A Conversation for The Moon and the Tides

Comments on the Tides

Post 1

Jimi X

Please add anything about the influence of the moon on the tides here. I think this entry is pretty complete, but if there's anything I missed, please tell me!

smiley - smiley

- X


Comments on the Tides

Post 2

Güthwinë

I had once read somewhere that the tides were caused by something completely different than what is quoted here, although this is certainly the more widely accepted of the two ideas. Frankly, both make sense, and I have no clue which is correct.

Basically, it goes like this: The moon orbits around the earth, right? Well, no... the moon and the Earth both orbit around the center of their collective gravities, or whatever. Well, if you think about the Earth rotating around this point (and the moon as well, which would have massive tides, were it to have oceans), you see that the near side of the Earth is much closer to this point than the far side (duh), by, surprisingly, the diameter of the Earth. Now, the farther side of the Earth moves around its path faster than the inside of the Earth does, in the same way that the outside of a spinning disc (such as a record or CD) moves much faster than its interior does. (Still with me? Well, there's not much I can do about that. smiley - smiley) Now, on to orbits. An object in orbit has to have a certain speed to maintain the height of its orbit. When it slows down, it falls to a lower orbit. If it speeds up, it rises to a higher orbit. Now, seeing that the 'out' side of the Earth (the one facing away from the moon and the center of gravity in the system) is moving *faster* than the inside, it tries to rise to a higher orbit, and would, but for that whole obnoxious gravity thing. In the same way, the near side of the Earth, moving slower, is trying to fall into a lower orbit.

From what I have read, this is what tide is *really* caused by. I believe the source was Larry Niven's short story "Neutron Star". I guess I probably get what's coming to me for learning about science from science fiction. Hmmm......

It made sense when I read it... and even though I did a bugger of a job explaining it, I hope you can understand. (It really helps when you can draw pictures!).

Yes, like I said, I don't know which one is true, and the explanation from the article is what is generally taught. Any astrophysicists feel like explaining once and for all?


Comments on the Tides

Post 3

Jimi X

What I wrote is what I learned in my astronomy courses in college. Of course, my professors and the textbook manufacturers could have been wrong.

Really.

They were kind of terrible.


Tides on the moon

Post 4

ommigosh

Great article Jimi-X !

I don't really understand what Guthwine was saying but it sounded interesting! Maybe it is not always too unhealthy to sometimes have a healthy disrespect for accepted scientific wisdom and textbooks and professors and suchlike. (I work at a university....I know).

Guthwine mentions tides on the moon which would occur if only it had some water.....well, I seem to remember reading that the body of the moon itself does actually get deformed a bit due to the gravitational pull of the earth below. So Guthwine, there are lunar tides of a sort.
Of course, they would be much smaller if the moon was made of rock instead of cheese.


Tides on the moon

Post 5

Jimi X

The lunar 'seas' were caused by the Earth's tidal influence on the Moon, that's why there are very few on the far side.


Tides on the moon

Post 6

jqr

I always thought one should sleep on the north-south axis to mitigate the effects of tidal forces on one's body during the night.


Tides on the moon

Post 7

Güthwinë

Like I tried to make clear in my post, I really don't completely understand what I was talking about either.

Common enough.

smiley - smiley


Comments on the Tides

Post 8

Astronomer

An intriguing concept. Though we should examine it a little longer to see if this model can be accepted.
By your description, the side of the Earth not facing the Moon will see a high tide, and the side facing the Moon will be on low tide. The earth takes about a day to spin enough so that the Moon is on the same place of the sky. This means we would see one high tide and one low tide each day, anywhere on Earth. Though, we have 2 events of each type by day.
The error in this line of reasoning lies on the supposition that the things on Earth would not be at the "right" distance from the center of rotation, according to their velocities. The point of the Earth surface facing the moon changes, along with its velocity. The path of any points at the same latitude in Earth surface is the same, meaning that their velocity changes during the day. If not, there would be places with an eternal moon, or eternally moonless.
I think this is not exactly clear, so if you guys could not understand, please ask for more explanations.
Allen


Comments on the Tides

Post 9

Cefpret

I learnt that the spin of the Earth is only important for the final 'counting' (how many tides per day), but not for the forces or for the tides effect as a whole.

Because I haven't really understood what you say (sorry) -- is that correct?


Comments on the Tides

Post 10

Jimi X

In the entry what I was trying to describe was that the moon pulls the water toward it on the side facing it and it pulls the land toward it on the opposite side of the earth from where it is located.

Having the land pulled would create a high tide because the land would recede under the water. Having the water pulled on the near side would also pile up the water over the land, creating a high tide.

So we've got two high tides spaced evenly and the impact on the locations exactly 90 degrees perpendicular to the high tides would experience low tides.


Comments on the Tides

Post 11

Cefpret

The thing that is responsible for the high tide on the opposite side is the centrifugal force.

a) Moon facing side: Centrifugal force away from the moon + big gravitational force towards the moon = force towards the moon.

b) Opposite side: Centrifugal force away from the moon + small gravitational force towards the moon = force away to the moon.

a) and b) are not totally equally strong, but almost. The two forces (grav. and centr.) are equal in the common center of gravity of the Earth--moon system which lies inside Earth. There the resulting force is zero.


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