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111

The premise of this question is wrong. If the moon is in between the earth and the sun (as shown on your diagram), and you can see the moon, then it is day, not night: If on the other hand, you are on the opposite side of the earth during that configuration (so that it is night), then you can't see the moon because the earth is blocking your view of it:

99

The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction. I hope everyone who reads ...

76

The orbital plane of the Moon around the Earth is at an angle to the orbital plane of the Earth about the Sun. This diagram was drawn to show why we do not have eclipses more often but is also shows how it is that a full Moon occurs. . Image downloaded from Taylor Science Geeks website

69

One of the reasons people often have bad intuitions like yours about the relationship between the Earth and the Moon is because they've never seen an accurate picture. The distance from the Earth to the Moon is often pictured something like this: The relative sizes of the Earth and the Moon are accurate but the distance is not. Given this picture it looks ...

65

The concept you're looking for is that of a planet's Hill sphere. If a planet of mass $m$ is in a roughly circular orbit of radius $a$ about a star of mass $M$, then the radius of this "sphere" is given by $$r_H = a \sqrt[3]{\frac{m}{3M}}.$$ For the Sun-Earth system, this yields $r_H \approx 0.01 \text{ AU}$, or about 1.5 million kilometers. The ...

63

The Jet Propulsion Laboratory has incorporated general relativistic effects in its numerical integration of the planets since the mid to late 1960s. For example, the JPL DE19 ephemeris, released in 1967, incorporated relativistic effects in its modeling of the solar system. This didn't help much. Had they ignored relativistic effects there would have been ...

51

Here is a picture that may help in describing this. First row: Sun (red), Earth (blue) and Moon (black) to scale (axes - in km). You can see just how far away the sun is... and on this scale the Earth and Moon are essentially invisible (they are inside the "zoom box"). Second row: zooming in (50x), you can barely make out the Earth and Moon, Zooming in ...

48

I refer you to the picture below, taken from Ciocca & Wang (2013). This clearly shows that the spectrum of the moon (normalised to have a similar overall strength as sunlight) is redder than sunlight and so has a lower "colour temperature". This is a fact, not a perception. EDIT: Just to clear up some confusion - the OP talks about "yellower" because ...

48

The sun doesn't just illuminate the moon directly. The moon is also illuminated by sunlight reflected from the earth. This is called earthshine. This makes the parts of the moon that face us visible even when the sun is on the other side. According to NASA, it was Leonardo da Vinci who first explained this. As an example, the brightly lit portion of ...

46

Suppose the Moon didn't orbit the Earth at all, so it just stayed at some fixed point while the Earth rotated underneath it: In this case every point on the equator would pass directly under the Moon every 24 hours, and we'd get a high tide every 24 hours. (There's another high tide when we're exactly on the opposite side of the earth to the moon, but let's ...

45

No, because of the sizes of their surfaces. Let's make these simplified assumptions: The Earth and the Moon are both spheres 1 AU from the Sun. The total amount of sunlight an object receives is proportional to the solid angle it takes up from the Sun's point of view. The Sun and the Moon are each visible from a hemisphere of the Earth. Then the total ...

44

It'd suck in very little mass. One thing to understand about black holes is that they have super-strong gravity, but only when you are very close to their event horizon. Otherwise they're just normal objects. If the Moon became a black hole, it would have a radius of about 0.1 millimeters. You need to get pretty close to this distance before you notice ...

40

You seem to be asking if the reflection of the sun from a spherical mirror, a convex surface would be the same as the reflection from a flat mirror. A convex mirror is dispersive The image in the diagram above is a virtual image. Light does not actually pass through the image location. It only appears to observers as though all the reflected light from ...

39

To answer the question, your friend needs to understand that the sun is not a little thing on the other side of the globe. It is very large, and it only looks small because it is far away. Also the moon is not small and close. So the light of the sun can easily shine past the earth and onto the moon. The other anwsers are also good, but you're not likely to ...

36

The fact of parallax in the observed position of the moon was known in ancient times. This makes possible a calculation of the distance of the moon in terms of Earth radii. No parallax was then observed for the Sun. The first known calculation of the distance to the moon is generally attributed to Hipparchus of Nicaea

34

Lunar occultations. Just missed the moon block Mars last month. Not sure when Mars will block a star, thereby proving the stars are further than the moon.

30

The relevant "100%" from which you should calculate the percentage isn't the depth of the ocean but the radius of the Earth $$R\sim 6,378,000\,{\rm m}$$ Multiply this $R$ by $10^{-7}$ and you will get $0.6$ meters, a reasonable estimate for average tides. You must understand that the surface of the ocean always tries to create an "equipotential surface" –...

27

A few sanity checks without actually computing anything: First, the error due to neglecting general relativity is so small that it didn't affect prediction of lunar eclipses and wasn't actually noticed anywhere except in Mercury's orbit (at least not until they purpose-built experiments to detect minor discrepancies). I know this doesn't give a completely ...

27

If the moon were a uniform sphere, it would indeed appear dimmer at the edges of a full moon. Surface roughness is the main reason the edges are not as dim as the sphere model predicts. When viewing a full moon from earth, the light is coming from nearly behind us. That means the edges of the full moon are illuminated by sun rays nearly parallel to the ...

26

I'll start the ball rolling on this one. My GR knowledge is probably not good enough to make this a truly satisfying answer... The gravitational acceleration for an object moving radially at non-relativistic velocities in the Schwarzschild metric is modified by a factor $(1 - r_s/r)(3[1-r_s/r] -2)$, where $r_s = 2GM/c^2 = 0.00885 m$ for the Earth. If we ...

26

The moon does have a night and a day, but this isn't as fully connected to your question as you might think. The moon is tidally locked with the earth, meaning that the same side always faces earth. Since the moon also orbits around the earth (with a period of a lunar month), this means each side changes, over the course of a lunar month, between facing ...

23

While excellent answers have already been provided (yes, it's Earthshine; yes, when the Moon is between the Sun and the Earth, you don't see the Moon at night, you see it from the daylit side of the Earth) given all the "artist's renderings" in the question and the answers, I thought it might be useful to include a diagram that demonstrates the actual scale ...

23

It would suck in basically nothing. If the mass of the moon were concentrated in a black hole, you could draw a sphere around that black hole the size of the moon, and for everything outside that sphere, nothing would have changed. The gravity field would be just the same as it is now. Now, black holes are tiny. At the mass of the moon, the radius is on the ...

23

A lot of debris has probably fallen back to earth. To stay in orbit you need enough angular momentum to overcome attraction. But if the collision happened at an angle a portion of the debris could have enough angular momentum to sustain orbit. Here is a nice video of how the collision could have happened. Here are some snapshots from the video in case the ...

22

The moon appears white from the Earth for two reasons. The first is that the reflected spectrum of sunlight is very broad and contains no very significant features. On this basis, the spectrum of the moon could be considered pinkish, as the reflectance of sunlight (which appears almost white to the human eye) is twice as effective at red wavelengths than ...

20

The moon is actually grey. You can see this if you look at images taken in space, or, preferably, on the moon itself. For example, this one, of Buzz Aldrin: (Courtesy of NASA) But, seeing as how at night you compare it to a black sky, it appears white.

20

I've wondered the same thing, why the edges of the full Moon don't look darker, and why the terminator for a quarter-Moon phase doesn't look dimmer than the point at the edge of the Moon that is opposite the Sun. It isn't a foreshortening effect, as in your last hypothesis, that actually doesn't work. If the Moon's surface were uniformly bright in all ...

20

For the full moon, the answer in the OP is clearly correct, as others have confirmed: the brightness is related to the solar energy per square foot, and when viewing from the front, the square feet increase as you go towards the edge at the same rate that the solar energy per foot fades. This assumes that reflections are omnidirectional (as with matte dust ...

19

The simple answer is that the sun's gravity produces the same acceleration on both the Earth and the Moon. The Sun is pulling both of them along, but they are falling together. You may imagine two skydiver jumping out of a plane at the same time (and we'd better ignore air resistance). They are subjected to gravitational forces from the Earth that vastly ...

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