A satellite is in a circular Low Earth Orbit (LEO) of radius $r_1 = 6{,}700\text{ km}$. It is transferred to a higher circular orbit of radius $r_2 = 26{,}800\text{ km}$ ($r_2 = 4 r_1$) via a coplanar Hohmann transfer ellipse. The standard gravitational parameter of Earth is $\mu = 3.986 \times 10^{14}\text{ m}^3/\text{s}^2$ ($398{,}600\text{ km}^3/\text{s}^2$). The total characteristic velocity change $\Delta V_{\text{total}} = \Delta V_1 + \Delta V_2$ required for the two impulsive burns (in $\text{km/s}$, rounded to two decimal places) is:

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### 1. Strategy & Theoretical Principles A Hohmann transfer utilizes a semi-elliptic orbit tangent to both circular orbits: - Initial circular velocity at $r_1$: $V_{c1} = \sqrt{\frac{\mu}{r_1}}$ - Semi-major axis of transfer ellipse: $a_{tx} = \frac{r_1 + r_2}{2}$ - Periapsis velocity of transfer orbit: $V_{tx,1} = \sqrt{\mu \left(\frac{2}{r_1} - \frac{1}{a_{tx}}\right)}$ - Apoapsis velocity of transfer orbit: $V_{tx,2} = \sqrt{\mu \left(\frac{2}{r_2} - \frac{1}{a_{tx}}\right)}$ - Final circular velocity at $r_2$: $V_{c2} = \sqrt{\frac{\mu}{r_2}}$ - Burn impulses: $\Delta V_1 = V_{tx,1} - V_{c1}$, and $\Delta V_2 = V_{c2} - V_{tx,2}$. ### 2. Analytical Formulation Given $r_2 / r_1 = 4$: $$a_{tx} = \frac{r_1 + 4r_1}{2} = 2.5 r_1$$ Circular velocities: $$V_{c1} = \sqrt{\frac{398{,}600}{6{,}700}} = \sqrt{59.4925} \approx 7.713\text{ km/s}$$ $$V_{c2} = \sqrt{\frac{398{,}600}{26{,}800}} = \frac{V_{c1}}{\sqrt{4}} = \frac{7.713}{2} \approx 3.857\text{ km/s}$$ ### 3. Step-by-Step Calculation Transfer orbit velocities: $$V_{tx,1} = \sqrt{398{,}600 \times \left(\frac{2}{6{,}700} - \frac{1}{2.5 \times 6{,}700}\right)} = \sqrt{\frac{398{,}600}{6{,}700} \times \left(2 - \frac{1}{2.5}\right)}$$ $$V_{tx,1} = 7.713 \times \sqrt{2 - 0.4} = 7.713 \times \sqrt{1.6} = 7.713 \times 1.2649 = 9.756\text{ km/s}$$ First burn impulse: $$\Delta V_1 = 9.756 - 7.713 = 2.043\text{ km/s}$$ Transfer apoapsis velocity: $$V_{tx,2} = \sqrt{\frac{398{,}600}{6{,}700} \times \left(\frac{2}{4} - \frac{1}{2.5}\right)} = 7.713 \times \sqrt{0.5 - 0.4} = 7.713 \times \sqrt{0.1} = 7.713 \times 0.3162 = 2.439\text{ km/s}$$ Second burn impulse: $$\Delta V_2 = V_{c2} - V_{tx,2} = 3.857 - 2.439 = 1.418\text{ km/s}$$ Total $\Delta V$: $$\Delta V_{\text{total}} = \Delta V_1 + \Delta V_2 = 2.043 + 1.418 = 3.461\text{ km/s}$$ *Correction in manual check*: Total is $2.043 + 1.418 = 3.46\text{ km/s}$. Let's verify: $$V_{tx,2} = V_{tx,1} \times \frac{r_1}{r_2} = 9.756 \times \frac{1}{4} = 2.439\text{ km/s}$$. Correct (angular momentum conservation $r_1 V_1 = r_2 V_2$). Then $\Delta V_2 = 3.857 - 2.439 = 1.418\text{ km/s}$. Total $\Delta V = 2.043 + 1.418 = 3.46\text{ km/s}$. ### 4. Distractor Forensics - Notice the tolerance range must match $3.46\text{ km/s}$. Let's set correct answer to 3.46 with tolerance 3.42 to 3.50. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always verify angular momentum conservation ($h = r_p V_p = r_a V_a$) for the transfer ellipse. - **Standard Reference:** Curtis, H.D., *Orbital Mechanics for Engineering Students*, 3rd Edition, Elsevier, Chapter 6 (Orbital Maneuvers).

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