Something I've been wondering about for a while, but it's only just now occurred to me to ask.

(No real reason, why? 8-[ Don't be so nosy. [-X ... :wink: )

In any case...

Using gravity wells to slingshot a probe. How does that add (or remove) velocity? It can't be just a simple effect of falling towards the planet, that velocity would be lost as you drift away from the planet.

I'm guessing it has something to do with transferring angular momentum from whichever direction a planet orbits (or rotates?)

It's true that whatever speed is picked up on approach to the planet is lost as it heads away from the planet, but that speed is relative to the planet. So the probe basically picks up the speed of the planet as it passes by.

More detailed explanations.

This page has diagrams, if you scroll down far enough.

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That makes sense

Thinking about it this morning, I'd come close, but I didn't quite see it. Now I think I do:

To speed up, you approach the planet from "behind" in it's orbit. As it is travelling "away" from you, you end up falling towards the planet longer than it takes to break away from it on a hypothetical return trajectory.

To slow down, you approach the planet from "ahead" of it's orbital direction and gravity acts on you longer as you break away "ahead" on the return trip.

Do I "gots" it?

Yes, you gots it. APL's MESSENGER mission to Mercury is actually going to be using reverse gravity assists to lose enough energy so that it can go into orbit.

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Didn't Apollo 13 use this in order to gain more momentun to return to earth? I'm not sure, :-/. (Assuming I know what you are even talking about). I think the planet you are going to use to sling shot a probe needs to be positioned in the focal point of the ellipse the probe is travelling on. Keplers 2nd law deals with this, right? (Dont kill me, I realize that most of this info is most likely terribly inaccurate).

I don't think any of your points are correct. Apollo 13 did use a "slingshot maneuver" to return to Earth, but that was just a matter of letting the Moon's gravity turn the spacecraft around and send it back. The return speed was about the same as the arrival speed. (The Apollo 11 and 12 missions were also set up for such a return but didn't need it.)

A spacecraft would be traveling on a hyperbola relative to the planet (if it were traveling in an ellipse, it would be orbiting the planet), and by definition one of the foci of the hyperbola would be the planet.

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oopss, sorry. I guess thats why I'm still a Bad Newbie and you're a Bad PostDoc.

Oh well, at least I didn't claim I knew anything I was talking about. I heard my teacher mention in once, so :-/...

EDIT: I guess I was picturing THIS in my head http://www.physics.sfasu.edu/markwor...s/image003.jpg
Where the time it took to go from A to B is equal to the time it took to go from C to D. The area of AB = DC. So I just got very confused.
I'm dumb

IMHO, dumb people can't explain Kepler's Second Law.

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EDIT: Wow, I misunderstood what was being said. So I changed it.

Monoxide Child, you're missing ToSeek's logic.

1. Dumb people can't explain Kepler's Second Law;

2. You explained Kepler's Second Law;

3. Therefore, you're not a dumb person.

CRAZY! My bad. sorry. lol

I appreciate your stepping in for me and making the clarification. That was exactly the point I wanted to make.

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