Gravity assist mimicking elastic collision

AI Thread Summary
The discussion revolves around calculating the speed of a satellite after an elastic collision with a planet, where the satellite initially travels at 12.3 km/s and the planet at 11.3 km/s in opposite directions. Participants emphasize the conservation of kinetic energy and momentum, noting that the planet's mass is significantly larger, which affects the collision dynamics. One approach involves shifting to the center of momentum frame, allowing for a simplified analysis of the collision as a bounce, where the satellite's final speed is derived from its initial speed plus twice the planet's speed. Ultimately, the calculated speed of the satellite after the interaction is 34.9 km/s. This method highlights the importance of understanding frame shifts in collision problems.
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Homework Statement

A satellite with a mass of 252 kg approaches a large planet at a speed vi,1 = 12.3 km/s. The planet is moving at a speed vi,2 = 11.3 km/s in the opposite direction. The satellite partially orbits the planet and then moves away from the planet in a direction opposite to its original direction (see the figure). If this interaction is assumed to approximate an elastic collision in one dimension, what is the speed of the satellite after the collision?

The answer is 34.9 km/s.

Homework Equations



The Attempt at a Solution



I think the key words here are "elastic collision." Kinetic energy will therefore be conserved, as should momentum.

I don't understand how to approach the problem since I have no information about the final momentums/velocities of either object. I can't imagine the satellite changing the velocity of the planet to any consequential amount, but using the same velocity (11.3 km/s) for its initial and final states will only result in the satellite having the same initial and final velocities (in opposite directions).

The satellite changes direction so I tried

M1V1+M2V2=M1fV1f+M2V2

252(12.3)+M(11.3)=252(-v)+M(11.3)

which again suggests that vf=-12.3

any assistance much appreciated.
 
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When one mass is very (very) much larger than the other (as in planet vs spacecraft ), an elastic collision ends up reflecting the smaller body's velocity (like bouncing off a stationary wall), plus adding its own velocity to the craft.
 
Ahhh... that makes sense!

I've redone the problem without the assumption that the planet's final velocity is the same and here's what I got (after a good deal of algebra):

conservation of energy:

Mp(63.848-Vp2)=126V2-19062.5

and for conservation of momentum:

Mp(11.3-Vp)=252V+3099.6

where Mp=mass of planet, Vp=planet's final velocity and V=satellites final velocity


I can't get the two formulas to cancel the unkowns (Mp and Vp) out. Have any ideas?
 
I must apologize, but I find it hard to follow numbers in a derivation when I haven't done it myself. They don't jump out at me with significance.

I think that my approach to the problem (and I haven't worked it though myself, so be warned!) might be to think of shifting it to the center of momentum frame by adding a suitable velocity to both of the given velocities so that the momenta of the planet and the spacecraft would be equal and opposite (!yes, equal!) Then the result of "collision" in that frame would be a perfect bounce, with their velocities simply being negated (this is a characteristic of an elastic collision where the momenta of the participants is equal and opposite).

Converting back to the original frame of reference would yield the resulting final velocities. The trick then would be in looking at the 'adjustment' velocity that was used to shift to the center of momentum frame. If the mass of the planet is vastly greater than that of the spacecraft , it should simply be the negative of the planet's velocity (so the planet becomes essentially stationary in the center of momentum frame, which makes sense since it has just about all of the momentum). So to shift to the center of momentum frame we basically subtract the velocity of the planet from each body. Note that because the spacecraft is traveling in the opposite direction to the planet, it adds the planets speed to that of the spacecraft .

When the 'bounce' occurs, the spacecraft retreats at its new improved speed.

Shifting back to the original frame of reference, the planet resumes (essentially) its original speed and the spacecraft speed gets adjusted accordingly with another addition of the speed of the planet.

So in the end, the spacecraft 's seed is equal to its original speed plus twice that of the planet!
 
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