Can you see man-made lights on the dark side of the Earth from the surface of the moon with the naked eye?
This question is prompted from a comment on this answer:
Could a person on the surface of the moon see man-made lights on the dark side of the Earth with the naked eye? If not, how much magnification would be required to do see them?
The closest evidence I could find from a quick search was this picture from Salt Lake City Schools' Website:
The dark part of Earth is a bit hard to see here, though it can be seen more easily in the full resolution image. You can certainly see something toward the bottom edge in the dark area, but I can't tell for sure if it is lights from Southeast Asia/Oceania or something else.Edit: As pointed out in the comments, the image of Earth in this photo wasn't actually taken from the moon, but from NASA's Terra and Aqua satellites and combined with the lunar imagery in a 2002 visualization created by NASA to commemorate the 30th anniversary of Apollo 17.
@HagenvonEitzen - agreed. It has to be an approximation intended for illustration. There are far too few clouds and the colors are too bright. It seems to be a very cleaned up and altered version of the classic photo from Apollo 17 - http://commons.wikimedia.org/wiki/File:The_Earth_seen_from_Apollo_17_with_transparent_background.png
@briligg The image might indeed be doctored, but it's not the one from that link. Aside from the rotation, substantially more of Eastern Asia is visible than in the one you linked. It's centered substantially further out to sea in the Indian Ocean, whereas that one is centered between Africa and Madagascar.
Unfortunately, the SLC schools' website doesn't state the original source of the image, so it is dubious. I would be interested if anyone can find the original source. It does not appear to be taken from the 'Earthrise' photo from Apollo 8, as that one is centered over the Atlantic off the opposite coast of Africa.
@reirab - Yes, i see your point. In which case it may be a complete invention. I don't know how many photos of the whole earth there are - i'm inclined to ask on Space Exploration - but i think there are only the ones taken on the Apollo missions. Maybe some from SELENE. But in looking, i found this - http://visibleearth.nasa.gov/view.php?id=63447 Which seems to settle the question.
What... what have I done?!
@briligg Ah, indeed. So, the photo of Earth in the question is just from satellite imagery. I forgot about Space Ex SE. If a mod feels that this question would be more appropriate there, feel free to move it.
@reirab - i think this is one of those cases that falls in the overlap between the two sites. I asked my question about photos of Earth on Space Exploration because it involves photos taken during actual missions. This case could go either way, i think.
Maybe the reflected sunlight from the lit parts of Earth dazzles the vision so that artificial lights are not visible? But if so, then at full Moon, the cities should suddenly become apparent, like the corona of the Sun during an eclipse. Chang'e 3 has a UV telescope observing Earth and should've done so also during full Moon.
The problem is that at full moon the sun will be in your line of sight when you look at the Earth. The Japanese Kaguya probe took a photo of the Earth obscuring the sun from the Moon during what on Earth would have been a lunar eclipse, but the photos I could find were low quality. It would also be as dangerous as staring at a solar eclipse.
It's highly doubtful you could see any normal light source on the surface of the earth.
Using $$\text{brightness} = \frac{\text{luminosity}}{4 \pi \times \text{distance}^2}$$
(with brightness in watts, and luminosity in watts per square meter.
and distance to moon of $3.84 \times 10^8$ meters.)Try a hypothetical light source 100 megawatts output, all visible light, no heat.
$$\text{brightness} = \frac{100 \times 10^6}{4 \pi \times 1.474 \times 10^{17}}$$
$$\text{brightness} = 5.4 \times 10^{-11} \text{ watts per square meter}$$ at the lunar surface.
That's pretty dim. By, contrast sunlight at earth's surface runs about 1300 watts per square meter.
In reality it'd take about a gigawatt to produce 100 megawatts of light in the visible range.
That's about what it takes to power a city of a million homes.
Cities also bounce most of the light they do produce off the ground, which'll have an albedo of somewhere around 0.3. So with ordinary city lights it'll take over 3 gigawatts to reach $5.4 \times 10^{-11}$ watts per square meter on the lunar surface.You might fare better with a big laser. The Apache Point Observatory Lunar Laser-ranging Operation picks up multi-photon signals from the Apollo retroreflectors using only a 1 gigawatt laser and a 3.5 meter telescope. As the article states, the laser beam only expands to 9.3 miles in diameter on the way to the moon, so you might see it wink at you.
At 10 parsecs, the sun has a magnitude of 4.83. It'd be visible on an average night. That magnitude corresponds to a brightness of 3X10e-10 watts per square meter, about 5.6 fold brighter than our hypothetical earth based light source. That puts our light at magnitude 6.5 to 7. Naked eye visibility runs to about 6.0
So for comparison, what is the brightness of a 6th-magnitude star?
Interesting. Thanks for doing the math on this! Honestly, I wasn't really thinking one particular light source, though. I was primarily wondering about being able to see the lights of cities at night, as can be done from low orbit, not necessarily resolving one particular light source.
@reirab - At the distance of the moon, a 100km diameter city will subtend an angle of about 54 arcseconds (0.015°), so I don't think there's *too* much error in the "consider a city as a point light source" assumption.
@Nate Eldredge 6.0 is about the magnitude of Uranus. It is barely visible to the naked eye in good conditions. The ancients could've discovered it as a planet before the telescope was invented, but it also moves quite slowly across the sky. Not even Tycho Brahe noticed it.
@reirab viewing from orbit is very different than from the moon, as you could relatively easily adjust your eyes to the dark. Further most night photos from Earth are longer exposures (around 1/10 of a second) at high sensitivity settings. They wouldn't be very representative of what the naked eye would see.
This answer should how you can apply all the math correctly and still not get the correct result. The question seemed to obviously address all artificial lights - all the cities combined.
@Tomáš Zato: “all the cities combined” obviously won’t pose as a single light source — 1′ angular resolution of a human becomes 110 km at Moon–Earth distance, or somewhat more near the limb due to an oblique view. Still agree with your first statement.
Even if we assume a dark-adjusted human eye could in principle discern artificial lights on the night side of the Earth, there's the problem that your eyes won't be dark-adjusted because the day side will be much, much brighter than the night side. There's not enough dynamic range in the eye to sense that sort of a brightness difference at once.
And if you'd be looking when the Earth is new, so there's no day side visible, you'd have an even worse problem: the Sun would be in your field of view. A solar eclipse (a lunar eclipse as seen from the Earth) would be your best bet - except that the atmosphere would form a brilliant ring around the dark Earth...
There is a considerable range of phases (not exactly the full moon) when Earth is seen mostly in darkness. And — always, but during just the full moon — some near-side lunar locations enjoy night. Even during exactly the ½ moon phase it’s possible to see Earth from *night-side* Moon if Sun and Earth are separated on Moon’s sky by a couple of degrees of angle (the minimal angle differs from one conjunction to another).
@IncnisMrsi that would still not remove the brilliant ring of the atmosphere.
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Hagen von Eitzen 8 years ago
What's the real source of the image? Presumably NASA, of course, but it almost looks too photoshopped to be true. - If real, it should be from Apollo 14, I guess, because it must be southern summer.