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/s2. 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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