*illuminate*it from. So why does the moon look uniformly bright all the way across its disk? Why does it look like a flat cut-out?

It occurs to me to ask the question: what if the moon were made of golf-balls? Each golf-ball would be a mini-moon: when the moon was at half-phase, each of the little golf-balls would look like half moons to us here on earth. Except they'd be too small to see. So we'd just see the average of all of them, and the local average would be the same wherever we looked. So maybe the moon would look uniformly bright everywhere.

Except the golf-balls low on the horizon would also be partly in the shadow of other golf-balls, so the fringes should still look darker. In other words, I'm not completely buying my own explanation. But it's a thought.

Meanwhile, I tried to apply some calculus to this question, and I'm not too sure of my technique, but I got an answer and I wonder if anyone out there would like to double-check it. I compared the real moon...that is, the ideal, bright-white Lambertian "real" moon...to a flat cut-out sheet of drywall. According to my calculation, if you compare these two models at midnight on a full moon, the flat drywall cut-out provides 50% more night-time illumination than my "ideal/real" Lambertian moon. I'm not going to try and repeat my calcuations here, but I'm just wondering if anyone out there wants to see if they get the same answer as me.

## 3 comments:

Hi Marty,

Sorry for this late comment (over a year now from your post), but I could not resist. Nice post, BTW.

Recently I was thinking about a similar problem - visual magnitude of the ideal/Lambertian moon/planet depending on its phase. After I did some calculations, I was trying to verify their correctness on internet. To my surprise it was really hard to find anything relevant, but after googling for couple of hours I found your post.

My result was that the spherical Lambertian full moon would provide 1/3 less reflected light than the flat drywall moon. So first I thought... 33% is different from your 50%, but after reading your post again I realized it's the same thing (I bet you know what I mean). You know, it's 2am now and it's hard to focus :). So good news for both of us - we got the same result.

Anyway, I went further with my calculations as I was interested in the amount of reflected light from the moon in certain phase relative to the full moon. Still talking about ideal Lambertian moon. Really interesting integrals, man ;) For example, the first/third quarter moon (half-circle) gives exactly 1/pi (approx. 32%) of light compared to the full moon. As I said, I could not verify the result and I was wondering if it's correct. If you are interested and find some time to do the calculations, would you mind to confirm you get the same result? Thanks.

Peter

Hi Peter

I thinks that's great that you ended up working on the same problem as me, and then you found me on the internet. Yes, your 33% appears to be the same as my 50%...in Talmudic Law that's called the difference between a "shtut milguph" and a "shtut milbar". (But that's another story.)

I'd like to try and work out your integrals for the brightness of the half-moon but right now I'm stacked up with some other obligations. But I definitely agree the half-moon should be less than 50% the brightness of the full moon, because you are seeing a greater proportion of the surface area which is only obliquely illuminated.

Hi Marty, thanks for replying.

Right, before I did the computations, I had the same thinking that the half-moon would yield less than 50% of the brightness of the full moon. In fact, the real (non-Lambertian) half-moon gives only 8% of the light of full moon. Check out this interesting article:

http://www.asterism.org/tutorials/tut26-1.htm

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