Reverse Gravity assist results in larger orbit?

[asgkatz]

Can someone explain what is happening here that a reverse gravity assist is ending in a higher energy orbit?

https://imgur.com/Cr1D5OP

Also, from https://wiki.kerbalspaceprogram.com/wiki/Tutorial:Free_Return_Trajectory can someone explain this line: "The aim is for your angular velocity to be as small as possible compared to the planet. This way you won't have enough angular velocity to make a full orbit around Kerbin when you are back in its Zone of Influence."

 

[OHara]

That is a forward gravity assist in your image.  KSP shows your orbit relative to the Mun, around the Mun in its current location.  KSP just shifts your orbit relative to the Mun to the current Mon position, with no rotation.   When you reach the Mun at the bottom of your image, your closest approach Pe to the Mun will be to the left of the Mun.  So, the Mun moving 500m/s or so to the right pulls you along with it and gives a forward gravity assist.

KSP offers other modes for drawing these partial orbits, in the in-game settings options.  Most are the obvious choices, but the last one 'dynamic' makes no physical sense.  The default 'local to body' is the most useful, once you get used to the idea that the body will have moved before you reach it.

The wiki on free-return trajectory is just saying that when you exit the Mun's influence you want to be falling nearly straight at Kerbin -- so they mean you need small 'angular velocity' around Kerbin which is probably obvious; I don't see the point of bringing it up right there.

That means that as you leave the Mun you need to be going left relative to the Mun at nearly its 500m/s orbital velocity.  If you meet the Mun at the bottom of the KSP map-screen, then, your orbit relative to the Mun will look like an upright very flat 'V' shape, with the Pe directly below the Mun.  I find the free-return rather tricky to get in KSP.

[king of nowhere]

In addition to that, i have the feeling that your orbit will be highly inclined. check that your mun insertion is equatorial rather than polar, because if you enter in a polar orbit you will exit with a high inclination in any case

[Snark]

9 hours ago, asgkatz said:

Can someone explain what is happening here that a reverse gravity assist is ending in a higher energy orbit?

As @OHara points out, what you're doing isn't actually a reverse gravity assist.  It's just that the somewhat misleading way it's displayed in map view is tricking you into thinking it is.

OHara gives a pretty good explanation above, but perhaps a picture or two might help illustrate it.  Here's the image you shared:

Spoiler

It really looks like a reverse assist, right?  "See, I approach the Mun heading from Kerbin in the direction relative to the Mun, and then it deflects me in the direction.  Reverse gravity assist, yay!"

...Except that that's not actually what happens.  Here's my attempt at rejiggering the image to give a better picture of the real story:

Spoiler

(I show the Mun in two places, as those are approximately where it is when you enter and exit its SoI.)

...In short:  You're not approaching the Mun and getting deflected .  Rather, what you're actually doing is approaching the Mun and getting deflected with a strong component.

The best way to visualize "what's the net velocity change that I'm going to get out of a gravity assist" is to picture a vector that splits the difference between the approach and departure directions, like this:

So to sum it all up in one picture,

As you can see... it's pretty clear that what you did was give your craft a pretty strong boost, thus the higher-energy orbit.

So, basically, there were two things that were probably tripping you up, here:

• as OHara explains, the map view shows the Mun where it is now, but you need to picture where the Mun will be at the time you encounter it, and
• you may be wrongly picturing the direction of the assist-- you're probably thinking of it in terms of the departure vector (red arrow I've drawn), whereas really it's the green arrow instead.

[Guest]

Also, when you have an encounter, if you click on the target planet (assuming it's targeted) and select “focus view” it switches the camera to the target planet but where the encounter will take place in the future rather than now. I've found this usually displays things in a friendly way so you can get a better sense of orientation etc.

[asgkatz]

16 hours ago, Snark said:

As @OHara points out, what you're doing isn't actually a reverse gravity assist.  It's just that the somewhat misleading way it's displayed in map view is tricking you into thinking it is.

OHara gives a pretty good explanation above, but perhaps a picture or two might help illustrate it.  Here's the image you shared:

  Hide contents

It really looks like a reverse assist, right?  "See, I approach the Mun heading from Kerbin in the direction relative to the Mun, and then it deflects me in the direction.  Reverse gravity assist, yay!"

...Except that that's not actually what happens.  Here's my attempt at rejiggering the image to give a better picture of the real story:

  Hide contents

(I show the Mun in two places, as those are approximately where it is when you enter and exit its SoI.)

...In short:  You're not approaching the Mun and getting deflected .  Rather, what you're actually doing is approaching the Mun and getting deflected with a strong component.

The best way to visualize "what's the net velocity change that I'm going to get out of a gravity assist" is to picture a vector that splits the difference between the approach and departure directions, like this:

So to sum it all up in one picture,

As you can see... it's pretty clear that what you did was give your craft a pretty strong boost, thus the higher-energy orbit.

So, basically, there were two things that were probably tripping you up, here:

• as OHara explains, the map view shows the Mun where it is now, but you need to picture where the Mun will be at the time you encounter it, and
• you may be wrongly picturing the direction of the assist-- you're probably thinking of it in terms of the departure vector (red arrow I've drawn), whereas really it's the green arrow instead.

Excellent explanations guys. I appreciate it. Any thoughts on the second question?

[Spricigo]

2 minutes ago, asgkatz said:

Any thoughts on the second question?

Well, I don't see a question. Only the quote from the wiki.

It seem you are, in a nutshell, attempting this: 

does it helps?