As everyone knows, the Moon is sometimes visible during the day, while the Sun is also in the sky. Suppose you look up sometime during the day and see a half-moon in the sky. The Sun is also in the sky, separated from the Moon by 45 degrees of arc. What can you conclude from this?
In the above diagram, the vertical ray (using that word in the geometric, not the optical, sense) represents all possible locations of the Moon. (Since we are supposing we do not know how far the Moon is from the Earth, it could in principle be at any point along the ray.) The diagonal ray represents all possible locations of the Sun when it appears from Earth to be 45 degrees distant from the Moon. The horizontal ray extending out from the Moon represents all possible locations of the Sun which would cause a half-moon to be visible from Earth. Therefore, if you see a half-moon 45 degrees from the Sun, you can conclude that the Sun is 1.414 (the square root of two) times as far from the Earth as the Moon is -- and that therefore everything you know about astronomy is wrong, since astronomers tells us the Sun is approximately 395.5 times as far from Earth as the Moon is.
If that figure is correct, what should be the angular distance between the Sun and a half-moon? Well, it must be less than 90 degrees, since the red ray (representing the Sun at 90 degrees from the Moon) is parallel to, and thus never intersects, the half-moon ray. But, since 395.5 is a very large number, it must be only a little less than 90 degrees. I've forgotten all my trigonometry, so I'll leave the exact figure as an exercise for the reader.
Update: I've just realized the flaw in this reasoning -- that it applies only when the Moon is directly overhead. The angular elevation of the Moon must be included in the equation, not only its angular distance from the Sun.
Update 2: No, on second thought, I think I was right the first time.
7 comments:
"it applies only when the Moon is directly overhead" - Ah, well it never is at my latitude - so. that's that...
On a similar theme - a chap called Tristan Gooley would be very interesting to you (similar kind of mind, I think)
https://www.naturalnavigator.com/
I have a copy of his: The Walker's Guide to Outdoor Clues and Signs
You have rediscovered an ancient method:
Aristarchus of Samos wrote a book called "On the Sizes and distances of the Sun and Moon"
https://infogalactic.com/info/On_the_Sizes_and_Distances_(Aristarchus), where he describes this construction.
As far as the correct angle, you take the arc secant or inverse secant of 395.5, which is about 89.855 degrees
@Bruce
Actually, I was wrong in thinking I had been wrong. It applies regardless of the angular elevation of the Moon.
I'll check out Tristan Gooley. Thanks for the lead.
@NLR
I assumed someone else must have thought of it long ago, but you never know. One of my other astronomical "discoveries" turned out to be original.
First, try to find a half-moon in the sky when its angular separation from the sun is only 45 degrees. I have never observed such a thing!
@Ben
Right, because it’s impossible in our solar system. The Sun will always be nearly 90 degrees from a half-moon.
Indeed. Not only that, but epistemologically these kinds of naked eye observations are key to how we know about this aspect of solar system geometry.
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