What I mean is, if I was standing smack in the middle of the Sea of Tranquility, right next to the flag and the footprint (that's where it is, right?), with Earth on the horizon, and the sun directly overhead, would the Milky Way also be visible, and if so, in what direction?

Also, how would it be oriented in the sky? Assuming the Earth is lit from the sun in such a way that it's "top half" is in daylight, since the sun is overhead, and the Milky Way is visible, how would it look? I live in the United States, and I've seen it crossing the sky in a somewhat horizontal orientation, so how would it look from the laditude of the moon where the Sea of Tranquility is?

Does anybody really know? Is there a picture of it from the Moon that didn't suffer from the low exposure that drowns out stars and makes people think it was all fake? I can't imagine they didn't take pictures of the sky from the moon, considering how many more stars there would be...

Thanks.

It'd be visible from the moon... at night.

I suspect that they couldn't take pictures of the sky from the moon (weren't the cameras fixed to the front of their suits, so they'd basically have to be lying down to take a picture of the sky?)

As for how it'd appear from there, get a simulator program like Celestia or Orbiter and go take a look . Right now, night has just fallen on the Apollo 11 landing site, and in Celestia the milky way is oriented diagonally (about 45 degrees) as seen from there.

No. It is far too dim to see under those conditions. After all, it is too dim on Earth with just a bit of light pollution at night, let alone daytime. I have to go well out of town before I can see it. On the moon, you would have to be well shaded from direct and reflected sunlight and let your eyes adjust.

Assuming it was visible, and assuming I understand your question, the angle relative the Earth's terminator would be pretty much the same the angle relative the Moon's terminator as seen from Earth.

You couldn't take an optical photo of stars on the moon with the Earth or the Lunar surface in the frame without massively overexposing the Earth or the lunar surface. If you avoided direct or reflected sunlight, the Milky Way would look about the same from the moon as from the Earth. There wouldn't be many more stars seen by the (protected) naked eye on the moon, versus the naked eye on Earth somewhere with little light pollution. At best, the brightness would be maybe 10-20% better. The real advantage of lack of atmosphere is not brightness, but lack of distortion. That's why you can get better images with a big telescope without using adaptive optics in space.

The upshot is that optical photos without a telescope wouldn't get you anything on the Moon you couldn't get on the Earth. However, there were some photos taken of the sky on the Apollo 16 mission in the Far UV (and note that this camera was carefully shaded). Unlike optical photos, these couldn't be taken through the Earth's atmosphere, so did have scientific merit.

Here's a post about a Far UV image taken on the moon:

Why not fake the stars?

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The Leif Ericson Cruiser

But on Earth, the real problem is light sccattering off of the atmosphere, and destroying contrast. On the moon, even in full daylight, the airless sky should be pitch black, and provide high contrast. I should think that the Milky Way should be visible, even with the sun in the sky, because there is no way to scatter the sunlight around the sky. Of course, it's a lot easier without the sun, but I think not impossible.

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Your difficulty then is with pupillary accommodation: bright light sources cause your pupils to constrict, significantly reducing the amount of light hitting the retina. This makes dim objects (which would be easily visible through a dilated pupil) impossible to see.
If you walk out of a building lit by electric light into a moonless night on Earth, it still takes a little time before you can see the Milky Way: your pupils need time to dilate properly.
Now, the surface of the moon in daylight is much brighter than most buildings lit by artificial light; so I'd suggest that to see the Milky Way from the moon while the sun was up, you'd need to stand somewhere where you had no view of any sunlit surfaces.

Grant Hutchison

On the basis of my own simulations with a good CRT monitor, I concluded that the night view from the Moon would be much improved over the best possible conditions on Earth's surface, because the natural airglow would be eliminated. I used a sensitive exposure meter to measure the screen brightness, and when it got too dark for the meter, I compared successively darker swatches with a neutral density filter over the brighter one. I was able to set it to the published value for the airglow, and that was clearly visible in a really dark room. When I turned the background down all the way, a swatch of the simulated airglow stood out boldly, and I could see swatches that were 1/8 as bright, as indicated by the filter.

I generated the sky samples with JASC (now Corel) PaintShopPro software. I calibrated the brightness scale by visually matching the brightness of one-pixel simulated stars with actual stars when I could see Orion out the window by my desk. The airglow values were obtained from Sky and Telescope.

My conclusion: Faint extremities of the Milky Way and the zodiacal light that are barely visible through the airglow would show much more contrast at night on the Moon.

I have only one response to that:

Lovely.

Grant Hutchison

It's also dim. Astronauts were able to see some stars when avoiding direct or reflected light from the sun and when their eyes had a chance to adjust. But I hadn't accounted for natural airglow, so when you could see it, it could well look better on the moon.

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The Leif Ericson Cruiser

And it isn't just your pupils. There is the general issue of dark adaptation:

The increase in sensitivity of the eye to detection of light that occurs in the dark is called "dark-adaptation." The pupil dilates, and both neural (largely unknown) and biochemical changes in the retina occur. In darkness, after exposure to bright light which bleaches the visual photopigments, there is an initial hundredfold increase in sensitivity following an exponential time course that reaches a plateau after 5 to 9 minutes. This initial phase is attributed to regeneration of photo-sensitive pigments in the cones. Thereafter, there is a 1000 to 100,000 times increase in sensitivity following a slower exponential time course that reaches a plateau in 30 to 40 minutes.

This second phase is attributed to regeneration of rhodopsin in the rods. In addition to rhodopsin regeneration, neural changes increase sensitivity further. Dark-adaptation is delayed by prolonged exposure to bright light. Thus, it does take longer to reach maximum adaptation at night after a day in bright sunshine. When fully dark-adapted, the retina is about 100,000 times more sensitive to light than when light-adapted. Exposure of the dark-adapted eye to bright light results in a marked decrease in sensitivity involving two changes:

(1) a neural process that is completed in about 0.05 second, and
(2) a slower process, apparently involving the uncoupling of retinal and opsin in rhodopsin, occurring in about 1 minute.

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The Leif Ericson Cruiser

Thanks everyone. Hornblower, you didn't need to go to all that trouble on my account, but thanks anyway. I guess that's the nature of this board that most of the members are naturally curious about stuff like that anyway, and all I had to do was give you the idea.

No trouble at all. I did the work for my own amusement two years ago. Reporting on it here took about five minutes.

A couple of everyday examples of the adaptation problem:

You're working outside on a bright day. You walk inside the house for some iced tea and find the interior astonishingly dim, even with light streaming in from the windows and ceiling lights on. The lights, if you look at them directly, seem oddly underpowered.

You walk into a darkened theater on a sunny day. You can see the relatively bright screen, but little else. So you wait a few minutes before attempting to find a seat. Even then, it's pretty dark, but when you go out for food or a bathroom break an hour later, you have absolutely no problem looking around.

That's pretty much what the astronauts would face. Properly shaded from direct and indirect sunlight, they could see some of the brighter stars in the sky without much adaptation time. After that, it gets harder.

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The Leif Ericson Cruiser

You're quite welcome. The general question about seeing stars on the Moon comes up quite a bit, but as you can see, there are still details that we can all learn about, as with the points Hornblower brought up. Anyway, this is the Q&A section, and it's here for space and astronomy questions.

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The Leif Ericson Cruiser

Thank you for the description too about the time delayed adaption of the human eye. With CCT photography it is the capture of individual photons but would a more organic approach using the cone and rod structure of eyes be a way to get a more instant picture?

I was thinking along the lines of some of the best night vision animals say cats that while they also use hearing and whiskers are very visual creatures.

Cheers

and if I were located on the lunar far side rather than smack in the middle of the Sea of Tranquility, the milky way would be easily visible aroud the time of full moon, when both the Sun and Earth would be absent from my sky. Just make sure the lights from the LM are out.

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Just tell the last one to leave to turn out the lights. Even if his name is Bart.




BTW, high altitude/dark sky conditions here on Earth are probably close to what one would see on the Moon at night. I recall such conditions in Arizona and Colorado where the seeing was close to seventh magnitude. What was really remarkable was not only the detail available in the Milky Way, especially down near Scorpius and Sagittarius, but the incredible number of faint satellites coursing overhead.

ReIt would be visible as pointed out above if one shielded one's eyes from the Sun and bright lunar surface.

However the direction would depend on the time of year (Earth) and local lunar time. There are times where the Milky Way lies along the lunar horizon and would be hard to see, just as it is hard to see from Earth during autumn in the northern hemisphere, when it displays a similar aspect.

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Okay, here's another one:


I'm assuming that, when it's a full moon as seen from Earth, and you were standing in the middle of the "full" part, in other words, the illuminated part, the sun would be directly overhead.

I am also assuming that from a particular vantage point on the lunar surface, such as the Sea of Tranquility, the Earth would always be in the same place in the sky, since the moon is tidally locked to the Earth.

I'm also assuming that most, if not all, photos from the moon were taken from the Sea of Tranquility.

In these pictures, unless I don't remember properly, the Earth is near the lunar horizon, and has it's "top half" lit.

The question is this:

If one were to stand in the sea of Tranquility, near the landing site, where would the earth be? And sinces it's "noon" on the moon, and a "full moon" as seen from Earth, where would the sun be?

Also, where on the moon exactly, is the Sea of Tranquility? Perhaps the sun would not be directly overhead because perhaps the Sea of Tranquility is not on the equator. I seem to recall it being somehwere on the northern hemisphere, so then, during a "full moon" would the sun be overhead, behind, or maybe even right over the Earth, in the sky?

More or less: the moon looks full over a moderate range of sun positions, but the sun would certainly be high in the sky.
(In fact, the moon can't be perfectly full as seen from the Earth, because then it would lie in Earth's shadow, and be eclipsed.)

More or less: the moon's orbit is not perfectly circular, and its axis isn't at right angles to its orbit, so the Earth will wobble a bit in the moon's sky.

Not so. There were six moon landings at various locations across the Earth-facing side of the moon, though all were within about 25º of the equator and about 30º of the central meridian. Pictures were also taken from orbit.

That sounds like one of the "earthrise" pictures taken from lunar orbit.

Earth would be a little west of overhead, since the Apollo 11 landing site lies east of the moon's central meridian but more or less on the equator.
The sun position depends on what you mean by "noon on the moon", since it's only noon along one meridian at a time. If it's noon on the lunar central meridian, then the sun is west of overhead at the Apollo 11 site. If it's noon at the Apollo 11 site, the sun is pretty much overhead. Because the moon's orbit is tilted at about five degrees relative to the plane of the Earth-Moon system's orbit around the sun, and because the moon's axis is tilted a degree or so relative to its orbit, the sun position will vary by a little over six degrees north or south with the lunar "seasons".

The Apollo 11 landing site, rather grandly called Statio Tranquillitatis by the USGS, is at 0.8ºN 23.5ºE. It's in the extreme southwestern corner of the Sea of Tranquility, which spans about 20º in latitude and longitude.
During a full moon the Sun will always be close to the Earth in the moon's sky.

(BTW: You might consider starting a new thread for a new question. Give it an appropriate title, and you'll more easily attract the attention of the people who are most able to answer it.)

Grant Hutchison

The main point is true, but I generally figure that the Milky Way lies along the horizon in early spring. This is how ToSeek explained his answer to question four of the third astronomy challenge.

BTW can I insert a shameless plug for the fifth annual astronomy challenge, registering in the Astronomy forum

Here is a map of the Moon showing all of the Apollo landing sites.

http://www.nasm.si.edu/collections/i...gSitesMaps.jpg

The Apollo 11 site is near the equator and about 25 degrees of longitude from the sub-Earth point. The Sun was low in the east at the time, as seen from the site. A exercise in geometry shows that the Earth was about 65 degrees above the western horizon (plus or minus a few, depending on libration), showing a gibbous phase. Clearly, it would not have been in the field of view of any photo of the moonscape. The familiar photos of the Earth near the lunar horizon were taken from the command module in orbit.

If we were there at the time of full Moon as seen from Earth, we would see the Sun and perhaps a hairline crescent Earth close together at that high elevation. The latter would be hard to see in the glare unless we screen the Sun off with a suitable barrier. If we happened to be there during a total lunar eclipse, it would look like a ring of fire.

Hmmm.
I just posted a daft thing about plane of the galaxy and the ecliptic, realized it was daft, and deleted it with a note to that effect. But the note doesn't seem to have appeared.
So this is just a message to anyone who read my post during its brief appearance or in their e-mail notification, and who has come to tell me that I was talking nonsense:

I know that!

Grant Hutchison

The Earth has phases much like the moon when Earth is observed from the moon. My guess is a wide angle exposure of several minutes would show the darkened Earth and nearby stars. The dark of Earth only occurs a few days per year, the astronauts likely were busy with more important matters at this opportunity and may not have had the equipment to make a quality time exposure. The Earth does appear as a very thin crescent once per month as viewed from the Moon. The sun is approximately the same direction as the Earth during the dark of Earth, so the camera optics would produce some artifacts, unless the camera could be shaded without blocking the view of the Earth, in my opinion. Neil