Can you see lit cities at night from the moon?

Factual Questions
21

OTOH, the concept of stars’ magnitude assumes atmospheric interference (twinkling).
Still, the figure for 1 lightbulb assumes looking straight at it, not its reflection off concrete or fog. That is a big distinction.

But I think we should start with what our criteria for “lit cities” is. If a city can’t be more than a point source, will groups of cities also be seen as one object? Take a look at this “earth from space” map: http://indistinctunion.files.wordpress.com/2008/04/p-earth-night.jpg It’s a bit old ('62), but what I like is that it’s not over-exposed or post-processed. If you zoom out in your browser or img program, you can get a real sense of how the cities bunch up. Western Europe, Japan-Korea-Beijing, Eastern North America, etc., bunch up into sizeable areas of light. These areas have huge populations, far more than 10 million. I think the outlines of civilization would be as visible as ridges on the moon.
Btw, anyone else find it odd that we don’t know such a basic fact about our Earth? That after trillions of dollars spent on the space program, none of us, the most nerdy of the human race, knows what it looks like from space?

22

Those elves sure use a lot of lights making them toys.

23

There are gazillions of photos of earth from space. Some of them have been taken by astronauts at various altitudes and many more have been taken from satellites at varying distances.

Just put earth from space into Google images.

24

Keep in mind that the Apollo landings occured during lunar morning, so that ambient sunlight would stop down their irises enough that they wouldn’t be able to detect lights on the dark side of the earth. Even without that, the brightness of the illuminated limb of the earth would have the same effect. (In the famous “Earthrise” photo - I use it for wallpaper on my computer - the earth is at about 5/8 phase.)

One might be able to see lights on earth during a terrestial eclipse (or whatever they call it), but the halo effect of sunlight passing through the earth’s atmosphere that someone mentioned above would probably play hob with that.
I just look at the link Exapno provided and came across this. Note the two white pixels on the dark side of the earth. Whether it’s surface lighting or mere artifacts, I don’t know. To judge from other photos I’ve looked at, probably the latter.
Well, I’m just a Googlin’ fool tonight. Not sure what to make of this image. I’d bet it’s a composite - check out the shadow angles on the lunar surface; the light source appears to be to the left of and slightly behind the photographer. But if the eclipse part of the photo is legit, and it seems to be, this will give you some idea of what it would look like.

25

Nah, it’s all retouched bs that’s faker than a magazine cover.

That’s the problem with NASA. It’s always too self-conscious to show us things as they really are. And in the end, we don’t know anything about what things really look like.

26

Wow. Just wow.

27

While most of those pixels may once have been in an actual photo, I don’t think there is anything remotely close to reality in the whole of it.

Indeed.

28

What the hell are you talking about?

I mean literally what I said. Every commercial “earth at night” picture is a digital composite of an enhanced image. Every picture of a planet has been modified to add in color. Even mars is made redder than it is (as this article points out).

What the fuck are YOU getting at?

29

There is, of course, absolutely no doubt that the picture in BJMoose’s link is a complete fake, for many reasons. First: the image of the earth is the famous Blue Marble photo taken by Apollo 17 in 1972.) Second, the image of the the moon’s surface is from Apollo 15, 16, or 17 (which we know because it shows a lunar rover), and none of those missions (or any of the other Apollo missions, for that matter) was on the moon during a lunar eclipse. Finally, if any other evidence were necessary, the surface of the moon is brightly lit, which obviously wouldn’t be the case during a lunar eclipse!

The Photoshop artist who created the image has assumed that the sun would illuminate a red ring all the way around the earth, similar to the one seen around the moon in a solar eclipse. But I doubt it would look anything like that. The atmosphere is only a couple of miles thick, compared to the Earth’s 8,000-mile diameter. I don’t think that once you, the viewer, were in the earth’s shadow, you would be able to see sunlight refracted through the earth’s atmosphere.

You’re a little confused. During a solar eclipse, there is no “‘black’ disk of the Earth.” It’s the disk of the moon that is casting a shadow on a small portion of the surface of the Earth.

The circumstances you describe would happen during a lunar, not a solar, eclipse. And you (like the Photoshop artist) seem to assume that a lunar eclipse as seen from the moon would appear similar to a solar eclipse as seen from the earth. But, as I said, the earth would not have the same illuminated ring that the moon displays during a solar eclipse because the earth is much bigger than the moon. It completely blots out the sun for a couple of hours, the length of the typical lunar eclipse.

30

national geographic recently did a piece on light pollution, by the way. it’s an interresting read.

31

Not confused, just word choice challenged/dyslexic.

I’ve talked and written about lunar and solar eclispes hundreds of times. This is probably the first time I’ve talked about a solar eclispe as seen from the moon, so the old phraseology habit kicked in.

If you read for content, you can tell I was talking about seeing the sun being eclipsed by the earth as SEEN from the moon, hence the “black” disk of the “earth”.

I think most of the other readers here were able to figure out what I meant :dubious:

32

Even as bright as the U.S. East Coast, Western Europe and Japan have gotten at nighttime nowadays, I still highly doubt that you could see those lights with the naked eye from the lunar surface. Still a long way away, with the entire Earth a relatively small object in the sky.

33

I sure screwed the pooch on this one, didn’t I? While the thing I had linked to was obviously a composite, I convinced myself that the weasel had done no more than cut and paste. Indeed, a few minutes before I had seen a thumbnail of the same “terrestial eclipse” photo (or should I say “artist’s rendering”?) and that helped convince me the eclipse element was legit.

Silly me. It is indeed a diddled-with version of the Blue Marble pic. So I’ve made a new rule for myself: if part of a photo is obviously fake, I have no business trusting any of it.

My apologies all around.

34

Okay, here are some numbers as to whether you could see a red ring of the earths atmosphere as seen from the moon during an eclipse of the sun by the earth.

Two major assumptions here.

Approximately 4 vertical miles of the earths atmosphere contribute significantly to scattered light that heads towards the moon. Yes, our atmosphere is a bit taller than that. My WAG is that the very lowest part scatters absorbs TOO much light, while the upper part doesn’t scatter enough. This number is probably right to within a factor or 2 or so either way.

The second assumption is that within that representative 4 mile tall/deep part of the atmosphere, 1 percent of the suns light gets scattered and makes it back out of the atmosphere. I doubt that number could be more than about 10 times higher, but it could be significantly lower (this is the part I am most unsure about).

As Squink? calculated a 100 Watt lightbulb would be magnitude 26 or so as seen from the moon.

For those that don’t know, a magnitude 4 star is 2.5 times brighter than a 5th magnitude star, and a 3rd is 2.5 times brighter than a 4th and so on and so on. The brightest stars in the sky are around 0 to 1st magnitude. The faintest stars seen naked eye in a decently dark sky away from a city are about 6th magnitude. A difference of 5 magnitudes is exactly a factor of a 100. A 10 magnitude difference is 100 times a 100 and so on an so on. The Sun and full moon are large negative numbers for reference.

The amount of solar insolation is roughly 1000 watts per square meter. 1 percent of that is 10 watts per square meter, or about 1 watt per square foot of scatter light headed back into space.

Earths circumference 24,000 miles. 5000 feet per mile. 4 miles “high” of atmosphere to do the scattering. Thats 2.5 x 10^12 square feet (or watts).

Dividing by the 100 watt bulb, thats 2.5 x 10^10 times brighter than the bulb as seen from the moon. 10^10 times brighter is 25 magnitudes. 2.5 is yet another magnitude still.

So 26 magnitudes more light than the 100 watt bulb. Magnitude 26 minus 26 equals 0.

So, the amount of light from the “earth ring” as seen from the moon would have an integrated magnitude of 0, or about the amount of light from the brightest star in the sky.

Thats fairly bright. Next question is, would that amount of light spread in a thin ring of two degree diameter still be visible? My WAG is probably, but barely.

Though it wouldnt be bright enough to cause problems seeing any cities/metro areas that might or might not be visible.

Did I mess up the math any here?

35

Just saying that if the shots of earth at night that we’ve all seen are* not* time exposures, if the city lights are* not* enhanced in those photos, then I’d guess that from the dark side of the moon, facing the earth during a total lunar eclipse (as seen from the earth), that some large areas of city lights would be visible as specks of light. After all, in the low earth orbit shots, there are plenty of metropolim that take up a good bit of area.

36

Oops

I see I lost a factor of 4 (miles) there in the calcs. So the integrated magnitude is 1.5 times “higher”, making it a total magnitude of -1.5 rather than zero.

Now, we need to look at the fact that the light is “spread” around the sky.

The smallest a naked eye can resolve is about 20 arc seconds (60 seconds in minute and 60 minutes in a degree). Which means even a point source like a star or an extremely narrow line won’t appear any smaller than 20 arc seconds.

With the earth appearing about 2 degrees in diameter. Lets assume the ring is 20 arc seconds wide. So 2 degrees pi 60 *60 equals 400,000 arc seconds.

So, now what about some other objects in the sky?

M13 is a globular cluster in Hecules. It has a magnitude of 6 . It has an apparent diameter of 23 arc seconds. It is just visible to the naked eye. Take 400,000 / pi /12 /12 equals about a 1000.

So, M13 has an angular area 1000 times smaller than the “earth ring”.

A factor of 1000 is 7.5 magnitudes.

Take the integrated magnitude of -1.5 and add 7.5. That equals 6. The magnitude of m13 we started with!

So, the ring has the same “area” brightness as M13. So the ring should be faintly but reliably visible in a dark sky.

Now if my 4 mile “high” atmosphere is too large or 1 percent scattering is too large then the ring might not be visible. Or change the numbers some in the other direction and it should be easily visible (but probably still not reallllllly bright).

So, like the cities themselves, this too apprears to be a borderline case that require more accurate measurements and calculations.

Uggh…I feel like the Mythbusters where they have to proclaim “possible” rather than confirmed or busted!

37

I don’t know. When the sun shines light on the earth, which then reflects the light to the moon, which reflects the light back to the earth, it is perfectly visible.

I’ll be damned if I can’t see city light.

Btw, does anyone know the W/m^2 (in lightbulb equivalents) of earthlight on the surface of the new moon?

38

And can someone explain why the moon’s surface is STILL visible during a lunar eclipse? Is it starlight? The earth’s atmospheric ring of light? Or the damn cities reflecting off?