Velocity after slingshot at Jupiter

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SUMMARY

The discussion focuses on calculating the velocity of the spacecraft Voyager 2 after a gravitational slingshot maneuver around Jupiter. Given the spacecraft's initial speed of 12 km/s and Jupiter's orbital speed of 13 km/s, the analysis applies the conservation of momentum and kinetic energy principles. By treating the mass of Jupiter as significantly larger than that of Voyager 2, the final velocity of the spacecraft can be approximated using the equation V_J(final) + V_J(initial) ≈ 2V_J(initial), leading to a final speed relative to the sun that reflects this interaction.

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  • Knowledge of kinetic energy conservation principles
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  • Basic understanding of gravitational slingshot maneuvers
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Spacecraft voyager 2 (of mass m and speed v relative to the sun) approaches the planet Jupiter (of mass M and speed v_J relative to the sun). The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the sun, after this slingshot encounter, which can be anylized as a collision? Assumer v = 12km/s and v_J = 13km/s (the orbital speed of jupiter). The mass of Jupiter is very much grater than the mass of the spacecraft .

Can someone point me in the right direction?

Thanks
 
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Remove the fancy terminology and focus on collision mechanics.

Apply the conservation of momentum equation.
Apply conservation of kinetic energy.

Replace M as a function of m, such that all the m's in the momentum equation cancel. Do some factoring and cancelling and you can make the approximation that V_J(final) + V_J(initial) \approx 2V_J(initial) since the mass is very very large compared to the space craft.
 

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