Let's break down the problem step-by-step as requested. --- ## **Step 1: Understanding theby Frame and Velocity** **:** - spacecraft's velocity the flyby (in the flyby frame) is: \(\vec{v}_{sc,in} = 25 \hat{u}_x + 2 \hat{u}_z \) km/s - Periapsis radius = 500 km. **Explanation:** - The flyby frame uses \(\hat{u}_x\) (along incoming asymptote) and \(\hat{u}_z\) (perpendicular, in the plane of the flyby). - The incoming velocity vector components are 25 km/s along \(\hat{u}_x\) and 2 km/s along \(\hat{u}_z\). --- ## **Step 2: Outgoing Velocity for Each Turn Direction** ### **2.1 Counter-Clockwise Turn** - **During a hyperbolic flyby:** The magnitude of the velocity relative to Mars (\(v_{\infty}\)) remains constant, but its direction changes (the vector is rotated by the turn angle). - **Counter-clockwise turn:** Rotate the incoming v-infinity vector by angle \(\delta\) counter-clockwise in the \((\hat{u}_x, \hat{u}_z)\) plane. \[ |\vec{v}_{\infty}| = \sqrt{25^2 + 2^2} = \sqrt{625 + 4} = \sqrt{629} \approx 25.1 \text{ km/s} \] - **Outgoing velocity (\(\vec{v}_{sc,out}\))** for a counter-clockwise turn: - The vector rotates by angle \(2\delta\) (total deflection angle). - The new vector: \[ \vec{v}_{sc,out} = 25 \cos(2\delta) \hat{u}_x + 2 \cos(2\delta) \hat{u}_z \] - More simply: after a counter-clockwise turn, the \(\hat{u}_z\) component changes sign (if the turn is 180°), so: \[ \vec{v}_{sc,out} = 25 \hat{u}_x - 2 \hat{u}_z \] **Explanation:** The counter-clockwise turn reverses the \(\hat{u}_z\) component, keeping the magnitude the same. This is called a "leading-edge" flyby because after the turn, the outgoing velocity is rotated in the direction of Mars’ motion. --- ### **2.2 Clockwise Turn** - **Outgoing velocity (\(\vec{v}_{sc,out}\))** for a clockwise turn: - The vector rotates by \(-2\delta\). - The new vector: \[ \vec{v}_{sc,out} = 25 \hat{u}_x + 2 \hat{u}_z \] (No change in direction; same as incoming). **Explanation:** For a clockwise turn, the outgoing vector remains the same as the incoming one. This would be a "trailing-edge" flyby, as the outgoing vector points away from the direction of Mars’ motion. --- ## **Final Answers** ### **2.1 Counter-clockwise turn:** \[ \vec{v}_{sc,out} = 25 \hat{u}_x - 2 \hat{u}_z \, \text{km/s} \] - **Type:** Leading-edge flyby. ### **2.2 Clockwise turn:** \[ \vec{v}_{sc,out} = 25 \hat{u}_x + 2 \hat{u}_z \, \text{km/s} \] - **Type:** Trailing-edge flyby. --- **Summary Table:** | Turn Direction | \(\vec{v}_{sc,out}\) (km/s) | Flyby Type | |---------------------|----------------------------------|-----------------| | Counter-clockwise | \(25 \hat{u}_x - 2 \hat{u}_z\) | Leading-edge | | Clockwise | \(25 \hat{u}_x + 2 \hat{u}_z\) | Trailing-edge |

Let's solve the problem step-by-step, providing explanations and final answers for each part. --- # Problem Breakdown ## **Given Data** - Incoming velocity of spacecraft: \[ \vec{v}_{sc,in} = 25 \hat{u}_x + 2 \hat{u}_z \text{ km/s} \] - Radius of periapsis: \( r_p = 5000 \text{ km} \) --- ## **Step 1: Calculate the Magnitude of Incoming Velocity** ### **Magnitude Calculation** To find the magnitude of the incoming velocity vector: \[ |\vec{v}_{\infty}| = \sqrt{(25)^2 + (2)^2} = \sqrt{625 + 4} = \sqrt{629} \approx 25.1 \text{ km/s} \] ### **Explanation** This calculation gives us the total speed of the spacecraft as it approaches Mars. The components along \(\hat{u}_x\) and \(\hat{u}_z\) contribute to this total speed. --- ## **Step 2: Analyze Velocity After Flyby** ### **2.1 Counter-Clockwise Turn** For a counter-clockwise turn, the outgoing velocity vector will be represented as: \[ \vec{v}_{sc,out} = 25 \hat{u}_x - 2 \hat{u}_z \text{ km/s} \] ### **Outgoing Velocity Calculation** The outgoing velocity is derived by rotating the incoming velocity vector by an angle in a counter-clockwise direction. The component along \(\hat{u}_z\) is negated, indicating a change in direction. ### **Flyby Type** - **Type:** Leading-edge flyby - **Explanation:** In a leading-edge flyby, the spacecraft's outgoing velocity vector points in the direction of Mars' motion. --- ### **2.2 Clockwise Turn** For a clockwise turn, the outgoing velocity vector remains: \[ \vec{v}_{sc,out} = 25 \hat{u}_x + 2 \hat{u}_z \text{ km/s} \] ### **Outgoing Velocity Calculation** In this case, the direction of the \(\hat{u}_z\) component does not change, which means the spacecraft exits the flyby maintaining its original velocity vector direction. ### **Flyby Type** - **Type:** Trailing-edge flyby - **Explanation:** In a trailing-edge flyby, the spacecraft's outgoing velocity vector points opposite to Mars' motion. --- ## **Final Answers** ### **2.1 Counter-Clockwise Turn:** \[ \vec{v}_{sc,out} = 25 \hat{u}_x - 2 \hat{u}_z \text{ km/s} \] - **Flyby Type:** Leading-edge ### **2.2 Clockwise Turn:** \[ \vec{v}_{sc,out} = 25 \hat{u}_x + 2 \hat{u}_z \text{ km/s} \] - **Flyby Type:** Trailing-edge --- ## **Summary Table** | Turn Direction | \(\vec{v}_{sc,out}\) (km/s) | Flyby Type | |---------------------|-----------------------------------|-----------------| | Counter-clockwise | \(25 \hat{u}_x - 2 \hat{u}_z\) | Leading-edge | | Clockwise | \(25 \hat{u}_x + 2 \hat{u}_z\) | Trailing-edge | This structured approach ensures a clear understanding of the spacecraft's behavior during the flyby of Mars.