A spacecraft departs Earth on an interplanetary trajectory. The hyperbolic excess speed relative to Earth is $V_\infty = 3.5\text{ km/s}$. The gravitational parameter of Earth is $\mu = 398{,}600\text{ km}^3/\text{s}^2$ and radius of low Earth parking orbit is $r_p = 6{,}600\text{ km}$. The trans-planetary injection burn velocity $V_p$ (in $\text{km/s}$) at periapsis of the departure hyperbola is closest to:

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### 1. Strategy & Theoretical Principles From energy conservation on a hyperbolic escape orbit (Vis-Viva energy equation): $$\mathcal{E} = \frac{V_p^2}{2} - \frac{\mu}{r_p} = \frac{V_\infty^2}{2}$$ where: - $V_p$ is the injection velocity at periapsis - $r_p$ is the periapsis radius - $V_\infty$ is the hyperbolic excess speed at infinity. ### 2. Analytical Formulation & Calculation Rearranging directly for periapsis velocity $V_p$: $$V_p = \sqrt{V_\infty^2 + \frac{2\mu}{r_p}} = \sqrt{V_\infty^2 + V_{\text{esc}}^2}$$ where $V_{\text{esc}} = \sqrt{\frac{2\mu}{r_p}}$ is the local parabolic escape speed. Given: - $V_\infty = 3.5\text{ km/s} \implies V_\infty^2 = (3.5)^2 = 12.25\text{ km}^2/\text{s}^2$ - $\mu = 398{,}600\text{ km}^3/\text{s}^2$ - $r_p = 6{,}600\text{ km}$ $$\frac{2\mu}{r_p} = \frac{2 \times 398{,}600}{6{,}600} = \frac{797{,}200}{6{,}600} \approx 120.78788\text{ km}^2/\text{s}^2$$ ### 3. Step-by-Step Calculation $$V_p^2 = 12.25 + 120.78788 = 133.03788\text{ km}^2/\text{s}^2$$ $$V_p = \sqrt{133.03788} \approx 11.5342\text{ km/s} \approx 11.53\text{ km/s}$$ ### 4. Distractor Forensics - Candidate adding speeds directly ($V_p = V_{\text{esc}} + V_\infty = \sqrt{120.79} + 3.5 = 10.99 + 3.5 = 14.49\text{ km/s}$) violates energy conservation. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always add energies ($V_p^2 = V_\infty^2 + V_{\text{esc}}^2$), never velocities linearly. - **Standard Reference:** Curtis, H.D., *Orbital Mechanics for Engineering Students*, Chapter 8 (Interplanetary Trajectories).

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