# What is the relationship between the periapsis altitude and the change in velocity in a gravity assist?

In a gravity assist maneuver if I were to decrease the distance from the rocket to the periapsis of the planet it is orbiting (I believe this is called the periapsis altitude but I may be wrong) it would cause a greater change in velocity (meaning a more effective gravity assist) assuming all other variables are constant.

Is this relationship between change in velocity and periapsis altitude linear? Because I originally assumed it would be linear but then I read something that made me think otherwise:

http://maths.dur.ac.uk/~dma0rcj/Psling/sling.pdf

(in the paragraph exactly after equation 10)

From what I understood it says that the change in velocity would be equal at a periapsis altitude of both 2 million km and 87500 km which would indicate that it would not be a linear relationship (maybe a parabola for example).

I am asking because for an essay I am writing I think it would be cool to find the optimal distance the rocket should fly from the planet (ignoring affects from the atmosphere), which I would be able to find if there is a clear maximum on the graph of change in velocity vs periapsis altitude.

• What do you mean with "the distance from the rocket to the periapsis of the planet it is orbiting"? Are you trying to figure out a difference if the planet is near its periapsis in its orbit around sun. Or do you mean the distance the probe passes the planet, where the swing by is made? Commented Jun 18, 2020 at 7:54
• @CallMeTom So by periapsis altitude I mean the distance from the rocket at periapsis to the planet (the closest distance the rocket is to the planet). And so I would make this distance smaller and see the effect it has on the acceleration of the rocket. In this image the distance is r and thats what I would decrease: cdn.discordapp.com/attachments/241626066693783554/… Commented Jun 18, 2020 at 20:57

The link in the question is 404 dead but I think this image from a different question could explain your observation:

(Source)

There is a maximum but this representation is not the scenario you propose in the question as this has $$v_{\infty}$$ as the independent variable, not periapsis distance. The equation is:

$$\Delta V = \frac{2 \cdot v_{\infty}}{1 + \frac{r_p \cdot v_{\infty}^2}{\mu}}$$

So, all else equal, the relationship between $$\Delta V$$ & $$r_p$$ is a $$\frac{1}{1+x}$$ relation. This means that $$\Delta V$$ is always greatest when $$r_p$$ is smallest and has no extra curvature:

(Personal work)

The above graph shows the $$\Delta V$$ (km/s) VS $$r_p$$ (km, height above surface) plot for a representative Earth gravity assist.

Arguably more important than the strength of a gravity assist maneuver is the direction of it. It must send you towards the destination/next planet or else no one will care how much $$\Delta V$$ you got from planet X!