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I believe travelling to Alpha Centauri at ~10 km/s would take of the order of 100 000 years (10 km/s is the order of speed of probes currently leaving the solar system). That seems 1. rather a long time to wait for a probe to arrive and 2. rather a difficult engineering task to produce a probe that will successfully operate for that long.

Various suggestions have been made for propulsion systems that would provide vastly faster speeds, but are hypothetical. Sling shot manoeuvres however are well established. Is there any limit to the velocity increase that can be achieved by repeatedly sling-shotting around a pair of bodies? I am assuming here that it is always possible to arrange the sling shot exit from one body so that it takes you on the correct course to sling shot around the second and then back to the first etc. (See https://en.wikipedia.org/wiki/Gravitational_slingshot for the principle of increasing speed by slingshotting).

Would it be possible to repeatedly sling shot to achieve a speed ~1000 times faster (~10000 km/s) so that the journey would "only" take 100 years?

My intuition says this will not be possible, because at high speeds you will need to get so close to each of the two bodies in order to redirect back to the other object that you will presumably crash into them. However, I don't know the maths and so wondering what the limit would be

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In theory, it might be possible, but in practice it is not. The problem is the sling shots (gravity assists) themselves take time.

Some years ago, we did a study of a possible flag ship mission to Saturn's moon Enceladus. (1) Enceladus is in a very tight orbit around Saturn, so is in a very deep gravitational well. Once our ship reached Saturn, it would take a lot of fuel to get down to Enceladus. Since we could only carry so much fuel, we planned three gravity assists. This meant that it would take more than 10 years to reach Enceladus.

Gravity assists took up more than half of that 10 years. Assuming the total gain was on the order of 10 km/sec, that's 20 km/sec per decade. I think you can see where this is going.

(1) https://www.lpi.usra.edu/opag/Enceladus_Public_Report.pdf

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  • $\begingroup$ Gravity assists to get to Saturn, or among its moons to lower the orbit once there? Or both, of course. I see Cassini used four assists to get to Saturn: Venus, Venus, Earth, Jupiter. $\endgroup$
    – jamesqf
    Aug 5 at 4:49
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    $\begingroup$ @jamesqf To get to Saturn with enough remaining fuel to get down to Enceladus. $\endgroup$
    – Vince 49
    Aug 5 at 4:53
  • $\begingroup$ It's not possible in theory. $\endgroup$ Aug 6 at 22:43
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Once you get going fast enough you will escape the system and won't be able to do any more slingshot maneuvers.

Edit: Yeah, you can still encounter bodies for slingshots. Once you are at escape velocity they will be few and far between, though. This isn't a theoretical limit but it is a practical one. Voyager is going to be dead long before it could do another slingshot.

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    $\begingroup$ If that was true Mariner / Voyager wouldn't have been able to sling-shot after they reached solar escape velocity, but they did. E.g. see en.wikipedia.org/wiki/Gravity_assist#/media/… $\endgroup$
    – Raffles
    Aug 5 at 7:35
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    $\begingroup$ They did those slingshot maneuvers on their way out, so to speak. Once you’ve reached escape velocity you’ve (by definition) left orbit and can’t stay in orbit to do more and more slingshot maneuvers. $\endgroup$
    – Michael
    Aug 5 at 8:07
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    $\begingroup$ You could slingshot around the next system! $\endgroup$
    – gerrit
    Aug 5 at 8:34
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    $\begingroup$ @Michael: The Voyagers are pretty good examples. Both had exceeded solar escape velocity by the time they passed Saturn. Voyager 2 was only able to go on to Uranus & Neptue because of a fortuitous orbital alignment. (See "Grand Tour".) It would have been just about impossible to go back to Jupiter. $\endgroup$
    – jamesqf
    Aug 5 at 19:57
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    $\begingroup$ @Peter-ReinstateMonica Two problems. First, bouncing back and forth requires planets with enough gravity that you can make basically a 180 during your turn. That's basically mega-Earths. Second, bouncing back and forth between two worlds isn't going to do much for you anyway. Head on, a 180 is the best gravity maneuver possible. Side on, a 180 just turns you around but you gain nothing. You need at least three bodies to play that dance. We don't have even one such planet here. $\endgroup$ Aug 7 at 1:39
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There are practical limitations besides just time. I don't know the whole math involved, but a very high speed, very close pass to a massive object like a planet will not result in much deflection, if any. The reason is that the amount of time that one is actually close enough to the massive body shrinks the faster one is going. I believe this is the Tisserand parameter.

Further reading:

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  • $\begingroup$ I don't think this is true. Gravity bends space, so you always get a deflection, even travelling at the speed of light (although in that case of course there is no increase in speed!). See en.wikipedia.org/wiki/Gravitational_lens. The question is, how deep down the gravity well do you need to go in order to come back out the way you need to? You can't go so deep that you actually crash into the planet, so what is the limit that you can (safely) go to, and what speed can you achieve? It's that limit that I'm after in asking the question. Thanks $\endgroup$
    – Raffles
    Aug 5 at 7:28
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    $\begingroup$ Assuming a spherical body, the lowest possible altitude is the radius of the object. At that altitude, there will be a certain gravitational acceleration. Above that altitude, the acceleration will be lower. For every encounter velocity, you will pass through the gravity well in a certain amount of time, the time being shorter for larger velocities. This gives you a very loose upper bound of time-spent-in-gravity-well * accceleration-at-object-radius = velocity-gained. This value will clearly be lower at higher encounter velocities. There's probably a website that can give you actual numbers. $\endgroup$
    – AI0867
    Aug 5 at 10:19
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    $\begingroup$ @Raffles The amount of the deflection decreases with your initial speed. Consider that a planet is so deflected by the sun that it stays in orbit though it's millions of miles away, while a light beam actually grazing the surface of the sun is barely deflected by less than a degree. $\endgroup$ Aug 5 at 13:11
  • $\begingroup$ @RossPresser you are right that it will decrease as speed increases if you maintain a fixed distance from the object each time you pass it. However if you move closer to the object each time you pass it, so that you exactly compensate for the speed increase, you will get the same deflection. This is borne out by the escape velocity formula which is inversely proportional to the distance... which I was not aware of when I posted the question, but thanks to you fine people I have now come across :-) en.wikipedia.org/wiki/Escape_velocity $\endgroup$
    – Raffles
    Aug 7 at 8:18
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    $\begingroup$ Do you think there is a limit of the speed increase if the planets are black holes? I mean, can lightspeed be achieved at a distance away from the horizons? $\endgroup$ Aug 7 at 8:28
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In principle you can' t gain much velocity. Only a change in direction. Even with an array of black holes your final velocity in space will be not much different from the velocity you had initially.

Suppose there is an arbitrary distribution of masses inside a volume. If you shoot a test mass inside this system then it will exit deflected with a velocity that is deflected but it will end up with a velocity that is always smaller than the initiak one.

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  • $\begingroup$ I think if you had black holes you could actually achieve a very high speed indeed (provided you can plot your course accurately enough $\endgroup$
    – Raffles
    Aug 6 at 16:45
  • $\begingroup$ @Raffles How can that be done? It was my first thought too. I even made the comment that almost lihjtspeed could be achieved. But then I thought diffrrently. $\endgroup$ Aug 6 at 17:07
  • $\begingroup$ Re "In principle you can' t gain much velocity. Only a change in direction.": So the Voyager probes didn't gain much velocity? $\endgroup$ Aug 6 at 22:27
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    $\begingroup$ @Raffles ...and provided you've purchased your spaceship hull from Pierson's Puppeteers General Products, of course. Otherwise, your mashed potatos (instead of your spaceship) would achieve such very high speeds... $\endgroup$ Aug 6 at 23:14
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    $\begingroup$ @Raffles Its more complicated than I thought!;) Can your velocity increase without bound (I mean approaching c)? $\endgroup$ Aug 7 at 8:33
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So if the idea is to ping-pong between two planets, each ping-pong in the series will give roughly the same acceleration, which is related to the orbital velocity of the body you are slingshotting.

For this to work you would need to leave each body in roughly the direction you arrived, and to do that you need to be travelling roughly at escape velocity (https://en.wikipedia.org/wiki/Escape_velocity). This is the same order of magnitude as the speed of the Voyager probes (actually slower than them) and makes the technique of no practical use for travelling to Proxima Centauri.

In fact by far the biggest useful sling-shot effect would be from the solar system itself, travelling around the galactic centre at ~220 km/s. The maximum slingshot effect is achieved by approaching in the opposite direction of the orbit and exiting in the same direction as the orbit, which adds double the orbital velocity to the start velocity, over 400 km/s in this case. Of course that means your destination better be somewhere along the tangent, which may mean picking a different destination, and it would still take ~2500 years even to get to Alpha Centauri.

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    $\begingroup$ you expect to approach Jupiter at 60 km/s and then not speed up while falling to down near the surface? $\endgroup$ Aug 6 at 17:01
  • $\begingroup$ @BrendanLuke15 60 km/s is the maximum velocity, which is at the surface. We reach this velocity after accelerating down towards the surface, then lose velocity on the way back out again. In the process we gain energy from Jupiter's orbital velocity i.e. sling shot. You are right that I haven't been very clear about that above, will try to tidy it up. Thanks $\endgroup$
    – Raffles
    Aug 7 at 7:26
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    $\begingroup$ I realy dont see why this is downvoted. Its the best answer... $\endgroup$ Aug 7 at 8:19
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    $\begingroup$ @Raffles how much speed do you lose on the way back out again? $\endgroup$ Aug 7 at 12:44
  • $\begingroup$ @BrendanLuke15 great question - this is the key. From the intertial frame of reference of the planet - you lose all of it i.e. it takes back everything it gave you on the way in. However from the frame of reference of the sun (which it is orbiting at velocity u) you gain up to 2u. See en.wikipedia.org/wiki/Gravitational_slingshot in particular this diagram: en.wikipedia.org/wiki/Gravity_assist#/media/File:GravAssis.gif Cheers $\endgroup$
    – Raffles
    Aug 7 at 13:47

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