A satellite moves in an elliptical orbit of eccentricity $e = 0.50$. When the eccentric anomaly is $E = 60^\circ$ ($E = \pi/3\text{ rad}$), the true anomaly $\theta$ of the satellite is:

(A) $\theta = 90^\circ$

(B) $\theta = 60^\circ$

(C) $\theta = 45^\circ$

(D) $\theta = 120^\circ$

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles The relation between true anomaly $\theta$ and eccentric anomaly $E$ in Keplerian orbital mechanics is: $$\tan\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + e}{1 - e}} \tan\left(\frac{E}{2}\right)$$ ### 2. Analytical Formulation & Calculation Given: - Eccentricity $e = 0.50$ - $E = 60^\circ \implies \frac{E}{2} = 30^\circ$ 1. Square root term: $$\sqrt{\frac{1 + e}{1 - e}} = \sqrt{\frac{1 + 0.50}{1 - 0.50}} = \sqrt{\frac{1.50}{0.50}} = \sqrt{3}$$ 2. Tangent term: $$\tan(30^\circ) = \frac{1}{\sqrt{3}}$$ 3. Evaluate $\tan(\theta/2)$: $$\tan\left(\frac{\theta}{2}\right) = \sqrt{3} \times \frac{1}{\sqrt{3}} = 1$$ 4. Solve for true anomaly $\theta$: $$\frac{\theta}{2} = \arctan(1) = 45^\circ \implies \theta = 90^\circ$$ ### 3. Distractor Forensics - **Option A ($90^\circ$):** Exactly satisfies the orbital anomaly conversion. - **Option C ($45^\circ$):** Half-angle $\theta/2$, forgot to double. - **Option B ($60^\circ$):** Assumes $\theta = E$ (circular orbit only). ### 4. Exam Trap & Standard Reference - **Trap Alert:** Notice $\theta > E$ in the first half of the orbit ($0 < \theta < 180^\circ$) because orbital speed is faster near periapsis. - **Standard Reference:** Curtis, H.D., *Orbital Mechanics for Engineering Students*, Chapter 3.

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