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suntrack2
2006-Apr-22, 04:45 PM
what will be the night seen, will it be helpful to lower the use of streetlights at night, will its light can shine the sea.

I know this is not possible, but we are just imagine here, how far the moon can reflect the light on earth more powerfully, will it give a light like the bulb of 60watts.

either mirror or fluid mercury on moon's surface.

TheBlackCat
2006-Apr-22, 05:19 PM
The problem is that if the moon is covered in mirrors, it will act like a giant spherical (roughly) mirror. Such a spherical mirror would reflect that vast majority of light away from the Earth, since only a very small fraction of the moon's surface would at the correct angle to reflect towards the Earth. Imagine pointing a flashlight at a mirror, unless the flashlight is aimed directly at your face in the mirror you will see no light (except a small amount of scattering, which would not occur in space). Additionally, if it doesn't hit your face it will only light up a very small part of the room where the beam directly hits a wall or object. It should also be noted that no mirror is perfect, all mirrors absorb part of the light that hits them (usually a suprising large percentage).

As the moon is now, it is a diffuse reflector. That means the light that hits it is scattered in all directions (although not uniformly). This means no matter where the sunlight hits the moon, some of it will be reflected back to the Earth. The is light an overhead projector board, if you shine a flashlight on it you will be able to see a big glowing circle where the flashlight beam hits it no matter what the angle is relative to the board. It won't be as bright as if it reflects off a mirror towards your face, but you can see it no matter where on the board it hits and it will provide some illumination to the entire room. Some of the light is also absorbed, of course.

So the question becomes, does the small percentage of the scattering that reaches us from the diffuse reflection of the entire moon give us more or less light than the large amount of reflection from a very small part of the moon if the moon was a mirror. I don't know the answer to that question.

The moon soil also has a special property of reflecting light more towards where it came from than anywhere else. The property means that more of the light is reflected back at the sun. However, it also means that as the moon and sun get closer to being on opposite sides of the Earth, the region that is not in shadow will reflect more light towards the Earth (since the Earth will be getting closer and closer to be in line with the sun). Whether this increases or decreases the light reaching us relative to a uniform scattering profile or some other non-uniform scattering profile (for instance reflecting most light at a 90 degree angle to the angle of incedence) is another question.

Gillianren
2006-Apr-22, 05:36 PM
Not only that, but I'm pretty sure mercury wouldn't be liquid the whole lunar day. (What is the freezing point of mercury, anyway?)

Infinity Watcher
2006-Apr-22, 06:09 PM
Well a this website (http://www.chemicool.com/elements/mercury.html) indicates 234.33K (-38 degrees celsius) as the freezing point of mercury (using googles calculator function indicates thats 37.9 Farenheit for those who prefer the older system of measurement)

TheBlackCat
2006-Apr-22, 09:14 PM
Not only that, but I'm pretty sure mercury wouldn't be liquid the whole lunar day. (What is the freezing point of mercury, anyway?)
It would probably still be reflective even when frozen.

ToSeek
2006-Apr-22, 10:10 PM
If the Moon could somehow achieve the albedo of Venus, it would be about five times brighter. But the full Moon would still only be the equivalent of about a 20-watt light bulb 12 feet away, still not enough to read by.

ToSeek
2006-Apr-22, 10:11 PM
Thread moved from BABBling to Q&A.

trinitree88
2006-Apr-22, 10:28 PM
Well a this website (http://www.chemicool.com/elements/mercury.html) indicates 234.33K (-38 degrees celsius) as the freezing point of mercury (using googles calculator function indicates thats 37.9 Farenheit for those who prefer the older system of measurement)

I get -36.4 degrees Fahrenheit.....F=1.8 (Celsius) + 32...when you're around -37...they're just about the same. Pete.

astromark
2006-Apr-22, 10:35 PM
If on the other hand the mirrors were to be free standing and adjustable. As the sunlight could be directed, focused. Could be interesting. But no. far to expensive. Not related to your thread but building a large optical telescope on the Moon would be of considerable use. Its night would be 28 days long. Is or does this idea have merit ?

Kaptain K
2006-Apr-23, 12:05 AM
The lunar night is only 14 days long, not 28.

astromark
2006-Apr-23, 02:13 AM
I knew that. . . .just testing ya. . . .Oops

Ken G
2006-Apr-23, 02:11 PM
So the question becomes, does the small percentage of the scattering that reaches us from the diffuse reflection of the entire moon give us more or less light than the large amount of reflection from a very small part of the moon if the moon was a mirror. I don't know the answer to that question.
An exact answer would require including many factors, but we can say roughly that a spherical mirror reflects light equally over all directions, as does the current Moon (except for the backscatter effect you mentioned). So the main difference between mirrors and the current situation is the higher degree of scattering, vs. absorption, for a mirror. Also, as you mention, instead of the Moon looking bright over its whole illuminated surface, it would only look bright at a very intense point. We would not have phases any more (the Moon would always be equally bright), but otherwise there'd be not much difference in illumination. Unless you went with astromark's adjustable mirrors, then you could turn night into day (i.e., you could make the Moon look like the Sun, but no brighter), but only for a very small region on Earth.

Infinity Watcher
2006-Apr-23, 02:12 PM
I get -36.4 degrees Fahrenheit.....F=1.8 (Celsius) + 32...when you're around -37...they're just about the same. Pete.
Could be. I never got my head aound faenheit and I could'nt remember the conversion formula so I just used the google calculator (typed 234.33 kelvin in Farenheit into google, it has some sort of conversion thing )

Relmuis
2006-Apr-23, 04:17 PM
If the Moon were to be somehow converted into a spherical mirror, we would see a reflection of the Sun.

As the Sun covers roughly 1/10,000 of the sky, this reflection would cover roughly 1/10,000 of the Moon's disk.

As the Moon's disk is roughly as large as the Sun's disk, the reflection of the Sun would be 1/10,000 as luminous as the Sun itself. It would therefore be ten magnitudes less bright than the Sun. It would be five magnitudes brighter than the full Moon, or 100 times as bright.

One would definitely be able to read by the light of one hundred full moons; and it might even seem more like day than night, for during a solar eclipse, it seems like day even if a small sliver of the Sun is still visible.

Ken G
2006-Apr-23, 08:09 PM
As the Sun covers roughly 1/10,000 of the sky, this reflection would cover roughly 1/10,000 of the Moon's disk.

As the Moon's disk is roughly as large as the Sun's disk, the reflection of the Sun would be 1/10,000 as luminous as the Sun itself. It would therefore be ten magnitudes less bright than the Sun.It would be five magnitudes brighter than the full Moon, or 100 times as bright.

You were doing well until the last sentence-- something got messed up in there somewhere. After all, a spherical mirror Moon does not have the Earth in any special place, so if it was 100 times brighter at the Earth, it would have to be 100 times brighter everywhere else too. Poof goes conservation of energy. But you are right that if the Moon scattered all incident light (i.e., had an albedo of 1), then the surface of the Moon would look 1/10,000 as bright as the surface of the Sun looks. The mirror would only concetrate that into a smaller point image of the Sun.

Clive Tester
2006-Apr-23, 09:39 PM
Ever heard of vacuum coating? – It is a technique where items such as ornaments and badges are placed in a vacuum chamber. Within the chamber a metal, such as aluminium is vaporised. The vaporised metal is attracted to the host surface, to form a reflective layer that is one molecule in thickness. Well, if we were talking hypothetically of a reflective moon, vacuum coating would seem like a good option.

Relmuis
2006-Apr-25, 01:59 PM
if it was 100 times brighter at the Earth, it would have to be 100 times brighter everywhere else too. Poof goes conservation of energy.

It would look 100 times brighter everywhere at the Moon-Earth distance. Farther away, the factor would be smaller, nearer to the Moon it would be even larger. After all, if I were standing on the surface of a Moon-sized spherical mirror, the reflection of the Sun would look as bright as the Sun itself, or 1 million times as bright as the Full Moon.

I may be wrong, but I don't think that the energy would be created from nothing. Most of the visible light which the mirrors would reflect, is now being absorbed by the Moon (which has a very low albedo) and then reradiated as infrared.

But I agree that a diffusely reflective Moon with albedo 1 would need to have the same total brightness as a perfectly mirror-reflective Moon.

Anyway, I have read somewhere that the Sun has apparent magnitude -26 while the Full Moon has apparent magnitude -11. That's 15 magnitudes difference, so if my calculations are correct (they might well be off by a factor 2 or 3), the Moon must have an average albedo of 0.01.

Roy Batty
2006-Apr-25, 05:07 PM
Anyway, I have read somewhere that the Sun has apparent magnitude -26 while the Full Moon has apparent magnitude -11. That's 15 magnitudes difference, so if my calculations are correct (they might well be off by a factor 2 or 3), the Moon must have an average albedo of 0.01.

Actually nearer 10 times that (http://www.asterism.org/tutorials/tut26-1.htm), but milage may vary depending on contexts (http://jeff.medkeff.com/astro/lunar/obs_tech/albedo.htm)

Relmuis
2006-Apr-25, 09:50 PM
In that case, we seem to have a paradox on our hands.

Roy Batty
2006-Apr-25, 10:12 PM
Doing a bit more digging I think the full Moon's apparent magnitude is nearer -12 or more, making the difference 14. Beware of rounding :)

Ken G
2006-Apr-26, 02:51 PM
It would look 100 times brighter everywhere at the Moon-Earth distance. Farther away, the factor would be smaller, nearer to the Moon it would be even larger. After all, if I were standing on the surface of a Moon-sized spherical mirror, the reflection of the Sun would look as bright as the Sun itself, or 1 million times as bright as the Full Moon.

Yes, but at the Earth-Moon distance, it could not be brighter than the Moon is now, except for a higher scattering efficiency, maybe a factor of 5-10 at most. Maybe a bit more, it depends on how good the mirror could be.


Most of the visible light which the mirrors would reflect, is now being absorbed by the Moon (which has a very low albedo) and then reradiated as infrared.

Yes, that's the scattering efficiency effect, but it could not improve by a factor of 100 (I think it's over 10%, and Roy Batty confirms that. Also, I'm not sure if mirrors can approach 100%, maybe they can.)


Anyway, I have read somewhere that the Sun has apparent magnitude -26 while the Full Moon has apparent magnitude -11. That's 15 magnitudes difference, so if my calculations are correct (they might well be off by a factor 2 or 3), the Moon must have an average albedo of 0.01.

That's a bit low, but I see that we are not disagreeing, just doing the calculation from a different angle and encountering different uncertainties. The discrepancy is worse than you think, because the full Moon is exceptionally bright, much more so than its average brightess from all angles. Did you square the ratio of the Moon's distance to the Sun's distance, and convert that to magnitudes?

Relmuis
2006-Apr-26, 03:29 PM
I got the magnitude values from the Encyclopaedia Brittannica; I did not calculate them myself.

And I envisioned a perfect mirror, reflecting 100 % of the visible light.

pghnative
2006-Apr-26, 07:41 PM
If the moon were mirrored and perfectly spherical, then it seems to me that an observer on earth would see a reflection of the Sun from only a very small portion of the moon's surface. That portion would be as bright as the Sun. The rest of the moon would be nearly invisible.

I don't know how to calculate the portion of the surface that would be lit up. Somewhere there are Apollo photographs showing the reflection of the Sun off of the Pacific Ocean --- that is an approximation of what I'd think you see with a spherical mirrored moon.

Ken G
2006-Apr-27, 01:03 AM
The lit up portion would have an angular size equal to the angular size of the Sun times the ratio of the Moon's distance to the Sun's distance, as Relmuis said. The problem was in the calculation of the resulting magnitude.

tony873004
2006-Apr-27, 08:02 AM
The lit up portion would have an angular size equal to the angular size of the Sun times the ratio of the Moon's distance to the Sun's distance, as Relmuis said. The problem was in the calculation of the resulting magnitude.
I don't this is right. The Moon's size should play a role as well. A shiny chrome ballbearing at the Moon's distance would have the same distance ratio as the Moon/Sun, but could not possibly have an appreciable angular size.

However, this method produces an answer similar to the method I try. Maybe a coincidence since the Moon's angular size and the Sun's angular size are about the same?

I think it would work like this:

The Sun subtends 1/2 degree in the sky.

The diameter of the Sun's reflection disk on the Moon will therefore subtend 1/4 degree of lunar surface in diameter.

The Moon's diameter is 3475km.

Therefore, 1/4 of a degree on the Moon is 3475 * pi / 1440 = 7.5 km

This will have an angular diameter of tan-1(7.5 km / 384000)= 0.001 degrees or 4 arcseconds.

Therefore, the angular diameter of the Moon's shiny spot would be 0.001 / 0.5 = 0.002 ( or 1 / 450 ) the angular diameter of the Sun as viewed from Earth.

Since this is an area problem, area (and hence brightness) will decrease as in inverse square.

0.001^2 is 1.25E-06, so the visual area of the solar disk is 1 / 1.25E-06 =~800,000 times greater than the area of the Sun's reflected disk on the Moon's surface. Hence, the actual Sun would be 800,000 times brighter than the hypothetical Moon's shiny spot, assuming 100% reflectivity. I believe that the Sun is about 400,000 times brighter than the real Moon, compared to 800,000 times brighter than the hypothetical moon. Therefore the hypothetical Moon's shiny spot would reflect about half as much light as the entire actual Moon.

M1-M2 = -2.5 log ( 800000 ) = -15. So there should be a 15 magnitude difference between the actual Sun, and its reflection on the hypothetical Moon's surface. So its appearant magnitude should be about -26.8 + 15 = -11.8, about that of a gibbous real Moon.

suntrack2
2006-Apr-28, 10:54 AM
thanks tony for the reply

Ken G
2006-Apr-28, 11:46 AM
I don't this is right. The Moon's size should play a role as well. A shiny chrome ballbearing at the Moon's distance would have the same distance ratio as the Moon/Sun, but could not possibly have an appreciable angular size.
However, this method produces an answer similar to the method I try. Maybe a coincidence since the Moon's angular size and the Sun's angular size are about the same?

You're right, there was a coincidence here, but it's not the angular size of the Moon, that was implicit in the calculation I suggested. The coincidence is in the fact that the Earth-Moon distance is not that different from the radius of the Sun (it's a little more than half the Sun's radius). For a Moon with the same angular size as the Sun, the total brightness of the Moon, relative to the Sun, scales like the square of the ratio of the Sun's radius to the Moon's distance to the Sun.

suntrack2
2006-Apr-29, 09:47 AM
Ken, as you quote above ( it's a little morethan half the sun's radius), means a small sunspot size of that equal distance can cover moon and earth fully ? it is true that the distance from earth-moon not that different from the sun's radius, the appearance of the moon's brightness is relent upon the distance, on the full moon day the dimlight like light falls on the earth, in this light the human vision grows upto a distance of some 50 meters not more than this. beyond this distance we can experience the dark there. In the above condition the moon's appearance will be look like a "sun's 3/4 th portion covered eclipse when we look from earth, and during this sort of eclipse the sun's light we experience here just like bulb of 40watts. this sort of light we can see when the moon covered for redirecting the light from sun to earth.