For the last 3 days, I've seen the crescent moon in the sky during the late morning.
2017/08/16, 10:16am EDT:
2017/08/17, 8:43am EDT:
(On 2017/08/18, I saw the moon around 10:30am, but didn't think of taking a photo.)
But I have been unable to find the moon in the sky around 2:30pm (during my lunch breaks). I've been wondering why I can't find it in the afternoon.
(No wonder I've never paid much attention to the path of the moon in the sky. At night, I'm usually inside or asleep. In the daytime, even when the moon is in the sky, it's hard to see.)
On all 3 days, it's been partly cloudy, with today being the least cloudy. So it's possible the moon was behind a cloud. But as much as I've searched the sky, it seems unlikely it's *always* been behind a cloud.
As of today (2017/08/18) at my location, per the NOAA solar calculator
(Find Sunrise, Sunset, Solar Noon and Solar Position for Any Place on Earth), solar noon is around 1:30pm. So at 2:30pm, the sun is still fairly high overhead.
On 8/16, it was 5 days before new moon and the eclipse, so the moon would have been about 5 * 13 = 65 degrees away from the sun. So that was most likely too near the horizon for me to see, as there are some trees and buildings around.
On 8/17, the moon would have been 4 * 13 = 52 degrees away from the sun. I think I should have been able to see it at that angle.
Today on 8/18, the moon would have been 3 * 13 = 39 degrees away from the sun. Surely I should have been able to see it at that angle.
The closer we get to the new moon, the thinner the crescent is. So the harder it is to see. It is hard to find a tiny arc of white in a light blue sky, and even more so when there are distracting white clouds around. But is that the only reason I haven't found it?
Per this page: Finding the Moon
, crescent moons are "not observable" except right before sunset or after dawn. But I've seen it at 10:30am which isn't right after dawn. So I think it would be more accurate to say "not easily observable".
If I can see it at 10:30am when the sun is already bright in the sky, why shouldn't I be able to see it at 2:30pm?
I got to wondering whether how I think of the angles in the sky is wrong. I am thinking of 45 degrees as being the distance from straight overhead to a point halfway to the horizon. But the 13 degrees that the moon moves per day is in relation to the center of the earth, not to my spot on the surface of the earth. Therefore, is how I'm visualizing the angles in the sky wrong?
When the moon orbits 45 degrees around the earth, is that a much greater distance than the distance I see from overhead to halfway to the horizon?
But... as can be seen in the diagram, the larger you draw the earth, the closer the 45 degrees gets to one's visible horizon
, and it would eventually even pass below the horizon
. Yet I've been able to see the moon in the mornings, and the distance between it and the sun hasn't seemed such a large angle. So surely the above diagram can't be right.
(Update #2, 2017/08/20
: I've figured it out. The diagram is basically correct, but my assumption about the 45 degree line eventually passing below the horizon was wrong (just because I don't draw the horizon line to infinity, doesn't mean it doesn't go to infinity). If the angle to the moon as measured from the center of the earth is 45 degrees (from directly overhead), then the angle as measured from the surface of the earth would be more than 45 degrees. But because the distance to the moon is so large in comparison to the size of the earth, the angle is only slightly more. See follow-up post
On the same topic, I got to wondering how much of the sky / celestial sphere am I actually capable of observing from a point on the earth, at any moment in time. Ie. if I turn all the way around, looking towards the horizon, and up above me, how much of the sphere of the sky which surrounds the earth, am I seeing?
Based on the diagram, the amount of sky seen would not be half the sphere, as I've previously assumed. Yet again, the larger one draws the earth, the less of the sky one would seem to see. Surely that can't be right?
Based on these answers
, it sounds like you should be able to see half of the sky at any time. But I don't understand the formulas and calculations listed.Update (afternoon of 2017/08/19):
Today, the morning of 8/19, around 7:40am and again at 10:20am, I wasn't able to find the moon in the sky, even though it was clear with no clouds. So as of 2 days before new moon, the crescent must be too small and faint to see in the daytime. Perhaps a clear sky being so much brighter than a partially cloudy sky, also makes it harder to see. MoonCalc.org
- shows you the current position of the moon in the sky, and moonrise/moonset directions, for any position you select on the map.Sun Locator Lite
- a free app which lets you find the sun and moon by pointing the phone at the sky (as long as the phone has an internal compass/magnetometer - mine doesn't, but Qiao's does). The Pro version
lets you get information for any day and time of the year.
Today, 2 days before the eclipse, the moon should be about 2 * 13 = 26 degrees from the sun. I used the above Sun Locator app to find the position of the moon and compare it to the sun's position, and estimated the angle between them. If anything, it seemed less than 26 degrees, not more. So that indicates that there's something wrong with my thinking in terms of the above diagram. But where have I gone wrong? I still haven't figured that out.
(And even with the app to show me its exact location, I still can't see the crescent moon in the afternoon sky.)
But I did have an epiphany on how much of the sky is visible from a point on earth at a single moment in time. It depends on what I'm calling the "sky". I think of the sky as a sphere centered around the earth, upon which I see moon, sun, stars, clouds, etc. But there are many such possible spheres around the earth, different distances from the center of the earth.
How much of the sky is seen depends on which of those spheres one considers. If one considers a sphere which is say, 10 kilometers above sea level, you can calculate the surface area of that sphere. The earth's diameter is 12,742 km. So the sphere's diameter would be 12,752 km, its radius (r) would be 6376 km, and it's surface area would be 4*pi*r^2.
[ another interesting thought... For an infinitely thin sphere, the size of the inside and outside surface areas should be the same, right? But how can that be? I can't visualize them being the same size. ]
Imagine that we cut a small slice, 10 km deep, from the top of that sphere. We can then calculate the surface area of that slice (with some formula, which I would have to look up.) That would tell us how much of the whole sphere we can see at a single moment, and it would be a fairly small portion.
But now, consider a sky-sphere with a much larger radius of 5 light-years - reaching the nearest stars - or even larger. At such distances, the diameter of the earth is minute in comparison - it can be considered negligible. A plane which touches the surface of the earth at one point is practically the same as another parallel plane which intersects both the center of the earth and the sphere. Either way, half of the sphere is above the plane, and half below. So the person can see half of that sky-sphere.
Now, what about a sphere with radius of 150 million km (about the distance from the earth to the sun
)? In comparison to that distance, the earth's diameter is roughly 0.01%.* So again, it's basically negligible, and we can see practically half of the sphere at any moment in time.
Other interesting tidbits:How far away is the horizon?
Short answer: About 4 to 5 kilometers away, at standing eye-level for an average-height adult.I see the moon: introducing our nearest neighbour
- has several good diagrams/images.
Per this page, the moon's orbital plane is tilted 5 degrees from the ecliptic. That's not as much as I imagined. But when you add in the 23.5 angle of the earth's axis, the moon can orbit up to 29 degrees above or below the earth's equator.Lunar Orbital LibrationLibration
definition: "a real or apparent oscillatory motion, especially of the moon."Altitude and Azimuth
* A lot of these numbers are rough calculations I've done, and they may have errors. Please don't rely on any numbers I've posted, without verifying them. If you find an error, please let me know so that I can correct it.