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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.

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  • $\begingroup$ 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? $\endgroup$
    – CallMeTom
    Commented Jun 18, 2020 at 7:54
  • $\begingroup$ @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/… $\endgroup$ Commented Jun 18, 2020 at 20:57

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

flyby delta V

(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:

delta v vs rp

(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!

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