A space probe is in a circular orbit around a planet with orbital speed $V_c = 4.0\text{ km/s}$. What is the escape velocity $V_{esc}$ (in km/s) from this exact orbital location? (Round off to two decimal places)

Answer Key:

5.66

### 5-Stage Derivation & Analytical Solution: 1. **Stage 1: Energy Conservation:** The specific mechanical energy for a parabolic escape trajectory is $\mathcal{E} = \frac{V_{esc}^2}{2} - \frac{\mu}{r} = 0$. 2. **Stage 2: Escape Speed Equation:** $V_{esc} = \sqrt{\frac{2\mu}{r}}$. 3. **Stage 3: Circular Speed Equation:** $V_c = \sqrt{\frac{\mu}{r}}$. 4. **Stage 4: Fundamental Ratio:** * $\frac{V_{esc}}{V_c} = \frac{\sqrt{2\mu/r}}{\sqrt{\mu/r}} = \sqrt{2}$. * $V_{esc} = \sqrt{2} \times 4.0\text{ km/s} = 1.414214 \times 4.0 = 5.65685\text{ km/s} \approx 5.66\text{ km/s}$. 5. **Stage 5: Official Reference:** Howard D. Curtis, *Orbital Mechanics for Engineering Students*, Chapter 2.

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