Recent content by goretteux

Thanks to everyone. I didn't quite get answers to all of my questions :P but at least I've got a few solid facts + documents to start figuring out from.

This is going to sound really stupid, but I still gotta ask... if you maintain just enough of a vector opposed to the Moon's equatorial rotation one, so you have constant zero inertial speed... do you just start falling vertically towards the surface?

No that's fine I actually got your point the first time :thumbup: That's the sort of thing I'm trying to avoid myself by posting here, albeit with much smaller circulation involved, but probably the same amount of self-esteem at risk ;) ---------- Post added at 12:44 AM ---------- Previous...

Thanks Andy44 Hmmm you answer brings more questions to my mind I'm afraid... First, I'm not sure I understand why you'd need to accelerate past zero ground speed in the other direction if you were to prepare for a landing? Second, the impact of a planet's rotation on launch makes perfect...

So that's what a retro-grade orbit means... thanks Andy44, I think I might finally be getting somewhere. Would you lose more energy in a retro-grade approach, and get an added energy boost in a pro-grade one, or does direction and intensity of rotation not matter and it's only the intensity of...

Awesome - that's the most useful answer so far (no offense to the other two), thanks Urumpwe. So you're saying the C-shaped trajectory *is* possible, just way way more hazardous than the S-shaped one given the chances of adding another crater on the Moon? One additional question: assuming you...

Well... not as far as I can see it myself: The first picture, if seen from above with the Moon at the top of the picture, depicts an S-shaped trajectory, so the capsule goes behind the Moon from left (leading hemisphere) to right (trailing hemisphere) The 2nd picture shows a "mirrored...

Yup I do remember Apollo 13 and the whole going-behind-the-Moon part... thanks for the information, at least I know now what a free-return trajectory is :) I have to say that still leaves me puzzled as to what that "C-shaped" trajectory is though :/ I've been wondering whether this C/S-shape...

Hi all, Please forgive this simple question if the answer is obvious to anyone around here who isn't a newbie like myself... I've been looking into information about lunar insertion for a few days now for a personal project, and it seems all the illustrations I find on the net about your...