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From what I understand, the TLI burn should be around the Moon's antipode projection on Earth. I get that. What I don't understand is why, according to this picture

https://history.nasa.gov/afj/launchwindow/figs/Fig%209.png

the antipode was at 10 degrees south for the 16th of July 1969? How was this calculated?

Thanks.

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  • $\begingroup$ Why do you think there is an issue with an antipode 10 degrees off the equator? The Moon has an orbital inclination of about 5 degrees to the ecliptic plane, Depending on where it is in it's nodal precession cycle, the antipode could be as much as 28 degrees off the equator. $\endgroup$
    – notovny
    Jul 20, 2023 at 21:01
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    $\begingroup$ I'm confused. According to history.nasa.gov/afj/launchwindow/lw1.html "the injection (TLI) must occur very close to the extension of the earth-moon line at the time of the spacecrafts lunar arrival". But lunar orbit insertion for Apollo 11 occured Jul-19 17:21 UTC (see here) and the lunar declination was 10°N ~Jul-18 8:00, see i.sstatic.net/lsVVj.png which I created using Horizons data. $\endgroup$
    – PM 2Ring
    Jul 21, 2023 at 13:18
  • $\begingroup$ Sorry, make that Jul-18 9:00 i.sstatic.net/5XG6T.png You can make declination plots using the link near the end of my answer space.stackexchange.com/a/61065/38535 The ID for the Moon is 301 $\endgroup$
    – PM 2Ring
    Jul 21, 2023 at 13:32
  • $\begingroup$ @PM2Ring thank you for the analysis! That's exactly the information I was trying to find. But I guess you looked the wrong time for TLI. The TLI occured on Jul-16 16:22 UTC (on the same link you posted). Unfortunately when I tried using the Horizons app, I could not make a Moon inclination plot like the one you did.. $\endgroup$
    – lester289
    Jul 22, 2023 at 17:33
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    $\begingroup$ Yes, the TLI was at Jul-16 16:22 UTC, but the important antipode is at the time of the spacecraft's lunar arrival, so somewhere around the time it inserts into the lunar orbit. The Moon's antipode latitude is just the opposite of the Moon's declination. So a 10°S antipode corresponds to a 10°N declination. $\endgroup$
    – PM 2Ring
    Jul 22, 2023 at 18:27

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