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Spacecraft Voyager 2 (mass m and speed l' relative to the Sun) approaches the planet Jupiter (mass M and speed V relative to the Sun) as shown in Fig. The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this "slingshot" encounter? Assume that v = 12 km/s and V = 13 km/s (the orbital speed of Jupiter), and that this is an elastic collision. The mass of Jupiter is very much greater than the mass of the spacecraft, M >> m. (See "The Slingshot Effect: Explanation and Analogies," by Albert A. Bartlett and Charles W. Hord, The Physics Teacher; November 1985, p. 466.)

Spacecraft Voyager 2 (mass m and speed l' relative to the Sun) approaches the planet Jupiter (mass M and speed V relative to the Sun) as shown in Fig. The spacecraft rounds the
planet and departs in the opposite direction. What is its speed, relative to the Sun, after this "slingshot" encounter? Assume that v = 12 km/s and V = 13 km/s (the orbital speed of Jupiter), and that this is an elastic collision. The mass of Jupiter is very much greater than the mass of the spacecraft, M >> m. (See "The Slingshot Effect: Explanation and Analogies," by Albert A. Bartlett and Charles W. Hord, The Physics Teacher; November 1985, p. 466.)

Grade:11

2 Answers

Kevin Nash
askIITians Faculty 332 Points
8 years ago
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Kozerone
11 Points
6 years ago
Kevin Nash was right.
$v_{2f}\approx2v_{1i}-v_{2i}
But how do we get there?
Well, we have to use conservation of momentum and conservation of kinetic energy to get there, because this is an inelastic collision (KE is conserved). You will get two equations from the two conservation laws. The information $M \gg m$ is useful when you try to solve for $v_{2f}$ using these two equations, because it will allow you to make this expression: $(m/M)^2(v_{2f}+v_{2i})^2$ go to zero.  So you will not need to know what the value of $M$ is, to solve for $v_{2f}$.
 
 

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