# How much delta-v can we squeeze out of a gravitational slingshot and what factors limit it?

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?

• 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! – user6738 Dec 25 '14 at 19:47
• I guess it was worth the waiting :) – user6738 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 𝑟=∞". I understand that TA = 0 at periapsis but I dont understand why this implies that the max bending is therefore twice the TA at 𝑟=∞. If you could explain this for me it would be greatly appreciated! – Alexander Ivanov Jul 9 '20 at 1:54

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.

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°.

• 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
• @JerardPuckett If it helps, $v_\infty$ is aka hyperbolic excess velocity. – TildalWave Dec 28 '14 at 1:15
• Can anyone explain what $v_\infty$, and $\mu$ are? – sampathsris Jun 12 '18 at 8:31
• $v_\infty$ 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

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.)

• @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 9:38
• @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