 0$ confirms speed exceeds parabolic escape speed ($v_{\text{esc}} = \sqrt{2\mu/r} \approx 8.14\text{ km/s} < 8.49\text{ km/s}$). - **(C) Parabola**: ❌ **INCORRECT**. Requires $\mathcal{E} = 0$ ($v = v_{\text{esc}}$). - **(D) Straight line**: ❌ **INCORRECT**. Specific angular momentum $\vec{h} \neq 0$, so the motion is planar and curved. **4. Official Master Key & Exam Takeaway**: - **Official Answer Key**: **B** - **Exam Takeaway**: Compute $\frac{v^2}{2}$ vs $\frac{\mu}{r}$. When $\frac{v^2}{2} > \frac{\mu}{r}$, the specific energy is positive, proving the trajectory is an unclosed hyperbolic flyby.
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