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How can an object move faster after leaving the gravity of a planet, should a gain in potential energy by moving away from the planet lead to a loss of kinetic energy? Or am I missing something essential like the a change in angular momentum of some body?

My question is similar to this question but I wish to discuss the more technical aspects.

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Tanmay Kulkarni
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  • You are basically correct on both counts. If we are looking at only two bodies, a larger planet and a small body, then there will indeed be no gain in kinetic energy. However, in the case of a slingshot both the large planet and the small body are in orbit around the sun. The total angular momentum has to be conserved, but there can be a transfer or orbital angular momentum between the two bodies. That transfer can go either way and missions to Venus and Mercury are using it to actually lose velocity in heliocentric coordinates. – CuriousOne Jul 21 '16 at 04:22
  • In the text book that had a footnote about this had told that the angular momentum of the planet increases if the speed of the other body increases ... – Tanmay Kulkarni Jul 21 '16 at 04:25
  • Sounds about right. The planet will not change velocity too much, though. If we contrast $10^{25}$kg (the order of magnitude of the mass of an Earth like planet) with the mass of a small body like one of our probes, the change in momentum/velocity turns out to be unmeasurably small. – CuriousOne Jul 21 '16 at 04:28
  • I referred to Kleppner for the explanation of gravity but I was unable to understand it fully.. – Tanmay Kulkarni Jul 21 '16 at 04:30
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    You could e.g. try "A gravity assist primer" by R.J. Cesarone. Looks sweet and short. I have seen book chapters in space navigation that go into much more detail, of course. – CuriousOne Jul 21 '16 at 04:33
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    If you hold the Sun as the frame of reference then it comes from the planet. However, the more I read it the more it looks like slingshot is used to change direction rather than gaining speed even though there is possibly some small gain in speed if the movement both direction and rotation of the main object can assist the probe. –  Jul 21 '16 at 05:24

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If you think of it in terms of conservation of momentum and collisions, the simplest version works just the same as tossing a handball at a on-coming freight train. The interaction is elastic, and the ball returns with the same speed it had going in in the center of momentum frame, but the center of momentum frame is moving in the ground frame, so the ball is going faster in the ground frame.

And it is that matter of there being multiple frames that you didn't pay enough attention to. The incoming spacecraft leaves the planet (or moon) with the same speed that it had coming in, but only in the center of momentum frame for the interaction. And because the planet (or moon) is moving in the frame of the solar system (i.e. the sun's frame) the spacecraft can be going faster in the solar system coordinates.

As CuriousOne noted in the comments, you can also lose energy in the solar-system frame. That's equivalent to throwing the ball at the back of the trai as it moves away from you.

Some basic math. We'll assume that our freight train is only moving at $+10 \,\mathrm{m/s}$ and that we can throw the ball at $20 \,\mathrm{m/s} with respect to the ground. Also that the collision is elastic and straight on.

So, you stand in front of the on-coming train and hurl the ball at it at $-20 \,\mathrm{m/s}$ in the ground frame. The center of momentum frame is essentially the trains frame, so in that frame the ball approaches the train at $-30 \,\mathrm{m/s}$. It hits and rebounds at $+30 \,\mathrm{m/s}$ in the train frame which is $+40 \,\mathrm{m/s}$ in the ground frame.

You'd better duck the ball even before dodging the train.

The details of the spacecraft-planet interaction are more complex in the sense that there will be motion across the line of motion instead of just in out, but they both represent nice elastic collisions, so the math remains very similar in spirit.

Note that I have not addressed the fancier version of the maneuver in which the space-craft fires its engines near periapsis which allows the maximally efficient use of fuel, but you should start understanding this basic effect.

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dmckee --- ex-moderator kitten
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  • Could you add some math and diagrams to give more clarity. Thanks for the quick response – Tanmay Kulkarni Jul 21 '16 at 04:21
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    Can you flesh this out of with more rigorous mathematic formulas or you also think this kind of handwaving plus some pedantic numbers are enough for it to be voted high? –  Jul 21 '16 at 04:59