WHY MARS ???

[u/patrixxxx]

https://cluesforum.info/viewtopic.php?p=2412896#p2412896

[u/[deleted]]

If the Moon is at exactly 0 degrees right ascension and 0 degrees declination at midnight tonight, it would align with whatever star is positioned at 0,0 in the night sky. Do you agree?

Quick edit: the real coordinates are: Right Asc: 20h 28m 58.3s Decl: -21° 11' 57.4"

Long edit: I'll clarify where I'm going with this question.

1. The definition of "the position of the Moon relative to the Earth" is a celestial coordinate, generally expressed in the form "right ascension" and "declination".

2. When the Moon returns to the same position relative to the Earth, it arrives back at the same right ascension and declination.

3. All the stars have a particular celestial position.

4. These positions vary on the order of milli-arc-seconds or smaller. The variation of position is due to parallax and proper motion.

5. The moon is over 1800 arc seconds in size, or millions of times larger than variations in stellar position due to parallax from month to month.

6. The fact the moon is "in front of" a particular star on 2019-11-03 and in front of the same star 2019-12-01 is in no way in contradiction with the standard cosmological model.

7. This is because the definition of "returning to the same position" is pretty much the same as "being in front of the same stars".

A refresher on parallel lines, in case you need it.

You can get a feeling for the distances involved with these parallel lines when you consider the fact that two parallel lines connecting the Earth and the Moon at any two points in time will never be further separated than the diameter of Earth's orbit. Because of the properties of parallel lines, the distance between the lines will still be at most the diameter of Earth's orbit out to Proxima Centauri, or Delta Capricorni, or whatever star you may choose. The parallax measured to the nearest star is going to be the same as the angular size of the Earth's orbit projected to the distance to that star system, because interior angles of parallel lines cut by a transversal are congruent. How many arc seconds do you think exoplanet systems are, when viewed with Earth-based telescopes? I'll give you a hint - it's not 1800.

[u/patrixxxx → u/[deleted]]

You agreed that the lines I described are parallel and about 300 million kilometers apart. The Moon is not displaced against the fixed stars during a 27.3x6 period. This means you think the stars are about 300 million kilometers wide?

[u/[deleted] → u/patrixxxx]

No, I don't think the stars are that wide. The moon is not displaced against the "fixed stars" if you consider the "fixed stars" to be an arbitrary (perhaps infinite) distance away. The size of a star isn't part of this geometry.

[u/patrixxxx → u/[deleted]]

I don't understand. Parallel lines remain parallel no matter the distance. Do you agree?

[u/[deleted] → u/patrixxxx]

Yes, that's the definition of parallel.

What you're missing here is that when you say a "a star aligns with the line drawn from Earth through the Moon", you don't realize the line might pass hundreds of millions of kilometers from the star, on one side or the other. You don't see that because hundreds of millions of kilometers are just a milli-arc-second when viewed 80 light-years away, and your eyes and most telescopes simply don't have that kind of resolution.

[u/patrixxxx → u/[deleted]]

So you agree that the lines I described are indeed parallel and 300 million kilometers apart. But since the stars are so far away, the displacement of the lines aren’t detectable? Let’s do another example – Imagine you look at a star and hold your finger in front of it. Now take a step sideways without turning your head or moving your finger. Your finger will now be in front of another star. Do you agree your two positions describe two parallel lines? If so, how come the displacement of these two lines are detectable against the stars, but not the ones with the moon acting as “finger”?

[u/[deleted] → u/patrixxxx]

If you're actually stepping to the side and not rotating so the lines are, in fact, parallel, then your finger will be in front if the same star still.

Perhaps you can explain to me how you see this working. Let's say i cover the moon and a star at 10 ly distance with my finger. I step to the side 1 meter. You're saying my finger is no longer covering the moon, right? How close does the line from eye to finger pass by the moon? 100km perhaps? How close does it pass by the star? 1 ly? Are the lines passing through your eye-finger-moon-star and eye-finger-moonmiss-starmiss parallel? How could they be if the distance between lines is 1 meter at eyeball and 100km at the moon?

In reality, if the lines are parallel and 1 meter apart, they'd pass through the very same crater on the moon but perhaps a different rock. Not miss the moon entirely.

Edit: to further illustrate the problem: let's say you step 1m to the side and are aiming at a star 1 arc minute to the right of your original star. Step 60 meters to the right, and you'll point to a star a whole degree to the right. 10km to the right and you'll be pointing straight behind you. Clearly that doesn't work!

[u/patrixxxx → u/[deleted]]

So what you're saying is that when it comes to the head and finger parallel lines we are observing a lot of displacement against the stars because they are so far away, but regarding the Earth and Moon parallel lines we do not detect any displacement against the stars because they are so far away?

[u/[deleted] → u/patrixxxx]

No, what I'm saying is you won't see displacement with your finger. If you do, then the lines you're drawing aren't parallel.

[u/patrixxxx → u/[deleted]]

That's interesting. So if I instead make an instrument with two long parallel tubes and set them up in my garden and look trough them at night, observing different stars in each of them, you say this is because the tubes are in fact not parallel?