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A lot of space probes use gravity assists from various space objects to boost their speed and save on propellant requirements. On what factors does the increase of delta-v depend? My guesses are:

  • Mass of object (the more massive, the better)
  • Distance from object to spacecraft (the closer, the better)
  • Time spent behind (or ahead, if you want to decelerate) object (the longer, the better)

However, I'm not sure about my speculations, as they come solely from free-time reading and experience gained playing the Kerbal Space Program game. Could someone more confident widen my perception of gravitational slingshot?

TildalWave
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    http://maths.dur.ac.uk/~dma0rcj/Psling/sling.pdf – Deer Hunter Dec 25 '14 at 19:06
  • @DeerHunter But it's not prohibited here and I will accept it because it's good! –  Dec 25 '14 at 19:47
  • I guess it was worth the waiting :) –  Dec 25 '14 at 21:04
  • A similar question was asked at the physics stackexchange. My answer there came to the same equation as the answer of Mark Adler, but contains a little bit more derivations, so might be might be helpful as well. – fibonatic Dec 28 '14 at 15:17
  • @fibonatic Hey, I had a question about your answer but since I do not have enough reputation to comment on physics stack exchange I will ask here. Firstly, I dont understand what you mean by this: "the true anomaly which is by definition equal to zero at periapsis and therefore the maximum amount of bending will be roughly twice the true anomaly at =∞". I understand that TA = 0 at periapsis but I dont understand why this implies that the max bending is therefore twice the TA at =∞. If you could explain this for me it would be greatly appreciated! – Alexander Ivanov Jul 09 '20 at 01:54
  • @fibonatic I was also wondering why the hyperbolic excess velocity is used instead of simply the incoming velocity of the spacecraft (as far as I know these are not the same thing). I have trouble understanding how v∞ makes sense in this context since it is the speed an infinite distance away from the celestial body and for your answer you seem to take it to be the incoming velocity of the spacecraft (this is essentially the relationship you are proving: what is the effect of the incoming velocity on the change in velocity). And so, if you can also explain this it would be really appreciated. – Alexander Ivanov Jul 09 '20 at 02:00
  • @AlexanderIvanov To answer both, I used that for sufficiently eccentric orbits with low enough periapsis the in- and outcoming trajectories are approximated well with a line and at the distance where the influence of the planet becomes negligible the cumulative change in velocity due to this planet is also negligible. In this case one can say that the incoming velocity of the spacecraft is roughly equal to hyperbolic excess velocity and the TA at the encounter is roughly TA at =∞ and thus when going to periapsis you sweep the angle of TA at =∞ and when leaving you sweep another TA at =∞. – fibonatic Jul 09 '20 at 04:01
  • @fibonatic I think I understand. So the TA at exact moment of entering the sphere or influence of the planet is the TA at =∞ (because is ∞ at this point). This then r goes to 0 at periapsis and then back to ∞ when leaving the planet. And then that is the maximum deflection angle, which would be equal to 180° (with high e). I was also wondering why your formula for deflection angle is different from the one here: https://space.stackexchange.com/questions/6504/to-what-extent-could-a-single-triton-flyby-slow-down-a-direct-hohmann-transfer-t/6510#6510 is it because yours is the max angle? – Alexander Ivanov Jul 09 '20 at 05:03
  • @AlexanderIvanov a small correction is that r doesn't necessarily go to 0 at periapsis (only if the periapsis height is zero, but then the space craft would crash into the planet). And the other deflection angle formula only differs in a constant $\pi$ thus 180 degrees, since that formula calculates the actual deflection of the velocity while mine calculates that angle swept when passing the planet (but was easier to derive). – fibonatic Jul 09 '20 at 12:58
  • @fibonatic Ok thank you for the explanations. I was also wondering if you could also explain to me how you came up with the formula for detla-v. I know how to express delta-v (v_out - v_in) in terms of v∞ , the heliocentric velocity of the planet, and the angle between these two vectors (using cosine law), but when I saw your formula it looks much simpler (you dont need to know heliocentric velocity at the correct time or the angle between v∞ and planet velocity). Unfortunately I have no idea what you did (I know this is a dumb question but I could not find any explanations online). – Alexander Ivanov Jul 10 '20 at 00:20
  • Or if you know of where I could find the derivation online because I have been looking but its hard to even find the formula mentioned somewhere and when I do there is no explanation as to where it comes from. – Alexander Ivanov Jul 10 '20 at 00:20
  • @AlexanderIvanov my equations already assumes that the velocity that you give is relative to the planet at the encounter. It also doesn't directly say in which direction the change in velocity points. So you can't use it to calculate gravity sling shots. However, it does show that one has diminishing returns in the resulting change in velocity as the encounter velocity (≈v∞) becomes bigger and bigger. – fibonatic Jul 11 '20 at 06:54

2 Answers2

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Yes, those are the three factors. Your third factor shows up as the $v_\infty$ of the spacecraft relative to the object. The first two are the GM of the object, $\mu$ and the closest approach distance $r$.

The $\Delta V$ you can get is:

$$2\,v_\infty\over 1+{r\,v_\infty^2\over\mu}$$

As you surmised, lower $v_\infty$ is good since you spend more time under the influence, so to speak. But not too low. The $\Delta V$ goes up as $v_\infty$ drops towards $\sqrt{\mu\over r}$, but below that, the $\Delta V$ starts going down again.

Delta V has a maximum

On the left side of the curve, there's not much velocity there to change. Note that the change in velocity comes entirely from a change in direction in the reference frame of the body doing the slinging. The magnitude of the $v_\infty$ going out is exactly the same as the magnitude going in. The change in direction is called the bend. The bend angle at the maximum $\Delta V$ of $\sqrt{\mu/r}$ is 60°.

Mark Adler
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  • Could you explicate the $v_\infty$ term a bit? You can't really measure it at infinity. $v$ at edge of sphere of influence? Hill sphere distance? How is $v_\infty$ determined? – Jerard Puckett Dec 26 '14 at 12:50
  • You can calculate $v_\infty$ from your current position, velocity, and $\mu$ of the body. – Mark Adler Dec 26 '14 at 19:10
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    @JerardPuckett If it helps, $v_\infty$ is aka hyperbolic excess velocity. – TildalWave Dec 28 '14 at 01:15
  • Can anyone explain what $v_\infty$, and $\mu$ are? – sampathsris Jun 12 '18 at 08:31
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    $v_\infty$ is the velocity of the object relative to the body at infinite distance, and $\mu$ is the mass of the body times Newton's gravitational constant, i.e. $GM$. – Mark Adler Jun 12 '18 at 14:20
  • @MarkAdler Could you explain how you got that the bend angle that will result in the maximum delta-v is 60 degrees? I am also wondering how you know that the optimum ∞ is at √/. It is not clear to me that the peak of the graph is √/. And finally this is probably a dumb question but I dont understand why you divided both axes by √/. If you could explain this to me it would be greatly appreciated! – Alexander Ivanov Jul 09 '20 at 02:24
  • See this answer for the bend angle. For $v_\infty=\sqrt{\mu/r}$, you get $e=2$. – Mark Adler Jul 09 '20 at 03:49
  • I divide by $\sqrt{\mu/r}$ on both axes to get a dimensionless plot that can be used for any flyby. – Mark Adler Jul 09 '20 at 03:50
  • It is quite apparent that the maximum on the plot is at $(1,1)$. If you want to prove it, take the derivative of $2x/\left(1+x^2\right)$, set that to $0$, and solve. You'll get $1$. (And $-1$, which is not of interest.) Plug that back in, and you get $1$. So it's at $(1,1)$. – Mark Adler Jul 09 '20 at 03:54
  • @MarkAdler I see, thank you. So for the angle, would that mean that 60° is the optimal bend angle such that delta-v is maximum for any gravity assist? In that case is it true that for a gravity assist it is the bend angle that determines the change in velocity? And if it is what would be the relationship? - Would an increase in the bend angle cause a higher delta-v for example? – Alexander Ivanov Jul 09 '20 at 06:21
  • You now have the equations you need to answer that. – Mark Adler Jul 09 '20 at 16:28
  • @MarkAdler Right, I got it thank you, I was also wondering where the equation for delta-v that you wrote in your answer comes from? I was trying to find the derivation online but it is very hard to find it mentioned anywhere, and when it is the derivation is not presented. – Alexander Ivanov Jul 13 '20 at 03:43
  • @AlexanderIvanov Since the incoming and outgoing velocity magnitudes are the same, all of the $\Delta V$ comes from the bend angle. Just take two velocity vectors with magnitude $v_\infty$ where the angle between them is $\delta$. Subtract those two vectors. The magnitude of that difference is the $\Delta V$. – Mark Adler Jul 13 '20 at 19:54
  • @MarkAdler Yes I understand it is a triangle but I do not understand how to represent delta-v in terms of the hyperbolic excess velocity, the periapsis altitude and gravitational parameter. This is what I tried to do in my answer but I was not successful. I also realize that I was using the formula for eccentricity in terms of ∞, r and but I also do not know where that comes from. – Alexander Ivanov Jul 14 '20 at 00:56
  • With the triangle you can get the $\Delta V$ in terms of $v_\infty$ and $\delta$. From the formulas in the answer I linked above, you can then get it in terms of of the other parameters you list. – Mark Adler Jul 14 '20 at 01:00
  • Too much for comments. You need to ask a new question. – Mark Adler Jul 14 '20 at 01:13
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As stated in Mark Adler's excellent answer, the maximum DeltaV possible occurs at the following condition:

$V_{\infty \text{ for maximum } \Delta \text{V}} = \sqrt{\mu/r}$.

Tabulated values for this quantity are difficult to find. But the escape velocity at distance $r$ is given by

$\text{Escape Velocity} = \sqrt{2\mu/r}$.

Tablulated values for escape velocities at the surface (although not directly relevant to slingshot) are much easier to find, and all we have to do to convert them is to divide by $\sqrt{2}$.

For example http://nssdc.gsfc.nasa.gov/planetary/factsheet/ gives escape velocities for all the planets of the solar system, plus the moon. It also gives their orbital velocities (around the sun for the planets and around the earth for the moon.)

For the terrestrial planets, the orbital velocity about the sun is several times greater than the planet's escape velocity, and it is possible to conceive a situation where $V_{\infty}$ is equal (or greater than) $\sqrt{\mu/r}$.

On the other hand, for the giant planets (Jupiter, Saturn, Uranus, Neptune) the orbital velocity about the sun is several times less than the planet's escape velocity. It is difficult to conceive a trajectory where a spacecraft from Earth would approach one of these planets with a relative velocity much greater than the orbital velocity of the planet[1]. In practice this may make it difficult for $V_{\infty}$ to get near the limit of $\sqrt{\mu/r}$[1], so it may be difficult to take advantage of all the $\Delta V$ available from the planet's gravity.

However we can get plenty of change of direction from the planet at lower $V_{\infty}$ (potentially up to nearly 180 degrees for the lowest $V_{\infty}$ values.)

[1] EDIT: to qualify further, add "at a convenient angle." See comments (obviously this depends on the exact mission, all missions are different.)

Level River St
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  • @MarkAdler 1. I'm talking about the orbital velocity of the planet around the sun, which is completely unrelated to the escape velocity of the planet. Sorry if that wasn't clear. – Level River St Dec 28 '14 at 09:38
  • @MarkAdler 2. According to the link Jupiter has Vescape=59.5km/s @ 1 bar. Let's use a larger radius where Vescape=49.5km/s to be clear of the atm. The orbital velocity is now 49.5/sqrt(2)=35km/s and the Vinf for max dV is also 49.5/sqrt(2)=35km/s. Now, it's certainly possible that a comet might approach Jupiter with that velocity, or a spacecraft in the future. But as Jupiter is orbiting the sun at 13.1km/s, a craft would have to approach at 35-13.1= 21.9km/s in retrograde orbit to get a Vinf of 35km/s. I cant imagine what kind of mission trajectory would do that with current technology. – Level River St Dec 28 '14 at 10:07
  • @MarkAdler New Horizons got about 4km/s from Jupiter. The biggest Jupiter gravity assist I can find is for Voyager 2, which looks about 10-15km/s (it got a bit more from Saturn too.) If there's any real or proposed spacecraft that got/would get close to 30km/s from Jupiter I'd love to hear about it (finding detailed trajectory info online is not that easy.) The savings are huge (exponential for chemical rockets) but only if the planet sends you in the direction you want to go. I wanted to get a feel for the numbers & compare what can be theoretically achieved with what can usefully be achieved – Level River St Dec 28 '14 at 11:53
  • That was not clear — the edit definitely helps. I have deleted the comments. – Mark Adler Dec 28 '14 at 17:49
  • It is not difficult to conceive of cases where an approach $v_\infty$ to a giant planet is greater than that planet's solar orbit velocity, since Voyager 2 did just that at least twice. At the Jupiter flyby Voyager 2 had already reached solar escape velocity, and was going about 21 km/s relative to the Sun by the time it reached Uranus. Uranus' orbit velocity about the Sun is about 6.8 km/s, so the $v_\infty$ had to be at least 14 km/s. Voyager 2 was going 20 km/s at Neptune, which has a solar orbit velocity of 5.4 km/s. – Mark Adler Dec 28 '14 at 18:39
  • @MarkAdler you are correct! Again the problem seems to be my wording. Voyager 2 was in a radially outward, prograde trajectory when it encountered Uranus. It seems unlikely from the info I've seen that extracting the full dV from Uranus would have added more heliocentric velocity than was actually added, (and it might have reduced it.) Voyager 2 gained about 5km/s from Uranus, about 1/3 of the theoretical dV, and it actually lost heliocentric velocity at Neptune, though there was a big change of plane (I'm not sure if they wanted an out of plane final trajectory or if it was to observe Triton) – Level River St Dec 29 '14 at 01:25
  • It was to observe Triton. That was the primary objective that determined the geometry of the Neptune flyby. – Mark Adler Dec 29 '14 at 03:40