A satellite is in a circular orbit around Earth with a speed $V_0 = 7.5\text{ km/s}$. A single impulsive maneuver is performed to change the inclination of the orbit by an angle $\Delta i = 60^\circ$ without altering the orbital radius or speed. The characteristic velocity change $\Delta V$ required for this plane change is:

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### 1. Strategy & Theoretical Principles For a pure orbital plane change without changing orbital radius or orbital speed ($V_1 = V_2 = V_0$): From the velocity vector triangle (cosine rule): $$\Delta V = 2 V_0 \sin\left(\frac{\Delta i}{2}\right)$$ ### 2. Analytical Formulation & Calculation Given: - $V_0 = 7.5\text{ km/s}$ - $\Delta i = 60^\circ \implies \frac{\Delta i}{2} = 30^\circ$ $$\sin(30^\circ) = 0.5$$ $$\Delta V = 2 \times 7.5 \times 0.5 = 7.5\text{ km/s}$$ ### 3. Distractor Forensics - Using $\sin(60^\circ) = 0.866$ without halving angle gives $\Delta V = 12.99\text{ km/s}$. ### 4. Exam Trap & Standard Reference - **Trap Alert:** Orbital plane changes are extremely fuel-expensive; a $60^\circ$ plane change costs a $\Delta V$ exactly equal to the entire circular velocity $V_0$. - **Standard Reference:** Curtis, H.D., *Orbital Mechanics for Engineering Students*, Chapter 6.

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