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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

Raffles
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    The known minimum for continuously operating electronics is probably Voyager 1 at around 44 years. Anything more than that is theoretical, Clock of the Long Now kind of stuff. – Chris B. Behrens Aug 04 '21 at 20:36
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    @ChrisB.Behrens Clock of the Long Now is also mentioned in What would be the (most difficult) challenge to make a 10,000 year satellite? – uhoh Aug 04 '21 at 21:02
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    fyi I've just asked Flyby puzzler; starting from Earth, how many times can you use Jupiter flybys in one century? – uhoh Aug 04 '21 at 22:03
  • In addition to the answers below, is this not limited by reaching the escape velocity of the body you are using for the assists? – Eric G Aug 05 '21 at 04:04
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    @Eric G: If you aren't travelling at more than the escape velocity, you'll crash into it. And if you start from a long distance away, with no significant velocity relative to the body, by the time you reach it you will be travelling at escape velocity. – jamesqf Aug 05 '21 at 04:45
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    The dream of the Long Now Foundation is to avert the collapse of civilization by slowing the pace of "progress." The Clock is not supposed to run by itself for 10,000 years. Rather, it is supposed to be maintainable by the kind of civilization that moves at a slow enough pace to survive for another 10,000 years. https://en.wikipedia.org/wiki/Clock_of_the_Long_Now – Solomon Slow Aug 05 '21 at 14:52
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    @uhoh, yeah that would be a fun project to work on. – Chris B. Behrens Aug 05 '21 at 16:28
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    @Solomon Slow, yeah, I think some of those premises don't exactly compile. – Chris B. Behrens Aug 05 '21 at 16:31
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    If the escape velocity of the solar system is exceeded, additional sling shot maneuvers might be difficult if not impossible. – Uwe Aug 05 '21 at 21:59
  • Somewhat related, but not practical: Z. Xia, “The Existence of Noncollision Singularities in Newtonian Systems,” Annals Math. 135, 411-468, 1992, https://sci-hub.do/10.2307/2946572 –  Aug 05 '21 at 22:18
  • Hey, I asked the same question but didn't get an answer: https://space.stackexchange.com/questions/51521/what-is-the-maximal-gravity-assist-boost-achievable-in-the-milky-way-galaxy-usin . My intuition was that it's possible using a pair of binary black holes (4 black holes total, and you ping pong between the systems). I think it's not possible for humans to survive it, but a robotic system should survive the ~100g worth of acceleration over a half hour. – csiz Aug 07 '21 at 03:12

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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

Vince 49
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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.

Loren Pechtel
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    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 https://en.wikipedia.org/wiki/Gravity_assist#/media/File:Voyager_2_velocity_vs_distance_from_sun.svg – Raffles Aug 05 '21 at 07:35
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    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. – Michael Aug 05 '21 at 08:07
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    You could slingshot around the next system! – gerrit Aug 05 '21 at 08:34
  • It just becomes even more inconvenient. You might need to slingshot off planets (or better, stars) of other systems. – CuteKItty_pleaseStopBArking Aug 05 '21 at 09:12
  • @gerrit: But the problem with that is that it takes 10K years or whatever before you reach the next system :-( – jamesqf Aug 05 '21 at 15:51
  • @jamesqf Exploring the galaxy is not a business for those in a hurry. It may take half a million years to get up to speed, but it will save time and fuel getting to the other end of the galaxy ;-) – gerrit Aug 05 '21 at 15:56
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    This is the best answer. You're limited by the number of slingshots before you reach the solar system escape velocity. You might be able to sneak one or two more in after that, but once you reach escape velocity you have nothing more to gain really. – zephyr Aug 05 '21 at 19:11
  • @gerrit: But the question's about just getting to Alpha Centauri. Presumably the OP wants to stop there, rather than just doing a quick flyby - Galileo/Cassini vs Voyage/New Horizons. – jamesqf Aug 05 '21 at 19:51
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    @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. – jamesqf Aug 05 '21 at 19:57
  • @jamesqf I know; my comments were not entirely serious for getting to Alpha Centauri. – gerrit Aug 06 '21 at 07:03
  • Your argument is good but I wonder whether it is absolute. If you maneuver accurately enough you could "sling bounce" with ever tighter encounters between planets inside the Solar System; an orbit around a planet that is tight enough should still change your vector. Enough to reach another planet? – Peter - Reinstate Monica Aug 06 '21 at 07:59
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    @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. – Loren Pechtel Aug 07 '21 at 01:39
  • @LorenPechtel How about a binary star system? Would it work around both suns there? – Mast Aug 07 '21 at 08:46
  • @Peter-ReinstateMonica on your point about needing mega-Earths, certainly a dense rocky planet would give a better max speed (surface escape velocity) than a gas giant of the same mass... however in order to ping-pong you only need that the planetary escape velocity exceeds the solar escape velocity. Once you get out as far as Saturn, the solar escape velocity is around 14 km/s, which is far overpowered by Saturn, which has a surface escape velocity of 36 km/s, allowing you to swing all the way round the planet and back again (or even drop into orbit) e.g. at 20 km/s - way above solar escape – Raffles Aug 07 '21 at 11:50
  • @Peter-ReinstateMonica It's a good point. I agree that three planets (suitably placed) would allow you to keep cycling round the entre system, and gain speed every on every slingshot. However, 2 planets is actually enough, because of the following two effects. Firstly, you can approach the inner planet roughly at a tangent, so maximum gain from that planet (i.e. more or less a 180 turn around, in the direction of orbit), whereas the corresponding loss from the outer planet will be smaller because you approach it at more of a right-angle. Secondly, the inner planet has a higher orbital velocity – Raffles Aug 07 '21 at 11:55
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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:

uhoh
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  • 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 https://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 – Raffles Aug 05 '21 at 07:28
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    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. – AI0867 Aug 05 '21 at 10:19
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    @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. – Ross Presser Aug 05 '21 at 13:11
  • @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 :-) https://en.wikipedia.org/wiki/Escape_velocity – Raffles Aug 07 '21 at 08:18
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    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? – Deschele Schilder Aug 07 '21 at 08:28
  • @DescheleSchilder No – but near-light-speed can be achieved a (tiny) distance away. – wizzwizz4 Aug 07 '21 at 14:50
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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.

Deschele Schilder
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  • I think if you had black holes you could actually achieve a very high speed indeed (provided you can plot your course accurately enough – Raffles Aug 06 '21 at 16:45
  • @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. – Deschele Schilder Aug 06 '21 at 17:07
  • Re "In principle you can' t gain much velocity. Only a change in direction.": So the Voyager probes didn't gain much velocity? – Peter Mortensen Aug 06 '21 at 22:27
  • @PeterMortensen I mean a cosmic gain. You can get a gain by falling in local wells, of course. But looking at the total solar system well its velocity must decrease. Unless a black hole passes. – Deschele Schilder Aug 06 '21 at 22:34
  • @PeterMortensen Of course Voyager got big boosts! – Deschele Schilder Aug 06 '21 at 22:41
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    @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... – Matija Nalis Aug 06 '21 at 23:14
  • @raffles Suppose you see a system of collected black holes circling around one another. There is a potential well connected with them. If you direct you ship with an initial velocity from an initial position, where the potential energy is high, you gain kinetic energy first. Then you fall directed. You have calculated all possible trajectories. How many times faster can you end up maximally when entering. – Deschele Schilder Aug 07 '21 at 04:25
  • @DescheleSchilder I think the limit in pratical terms, as mentioned by MajitaNalis would be a craft able to withstand the forces, but suppose we had that, (and can navigate accurately enough)... as I understand it, the event horizon is the distance from the gravitational centre at which the escape velocity is the speed of light. For black holes the radius of the object is less than the radius of the event horizon. Therefore you accelerate at a hyperbolic rate between the two bodies an arbitrary number of times - you will never crash into the object - the navigation better be spot on though ;-) – Raffles Aug 07 '21 at 07:56
  • @rafflesWhat object do you mean?:) That gains velocity? – Deschele Schilder Aug 07 '21 at 08:02
  • @Raffles Do you stay (if you ship would hold indeed, but lets assume a pointmass) outside the hole? – Deschele Schilder Aug 07 '21 at 08:07
  • @DescheleSchilder as per https://en.wikipedia.org/wiki/Gravity_assist – Raffles Aug 07 '21 at 08:08
  • @DescheleSchilder yes you stay outside, just the same as when you gravity slingshot around a planet - the only difference with slingshotting round a black hole is it is much denser and therefore smaller, so you can get closer, which means you can go faster :-) – Raffles Aug 07 '21 at 08:26
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    @Raffles Its more complicated than I thought!;) Can your velocity increase without bound (I mean approaching c)? – Deschele Schilder Aug 07 '21 at 08:33
  • @DescheleSchilder if there weren't any further complications, yes, I believe it could. However you would need two orbiting black holes to achieve that, so definitely in the realms of theory here, at least as far as our own solar system is concerned. One complication that would arise as you get closer to c - the slingshot maths assumes that the mass of the object that is doing the slingshot is much smaller than the object you are sling shotting around.. As you get faster, your mass starts to increase, and as you approach c it tends to infinity i.e. more than the black hole! So another limit. – Raffles Aug 10 '21 at 07:54
  • @Raffles Good point! Could you end up in the photonsphere? A photon can get a slingshot from there can we too? Assuming pointlike spaceship. – Deschele Schilder Aug 10 '21 at 07:59
  • @DescheleSchilder honestly I have no idea whether you could get fast enough to cross it, hit perigee and come back out again. You couldn't end up in it though - you'd have to be travelling at the speed of light to do that I think (my understanding is the photonsphere is composed entirely of photons in circular orbits). The time dilation would be enormous if you did go through it - you'd be travelling very close to c AND very deep in a gravity well. Maybe it would make a good question to post? "Is it theoretically possible to slingshot round a black hole and cross the photonsphere?" – Raffles Aug 11 '21 at 07:56
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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.

Raffles
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    you expect to approach Jupiter at 60 km/s and then not speed up while falling to down near the surface? – BrendanLuke15 Aug 06 '21 at 17:01
  • @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 – Raffles Aug 07 '21 at 07:26
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    I realy dont see why this is downvoted. Its the best answer... – Deschele Schilder Aug 07 '21 at 08:19
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    @Raffles how much speed do you lose on the way back out again? – BrendanLuke15 Aug 07 '21 at 12:44
  • @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 https://en.wikipedia.org/wiki/Gravitational_slingshot in particular this diagram: https://en.wikipedia.org/wiki/Gravity_assist#/media/File:GravAssis.gif Cheers – Raffles Aug 07 '21 at 13:47
  • @Raffles If you start and end with 0 speed w.r.t the planet what have you really gained? i.e., U=0 in this diagram – BrendanLuke15 Aug 07 '21 at 14:25
  • @BrendanLuke15 great diagram! From the frame of reference of the planet (top half of the diagram) you have gained nothing... however the planet is not the frame of reference we are interested in since we don't stay with it, we leave the planet behind. What we have really gained is shown in the bottom half of the diagram. This is what real slingshot manoeuvres are based on, not just of our own spacecraft, but also how we calculate the future trajectories of near earth asteroids.. E.g. Famously 99942 Apophis will slingshot past Earth in 2029 - https://en.wikipedia.org/wiki/99942_Apophis Cheers – Raffles Aug 09 '21 at 07:39
  • @BrendanLuke15 PS just in case I have misunderstood what you were asking, if the orbitial velocity itself was 0 - there would be no point in trying to slingshot, round it... but then if u was zero, the planet wouldn't be orbiting, it would be about to start falling into the sun I think? – Raffles Aug 09 '21 at 07:48
  • @Raffles the planet's orbital velocity is V in that diagram (and non-zero). Can we agree that you must leave the flyby planet at some non-zero relative speed to reach the next planet? If so, how is that possible when the proposed escape velocity flyby means an object "traveling along an escape orbit will coast along a parabolic trajectory to infinity, with velocity relative to the central [flyby] body tending to zero" (parabolic trajectory wikipedia)? – BrendanLuke15 Aug 09 '21 at 14:21
  • @BrendanLuke15 apologies you are right the v and u in the diagram are the opposite way round to elsewhere. It is possible, because the next planet you are going to is closer than infinity. The key thing I learned in order to get the answer to my question is that the voyager et.c slingshots were above escape velocity and so followed a hyperbolic trajectory, wheras if you travel close to escape velocity you can go into a parabolic trajectory or even a highly elliptical orbit, which takes you back to the other planet. However this does limit the maximum speed you can achieve, which is the answer. – Raffles Aug 10 '21 at 07:39
  • "Cheers"... (in your comment above) Good one! – Deschele Schilder Aug 10 '21 at 08:27