Position of LM Aquarius above Earth at Separation

[onebigmonkey]

collectSpace user LM-12 spotted that in magazine 59 images of the LM separation of Apollo 13 at 141:30, you can see Earth reflected in the LM window.

Naturally I want to see if I can pinpoint what is in shot to add it to my site, but I'm having a hard time getting my head round the view.

I know that the re-entry trajectory to splashdown was all in the southern hemisphere, and I also know that the only things in daylight would be western Africa, western Europe and Scandinavia, but exactly how clear the view is of those depends on where exactly Aquarius was above the Earth.

For info, here is the view from the moon of Earth at separation



and here is Earth in the LM window, reversed to correct the mirror image:



Anyone have any information that can add to the story?

[QuietElite]

I found a Apollo 13 Mission Report:
https://www.hq.nasa.gov/alsj/a13/A13_MissionReport.pdf
According to the numbers in Table 4-II. the position during the LM undocking was 1.23 S 77.55 E at an altitude of 11257.48 miles (18117,16 km)

[onebigmonkey]

I found a Apollo 13 Mission Report:
https://www.hq.nasa.gov/alsj/a13/A13_MissionReport.pdf
According to the numbers in Table 4-II. the position during the LM undocking was 1.23 S 77.55 E at an altitude of 11257.48 miles (18117,16 km)

Perfect, thanks

[bknight]

I found a Apollo 13 Mission Report:
https://www.hq.nasa.gov/alsj/a13/A13_MissionReport.pdf
According to the numbers in Table 4-II. the position during the LM undocking was 1.23 S 77.55 E at an altitude of 11257.48 miles (18117,16 km)

Perfect, thanks

Not going into the tables, was the trajectory from west to east relative to the Earth?

[gwiz]

Might have more luck getting a match with a weather map for the day rather than the geography.

[onebigmonkey]

Might have more luck getting a match with a weather map for the day rather than the geography.

I have that, but it helps to narrow things down a lot if you know where you are looking Looking at the content of the image, it looks to me as if we are dealnig with somewhere around equatorial Africa.

The real issue I have now is how large an area that brief glimpse of Earth covers!

Not going into the tables, was the trajectory from west to east relative to the Earth?

That's the other thing I'm trying to wrap my head around! The previous 2 sets of images showing earth were taken at around 06:30 and 11:00 on the same day, showing Australia and then the Indian subcontinent respectively, so Africa appearing in view is entirely consistent with that.

However, given that the actual re-entry path was from west to east, and the CM should have been facing blunt end backwards (unless they spun around later), it also seems logical that the view from Aquarius is actually towards the East. I'm now checking for likely looking weather in the mid-Pacific!, but unfortunately the timings of the satellite passes make this area about 6 to7 hours out of date relative to the time of the photograph.

[bknight]

From your first image the target location looked about 4-6 hours in darkness.  The idea of viewing Africa and then India would be reasonable.  Then there should be no more images as the re-entry shield would be positioned in the direction of the trajectory.

[BazBear]

An aside, and possibly a silly question, but are all human spaceflight (re)entries, more or less, west to east?

[12oh2alarm]

An aside, and possibly a silly question, but are all human spaceflight (re)entries, more or less, west to east?

The reentry direction is always the same as the lift-off direction simply because you can't turn a prograde orbit into a retrograde orbit.
For orbits, that is. If you stop somewhere to pick up rocks, things could in theory be different, since you have another choice for the lift-off direction.
However, even the Moon is in an Earth orbit, so I'd expect delta-v (fuel) requirements to be higher.
Think of it this way: lifting off to the east you have a free initial delta-v of several 100m/s because Earth spins that way.
Re-entering to the east also has the benefit of the same delta-v less to get rid of.

In other words, for lift-off and reentry you say "Thank you for spinning, dear planet, makes my life easier."

Gedankenexperiment: What if the Earth at the Equator spun with almost escape velocity? You could enter orbit with your car. But seeing the whole equator, even at 8km/sec relative to CoE, would take as long as driving 40.000km in a car. Now if you were to go into a retrograde orbit, you'd need 16km/s, and that would be your reentry speed. Once around Earth would take only some 40 minutes.
 

[bknight]

An aside, and possibly a silly question, but are all human spaceflight (re)entries, more or less, west to east?

I wasn't thinking to a conclusion when I asked the question, and yes I agree with your analysis of re-entries.
My main reason for asking the question was, again, looking at the Earth's image the landing target in the Pacific was 4-6 hours from daylight.

Silly of me to not think about the requirements for landing in daylight.

[ka9q]

Yes, relative to the earth all crewed flight trajectories have been prograde (west to east).

All Apollo lunar trajectories were retrograde at the moon (east to west). This made the free return trajectory possible; had a prograde approach been followed, then without a burn the spacecraft would have been slingshotted into interplanetary space. This was in fact done intentionally with the S-IVBs from Apollos 8, 10, 11 and 12 by slowing them down after separation to give the moon time to cross their path before they got there.

Interplanetary trajectories that return to the earth for a slingshot may follow either prograde or retrograde passes depending on the direction they're approaching and whether they need energy from the earth to reach a superior planet (Mars, Jupiter and beyond) or to dump energy into the earth to reach an inferior planet like Venus or Mercury.

[VQ]

An aside, and possibly a silly question, but are all human spaceflight (re)entries, more or less, west to east?

The reentry direction is always the same as the lift-off direction simply because you can't turn a prograde orbit into a retrograde orbit.

Unless, of course, you leave earth's sphere of influence. I believe a lunar free-return trajectory to a retrograde earth reentry is possible. Also, reversing a highly eccentric orbit at apogee, even without a flyby, has a relatively low dV requirement.