# Repeated slingshot encounters with the same planet

I am curious to know how you calculate a slingshot direction change

In both angle you are able to change per encounter (degrees) & the velocity ($m/s$) When $V_{inf}$ is higher than the planets escape velocity by a significant amount.

On the first encounter, the craft is climbing up to the planets orbit from a perfect transfer orbit from a planet closer to the sun (pure retrograde).

The end goal is to be in a pure prograde trajectory after the last encounter with the planet.

I want to know, if the planet escape velocity is say only half of the Infinite encounter Velocity

$$V_{inf}= 10 km/s$$ $$V_{esc}= 5 km/s$$

Prograde initial angle = $180 \deg$

Prograde final angle = $0 \deg$

How many $m/s$ are we allowed to change at a time?

How many encounters with the same planet would be nesscary to completely change the direction of the Encounter velocity from retrograde to prograde assuming we change direction from Retrograde through Radial out to Prograde?

Would the last encounter have to be done farther away with an even lower escape velocity at closest approach to avoid going into the radial in direction?

If so; the escape velocity on the final encounter would be close to how many $m/s$?

Initial velocities;

$$V_{Pro}= -10km/s$$ $$V_{Rad}= 0km/s$$

$$RadAngle= 180 degrees$$

velocity after 1st encounter with planet

$$V_{1Pro}= ?$$ $$V_{1Rad}= ?$$

$$RadAngle= ?$$

• going from retrograde to prograde will be a big problem because in the interim you'll have very thin, elongated orbits, with very low perihelium. And that means very high temperatures. – SF. Dec 26 '15 at 17:38
• I Think I can word the problem slightly better. You have a spacecraft coming from an Inner planet to the outer planet (outer planet in a circular orbit around the sun) , when the craft get's to it's furthest point from the sun; It is at my planet traveling 10 km/sec slower with respect to the sun than the outer planet, my goal is to get the craft at the end of my final encounter with it; traveling 10 km/sec faster with respect to the sun (this is intended to not be an escape trajectory from the sun). Given my planet has an escape velocity of 5 km/sec at the closest encounter point {char limit} – Tone Dec 27 '15 at 0:33
• How many encounters would I need to change the velocity of the space craft by that much; given that it does not encounter anything else during it's voyage, and would the final encounter have to be at a point where the escape velocity is lower for it not to be overdone? How much would the angle of deflection and/or velocity change in a single encounter – Tone Dec 27 '15 at 0:39
• Oh, this is better. So you're essentially performing a Hohmann transfer but intending to end it with gravity assist instead of a capture burn (of 10km/s), and then keep repeating gravity assists with the same planet until you have an equivalent of 10km/s ejection burn in solar-prograde direction. This becomes more doable. Highly dependent on planet size, density and orbit - theoretically you can "steal" half of planet's orbital speed with one gravity assist, if the planet is a point mass and you perform a gravity assist of almost zero radius. This is obviously impossible in reality. – SF. Dec 27 '15 at 1:06

If I understand your question right, you want to calculate how many fly-bys of the planet you must perform to turn your velocity vector 180 degrees around. If the spacecraft does not interact with any other object between the fly-bys, the $v_{inf}$ (encounter velocity) stays constant. The minimum number of encounters is therefore calculated by dividing 180 degrees on the maximum turn angle. (Ceil the resulting number)
$$Angle =\mathbf{2 \; \operatorname{sin}^{-1} \left( \frac{GM}{P \; v_{inf}^{2} + GM} \right)}$$
Where $P$ is the distance of periapsis, often limited by the radius of the planet or its atmosphere, $GM$ is the product of the planets mass and the gravitational constant, and $v_{inf}$ is the velocity at infinity.