Returning from retrograde laythe orbit?

[Space Nerd]

My ship is in a high retrograde elliptical laythe orbit,and it has around 1200 dv.

I checked before, getting into a kerbin intersection only takes around 900 dv, but would I got into a retrograde solar orbit and hit kerbin too fast?

[VoidSquid]

No, you won't, no worries  

As per the wiki, the orbital velocity of the Jool system around the sun is between 3927 and 4341 m/s, that's quite a number to break  

Just make sure you enter Kerbin's SOI in the correct direction, with it's own rotation, not against.

[Fraktal]

If I recall correctly, going inwards in the system is most efficient if you:

1. Drop out of your current SOI at the trailing end (ie. the part opposite of the direction the body is moving in).
2. Start your burn about 90° away from your orbital position being directly between the target and the body you're currently orbiting.

I think it's best to wait until Laythe is eclipsing Jool and then start your burn when your orbital position around Laythe is about 90° away from being directly between Laythe and Jool so that you only need to burn once, but I don't know if it's most efficient like that (basically, nudge your maneuver node back and forth until you get the lowest periapse around the sun with the dV you're burning). All I know is that once you're out of Jool's SOI, you are now orbiting the sun in a prograde direction and slowly falling inwards, so you only need to burn retrograde to drop down to Kerbin orbit. But as @VoidSquid says, the difference between hitting Kerbin's atmosphere prograde or retrograde is only a few tenths, if not hundredths, of a m/s at that point.

If I'm wrong with any of this, everyone can feel free to correct me because I've never been to Jool myself and am extrapolating this from a Kerbin->Eve rendezvous.

[Geschosskopf]

29 minutes ago, Space Nerd said:

My ship is in a high retrograde elliptical laythe orbit,and it has around 1200 dv.

I checked before, getting into a kerbin intersection only takes around 900 dv, but would I got into a retrograde solar orbit and hit kerbin too fast?

When you're in orbit around a secondary or tertiary body, NONE of your orbital parameters there have any significant impact on your subsequent orbit around the primary, regardless of which direction you leave your current orbit in.  Because you're orbiting some body, your ship is riding around the primary with that body, so already has all that orbital velocity in that direction.  Escape velocity from that body is insignificant compared to this.  Thus, leaving a planet or moon's SOI in any direction puts you into orbit around the primary going in the same direction and with nearly the same inclination as the body you just left.

You can see this easily at Mun.  Get into a retrograde Munar orbit and return to Kerbin.   You'll still have a prograde orbit around Kerbin on the way home.  Or note the trajectories of asteroids that will enter Kerbin's SOI.  While they're in solar orbit, they have nearly the same inclination as Kerbin but usually have a huge inclination as they fly through Kerbin's SOI.

 

[VoidSquid]

The reason I mentioned to enter the Kerbin SOI in the "correct" direction at all: Once I had a mission returning from Eve, perfectly went to a 120x120km orbit around Kerbin. Nex step: rendezvous with my orbital station to transfer the science to be processed in the labs. Then: WHAT??? Almost 5000 m/s required???

Stupid me had gone to a clockwise orbit...  

[Zhetaan]

1 hour ago, Space Nerd said:

My ship is in a high retrograde elliptical laythe orbit,and it has around 1200 dv.

I checked before, getting into a kerbin intersection only takes around 900 dv, but would I got into a retrograde solar orbit and hit kerbin too fast?

Retrograde Laythe orbit is still prograde Jool orbit.  Because your vessel is in orbit of Laythe, and Laythe is in orbit of Jool, your vessel therefore must orbit Jool at an average orbital velocity equal to Laythe's orbital velocity.  If you were to plot it, then it would look like a slightly odd sine wave orbiting Jool.  If you were to plot it relative to the sun, then it would look like a regular orbit because the game doesn't have enough resolution to show the variations.  Which, now that I read it again, is exactly what @Geschosskopf already said.

Remember that if you look at Laythe from the south pole instead of the north, a retrograde orbit looks like a prograde one.  The only practical difference is that it costs a few extra m/s to get into that orbit because you are moving counter to Laythe's rather miserable 60 m/s rotational velocity.

Everything is relative.

This video explains the idea better than I can, and with pictures:

The relevant part begins at 1:57.

Edited August 7, 2019 by Zhetaan

[VoidSquid]

21 minutes ago, Zhetaan said:

Because your vessel is in orbit of Laythe, and Laythe is in orbit of Jool, your vessel therefore must orbit Jool at an average orbital velocity equal to Laythe's orbital velocity. 

...and because your vessel is orbiting Jool, your vessel therefore must orbit Kerbol at an average orbital velocity equal to Jool's orbital velocity

[Space Nerd]

Thanks!