Orbital Slingshot and Conservation of Momentum Confusion

[quantumpandabear]

Hello,

I am an undergrad and am in an introductory level astrophysics course. I have a bit of confusion that I didn't know where to get help from so I made an account here. Please let me know if I miss some common etiquette or something... I don't understand how the slingshot maneuver provides a net gain in velocity. I just spoke with my professor about this, but his terminology left me confused. I understand that the target body for the flyby loses some of its momentum and that the satellite executing the flyby takes some of that velocity. That makes sense to me. What I do not understand is why that same interaction isn't undone after the satellite is adjacent to the body. If the satellite arrives with some V_1 and gains some ΔV from the planet in addition to having its velocity vector rotated some angle, why doesn't the satellite decelerate and the planet accelerate back to its previous velocity? For instance, take the classic example of a satellite with no velocity being passed by a planet executing a 180 degree flyby. From the planets perspective the satellite approaches the planet with the planet's velocity, call it V_1. After the planet passes, the satellite has a velocity of 2V_1 in the same direction of the planet rather than just v_1, because part of the planet's momentum was stolen by the satellite. Why does the kinetic energy of the satellite at the start of the flyby not equal the kinetic energy of the satellite after the flyby? It's as if the planet is capable of losing momentum from the satellite but is unable to take that same momentum back from the satellite.

I've read a lot of explanations and many analogies, but nobody seems to explain why the acceleration induced on approach isn't reversed on exit. Why can't the planet that loses some velocity to the satellite steal that velocity back from the satellite as the satellite exit's its sphere of influence? Thanks so much for spending the time to read. I look forward to any replies :)

 

[phyzguy]

If it were just a two-body problem (the satellite and the planet), then your intuition would be correct. That problem is completely symmetrical, and the velocity gain on the way in is lost on the way out. But it isn't a two-body problem, because both bodies are also orbiting the sun. So it becomes a three body problem, which is more complicated. Since the planet has orbital velocity around the sun, it's like bouncing off of a moving object. In this case you can gain velocity.

 

[Bandersnatch]

You said you've done some reading already. Did it include the Wikipedia article on gravity assist? I'm asking so that we can have a better understanding of where your confusion lies.
In that article, did you understand the representations of the different outcomes when viewed from the frame of reference of the planet (where the situation is indeed symmetric - i.e. inbound speed = outbound speed) and of the Sun? Was the analogy with the train and a tennis ball unclear?

 

[A.T.]

but nobody seems to explain why the acceleration induced on approach isn't reversed on exit.

It is reversed in the frame of the planet, but not in a frame where planet moves. Keep in mind that velocity and thus momentum and kinetic energy are frame dependent.

Here is a video about dynamic soaring, but at the begin it explains the frame dependent velocity gain from interaction with a bigger mass. Just replace the wall with a planet, and the bounce with a turn due to gravity.

 

[jartsa]

I understand that the target body for the flyby loses some of its momentum and that the satellite executing the flyby takes some of that velocity. That makes sense to me. What I do not understand is why that same interaction isn't undone after the satellite is adjacent to the body.

Can we replace satellite with a micro black hole?

Case 1: Planet approaches the black hole along a straight line until the black hole is at the center of the planet. Then the planet recedes from the black hole along a straight line.

This is a symmetric case. Black hole's momentum is unchanged at the end.

Case 2: The black hole does not move along a straight line.

This is not a symmetric case. Black hole's momentum is changed at the end. (If you say that this case seems symmetric too, then I would like to know exactly which two parts of the trajectory are symmetric? )

 

[sophiecentaur]

why that same interaction isn't undone after the satellite is adjacent to the body.

The maximum speed, when nearest to the target planet is not what is gained. (That could be much much faster than the planet when it is at its closest). The net increase in speed of the craft will be more related to differences between the orbital speed of the planet and the speed of the craft early on in the manoeuvre. The mass of the planet will not be very relevant - as long as it's a lot bigger than the craft.

 

[quarkstar]

If it were just a two-body problem (the satellite and the planet), then your intuition would be correct. That problem is completely symmetrical, and the velocity gain on the way in is lost on the way out. But it isn't a two-body problem, because both bodies are also orbiting the sun. So it becomes a three body problem, which is more complicated. Since the planet has orbital velocity around the sun, it's like bouncing off of a moving object. In this case you can gain velocity.

I wish I could still explain it that clearly and casually.

 

[phyzguy]

I wish I could still explain it that clearly and casually.

Thank you!

 

[sophiecentaur]

the velocity gain on the way in is lost on the way out.

This could be taken in the wrong way. It is the relative speed between ship and planet that is acquired and lost but the velocity change is what matters. But you do not need a third body in order to get momentum transfer from one body to another in an elastic interaction. The simplest example of this is when a ball bearing, traveling towards a locomotive, bounces off and its speed can increase by the closing speed between the two. The relative speeds will be the same before and after but the speed relative to the track is increased. The same happens, but in a more complex way, with a slingshot. Momentum (in the Sun's frame) is gained - which is the point of the exercise. You don't just want to be going faster after the encounter, of course, but the new direction (relative to the Sun's frame) is important . The Third Body just provides the problem and makes it harder.
I found this link which is chock full of useful stuff. I don't remember the name of the smart young man who first solved the practical problem for long distance space flight but he certainly made a difference to deep space exploration.

 

[A.T.]

... you do not need a third body in order to get momentum ...

Yes, the third body is just there to define a reference frame. But you can just as well say "a frame where the planet moves".