Because the moon’s orbit is tilted by about 5 degrees relative to the Earth’s, we don’t get a solar eclipse every new moon: usually the new moon passes either above or below the sun from our POV.
But if the moon is slightly out of the plane between the sun and the Earth, does that mean that there’s a very thin sliver on the bottom or the top of the moon that is actually illuminated, even at the point of minimum illumination?
“New moon” is defined as the moment when the Sun and Moon are at the same ecliptic longitude, so we do have a new moon every time it passes by the Sun.
Yes, usually the Moon is not directly in line with the Sun/Earth at new moon. The thin sliver is too thin & too dim to be seen through the glare of the Sun (i.e. scattered sunlight). If we had a telescope that was immune to glare of the Sun (like a coronagraph) we should be able to see it, but I don’t recall seeing a photo like that. I think the sunlit portion is just too narrow to be seen even with a good coronagraph.
Yes that is true.
Here is a link to the effects of “libration” - the apparent shifting of the lunar disc as perceived from Earth. These expose the edges of the physical disc, although I believe the diurnal libration would have the effect of showing parts of the sunlit disc to observers on different sides of the Earth.
Libration is completely irrelevant to this question, because the lit and dark portions of the Moon aren’t painted onto the surface. A spherical moon could twist and turn in place every which way, and not change at all how much of the lit portion is visible from Earth.
Libation has made the moon look pretty strange to me at times. Oh wait, libration, nm.
Having thought about this more - as the OP said, the new moon can be up to 5 degrees away from the Sun, and that should be enough to observe as a crescent. And I found this article that has a photo of the Moon at the moment of new moon. It’s an infrared photo though, which is less susceptible to atmospheric scattering than visible light.
To figure out how much of the lit portion is visible from Earth, first figure out the phase angle which will be equal to the angle between the observer and the Sun, viewed from the Moon. Call this i. The proportion of the illuminated disc visible is then (1 + cos i) / 2.
I saw a picture that claimed to be the Moon exactly at “new moon” phase with the barest hint of a crescent. I’ll have to see if I can find it again.
I posted one in a followup post (#6).
If I’m reading it correctly, that was 24 hours after the new moon. The one I saw claimed to be exactly at the new moon.
Nvmd scr4. I scrolled down and that’s the one I saw. Derp
It says “the moment of New Moon on July 8, 2013”. Now let’s say the angle formed between the Sun, Moon, and photographer was about 175.55 degrees. According to the formula I gave, 0.0015 of the disc was illuminated (0.15%). Still, the photo shows some light clearly visible.
Bad Astronomy: “Newest New Moon”.
From the article: “He got the timing to within a minute of the actual moment of newest Moon…”