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compton
2006-Aug-30, 08:11 PM
Is there a reason the moon spins in perfect sync with the Earth? It seems a strange co-incidence if not - I don't believe other planets' moons spin in this way, with one side is always visible, and the other never seen, to someone on the surface.

Surely at least it must be slightly out of sync? Or did the Romans, and dinosaurs before them, see the same side that we do?

Tog
2006-Aug-30, 08:22 PM
Is there a reason the moon spins in perfect sync with the Earth? It seems a strange co-incidence if not - I don't believe other planets' moons spin in this way, with one side is always visible, and the other never seen, to someone on the surface.

Surely at least it must be slightly out of sync? Or did the Romans, and dinosaurs before them, see the same side that we do?

It's called Tidal Locking, and it is pretty common. Other's can explain it far better than I can, but here's the Wiki. (http://en.wikipedia.org/wiki/Tidal_lock)

Celestial Mechanic
2006-Aug-30, 08:34 PM
Is there a reason the moon spins in perfect sync with the Earth? [Snip!] Surely at least it must be slightly out of sync? Or did the Romans, and dinosaurs before them, see the same side that we do?
Tides on the Moon caused by the Earth and Sun slowed down the Moon's rotation until the rotational period equalled the revolution. The synchronization is not perfect, there is a slight nodding from side to side called "nutation". Here is an excellent little movie of it here (http://antwrp.gsfc.nasa.gov/apod/ap010218.html). Love that little movie! :)

I can't speak for the dinosaurs, but the Romans saw the Moon pretty much as we do.

Jason Thompson
2006-Aug-30, 08:57 PM
Is there a reason the moon spins in perfect sync with the Earth?

Yes, it's called tidal locking. Others have linked to places that will explain it.

I don't believe other planets' moons spin in this way, with one side is always visible, and the other never seen, to someone on the surface.

Actually that is the norm. Almost every moon in the solar sysytem exhibits this behaviour. The only exception I can think of right now is Hyperion, a minor moon of Saturn. Thanks to the gravitational interactions with Saturn and other satellites, Hyperion shows chaotic rotation.

In the case of Pluto and Charon, the effect has gone even further. Not only is Charon tidally locked to Pluto so it always shows the same face to it, Pluto is tidally locked to Charon as well. Charon is hence only visible at all from one side of Pluto, where it always hangs in the same spot, never rising or setting.

Surely at least it must be slightly out of sync? Or did the Romans, and dinosaurs before them, see the same side that we do?

There is a wobble in the Moon's face as seen from Earth, caused by the fact that the lunar robit is not perfectly circular. But certainly the Romans and probably the dinosaurs as well saw the Moon the same as we do. The Moon has had about 4 billion years to settle to a tidally locked state. The dinosaurs were only wiped out about 65 million years ago, so the Moon had had plenty of time to get tidally locked by then.

grant hutchison
2006-Aug-30, 09:29 PM
I don't believe other planets' moons spin in this way, with one side is always visible, and the other never seen, to someone on the surface.

Actually that is the norm. Almost every moon in the solar sysytem exhibits this behaviour.
Quibble:
The majority of the inner, regular moons of the solar system show synchronous rotation like our own moon's. All those outer lumps of distant rock in irregular orbits probably don't, since tides are weak at that distance.
Known synchronous rotators include: Phobos, Deimos, Metis, Adrastea, Amalthea, Io, Europa, Ganymede, Callisto, Epimetheus, Janus, Mimas, Enceladus, Tethys, Dione, Rhea, Titan, Iapetus, Miranda, Ariel, Umbriel, Titania, Oberon, Triton, Charon. I suspect some other inner Saturnians and Jovians have been confirmed by Galileo and Cassini since I last looked, and theory suggests that other inner satellites are also in synchronous rotation.

Grant Hutchison

neilzero
2006-Aug-30, 09:30 PM
Good answers except: I suspect many moons are not tide locked. Neil

pghnative
2006-Aug-30, 09:42 PM
Good answers except: I suspect many moons are not tide locked. NeilI don't suppose you'd care to elaborate. Your statement appears at odds with Grant's.

grant hutchison
2006-Aug-30, 10:32 PM
I don't suppose you'd care to elaborate. Your statement appears at odds with Grant's.I don't see a conflict. For instance: of Jupiter's 63 known satellites, seven are known to be synchronous, and one other inner moon (Thebe) seems theoretically likely to be synchronous (and may have been confirmed since I last looked); two of the innermost irregulars (Himalia and Elara) are known to be asynchronous, and almost all the rest occupy irregular retrograde orbits considerably further out than Himalia and Elara.
So that's maybe eight known to be or strongly suspected to be synchronous, four or five borderlines, and the remaining 50 very likely to be asynchronous (though only two confirmed).

Grant Hutchison

compton
2006-Aug-30, 11:03 PM
thanks for the explanations. What I gather is that the rotation of an object orbiting a significantly larger one will stabilise so that its heaviest side is closest the larger body.

the name, tidal locking, plus the description on wiki indicate that a larger planet's gravity will influence where the heaviest point on the moon is, by distorting it.

doesn't this give an easy way to determine how the moon was formed, as (according to wiki) tidal locking occurs very quickly (1000 years was mentioned)? If the moon was captured as a pre-existing entity, heavy elements should be pretty much equally spread throughout the crust, while if the moon formed from magma whilst in orbit, wouldn't heavier elements tend to gravitate to the near side?

The wiki mentioned something about the maria, the seas on the moon, as being heavier rock, but it also suggested that wouldn't have been a factor and that it's co-incidental that they exist almost solely on the near side.

grant hutchison
2006-Aug-30, 11:11 PM
Even a completely uniform body with no "heaviest side" would be stabilized by tides.
Tides pull an orbiting moon into a slight ellipsoid, with its long axis pointing at the parent body. so there's a bulge facing the parent body, and a bulge facing away from the parent body. If such a moon rotates, these bulges will be dragged across its surface by the parent body's gravity (just as the tidal bulges in our oceans are dragged around the Earth by the moon's gravity). The friction this creates will slow down a rapidly rotating moon, or speed up a slowly rotating moon, until an equilibrium is reached when there is no more dragging: the moon rotates in the same time period as it takes to revolve around the parent body.
If it happens to have a "heaviest side", this will tend to align with the tidal bulges: but it might end up facing the parent body, or on the opposite side from it. ("Heaviest side away" is another low energy configuration which might trap the rotating moon, though I think not as low-energy as "heaviest side towards".)

Grant Hutchison

compton
2006-Aug-30, 11:35 PM
...
Tides pull an orbiting moon into a slight ellipsoid, with its long axis pointing at the parent body. so there's a bulge facing the parent body, and a bulge facing away from the parent body. ...

OK I think I get you. Is it right to say that, as the moon spins, it is constantly distorted, with a bulge moving round, like a mouse under a carpet. This process will take energy from its rotation, until it is in sync with the planet. (BTW this is what I meant when I said "a larger planet's gravity will influence where the heaviest point on the moon is, by distorting it" in my previous post)

However I don't see why the moon becomes ellipsoid - why not more of a tear-drop shape, with the bulging end facing the parent planet?

grant hutchison
2006-Aug-30, 11:48 PM
However I don't see why the moon becomes ellipsoid - why not more of a tear-drop shape, with the bulging end facing the parent planet?There is a gravitational gradient across the moon, created by its parent body. The force pulling on its nearside is greater than the force pulling on its centre of mass, which is greater again than the force pulling on its farside. So there is a bulge away from the parent body on the farside as well as a bulge towards the parent body on the nearside. That's why we have two tides a day most places on Earth: because of the tidal bulge that faces away from the moon.

Grant Hutchison

hhEb09'1
2006-Aug-31, 05:35 AM
If it happens to have a "heaviest side", this will tend to align with the tidal bulges: but it might end up facing the parent body, or on the opposite side from it. ("Heaviest side away" is another low energy configuration which might trap the rotating moon, though I think not as low-energy as "heaviest side towards".)I'm was under the impression, in the moon's case, that the mass concentration was not aligned with the earth, that it was about thirty degrees off. That seems like a lot to me, after all the most it could be would be ninety.

grant hutchison
2006-Aug-31, 08:30 AM
I'm was under the impression, in the moon's case, that the mass concentration was not aligned with the earth, that it was about thirty degrees off. That seems like a lot to me, after all the most it could be would be ninety.I'd guess there's a moment of inertia consideration involved. A blob of heavy crust that did not deform well might be displaced from the axis of the low energy position if it were in competition with an area of lighter rock that formed a larger tidal bulge. Or so it seems to me on first thought.

Grant Hutchison

compton
2006-Aug-31, 09:23 PM
There is a gravitational gradient across the moon, created by its parent body. The force pulling on its nearside is greater than the force pulling on its centre of mass, which is greater again than the force pulling on its farside. So there is a bulge away from the parent body on the farside as well as a bulge towards the parent body on the nearside.

Ah... so if it bulges on the opposite side because the earth's gravity is weaker there, there must be a force pulling it away from earth - which must be the centrifugal force of the spinning moon I guess. That pretty much makes sense (although I have a distinct feeling of stretching a rudimentary knowledge to cover some big gaps here and there!)

thanks for the explanations and insights!

grant hutchison
2006-Aug-31, 10:11 PM
Ah... so if it bulges on the opposite side because the earth's gravity is weaker there, there must be a force pulling it away from earth - which must be the centrifugal force of the spinning moon I guess.Yeah ...
There are a couple of different ways of looking at this. Some folk like to look at the moon in the reference frame that revolves with it around the Earth, and point out that gravity and centrifugal force balance at the moon's centre (which is why it's in orbit). Closer to the Earth, on the nearside, the Earth's gravity is stronger and the centrifugal force due to the moon's movement around the Earth is weaker: net force towards Earth. On the farside, gravity is weaker and centrifugal force stronger: net force away from Earth.
(Ken G likes to point out that tidal bulges would happen even if the gravity gradient was absent: the centrifugal force gradient would produce the bulges on its own.)
But the gravity gradient across the moon would also produce both bulges on its own, because it distorts the gravitational potential of the moon into an ellipsoid shape: instead of "wanting" to settle into a perfect sphere under its own gravity, the moon "wants" to bulge its nearside one way, its farside the other way (and to make a compensatory constriction in a ring that runs around the edge of the visible disc as seen from Earth).

So the tidal bulges are a summation of two effects: the important thing to realize is that there isn't one bulge on the nearside entirely due to gravity, and one on the farside entirely due to centrifugal force, as is sometimes suggested. Both bulges arise from both forces.

Grant Hutchison

DALeffler
2006-Aug-31, 11:06 PM
How much of the bulge is due to the gravity gradient and how much is due to centrifugal force?

hhEb09'1
2006-Sep-01, 03:14 AM
How much of the bulge is due to the gravity gradient and how much is due to centrifugal force?None of it is due to centrifugal force. :)

A nonrotating body in a circular orbit experiences a centrifugal force at each point that is the same in magnitude and direction. That sort of situation cannot produce a bulge.

The moon also has a once per month rotation that also produces a centrifugal force, but that produces a bulge all around the equator, not a tidal type bulge at all.

PS: we've discussed this before (http://www.bautforum.com/showthread.php?p=118617#post118617) :)

phunk
2006-Sep-01, 03:25 AM
Ah... so if it bulges on the opposite side because the earth's gravity is weaker there, there must be a force pulling it away from earth

It's not that there's something pulling the far side bulge away, it's just that it's not being pulled towards the earth as much as the rest of the moon. Same thing with the tides, there's nothing lifting the sea on the side opposite the moon, it's actually the earth being pulled out from under the sea.

grant hutchison
2006-Sep-01, 11:08 AM
None of it is due to centrifugal force. :)See, this is the problem with the whole "centrifugal force" argument. People get declarative about it.
hhEb09'1, are you suggesting that if the moon were following a circular orbit, in its current state of rotation, but in a gravity field that did not decline with distance from the Earth, that there would be no tidal bulges?

Grant Hutchison

grant hutchison
2006-Sep-01, 11:29 AM
Supplementary to my previous post. If you want to read more about the whole "centrifugal problem" as applied to thinking about tides, there's an exchange between Ken G and myself on page 2 of a previous "Spinning Moon" thread (http://www.bautforum.com/showthread.php?t=43086&page=2) that walks through interpretations with and without centrifugal force.
There's also a website dealing with the same material here (http://www.lhup.edu/~DSIMANEK/scenario/tides.htm); I think it's fair to say that the author isn't that taken with deploying centrifugal forces in the explanation of tides, because of the potential for confusion.

Grant Hutchison

hhEb09'1
2006-Sep-01, 04:50 PM
See, this is the problem with the whole "centrifugal force" argument. People get declarative about it.I had to look the word up to see why you were using it against me! :)

But it doesn't seem to apply--I did offer an explanation in my post. Not as complicated as the discussions we've had in the past, but it was a start.
hhEb09'1, are you suggesting that if the moon were following a circular orbit, in its current state of rotation, but in a gravity field that did not decline with distance from the Earth, that there would be no tidal bulges?Maybe I'm misunderstanding what you're asking, but I don't see how that follows from what I said. I was talking about the usual sort of gravity, but I would entertain generalizations--I'm just not sure why that makes a difference in the discussion. Could you elaborate on the question?
Supplementary to my previous post. If you want to read more about the whole "centrifugal problem" as applied to thinking about tides, there's an exchange between Ken G and myself on page 2 of a previous "Spinning Moon" thread (http://www.bautforum.com/showthread.php?t=43086&page=2) that walks through interpretations with and without centrifugal force.Ken G's second post (http://www.bautforum.com/showthread.php?p=576632#post576632) to this forum concerned the problem, so there's a lot of reading there too. :)

OTOH, his tenth post (http://www.bautforum.com/showthread.php?p=576862#post576862), to the same thread later that day, was editted to remove his full name and university affiliation information (at my suggestion :) ), and he added a postscript that basically says that centrifugal force can be a convenient device, but plays no essential role.

But I don't want to speak for Ken G, I'm sure he'll stop by.

There's also a website dealing with the same material here (http://www.lhup.edu/~DSIMANEK/scenario/tides.htm); I think it's fair to say that the author isn't that taken with deploying centrifugal forces in the explanation of tides, because of the potential for confusion.I have nothing whatsoever against centrifictional force. I'm with the BA (http://www.badastronomy.com/bablog/2006/08/30/when-i-say-centrifugal-i-mean-centrifugal/). :)

neilzero
2006-Sep-01, 05:16 PM
2:3 and 3:4 resonances seem quite common. Is it appropriate to call these bodies "tide locked" ? Neil

hhEb09'1
2006-Sep-01, 05:36 PM
2:3 and 3:4 resonances seem quite common. Is it appropriate to call these bodies "tide locked" ? NeilI would say no.

There may be references that say otherwise--the mechanism of tidal braking may have altered the rotation so that it drifted into the resonance.

But since the tides would still vary over the surface of the object, I don't think that they can be said to be locked.

Tidal braking would still occur in those instances, I believe. The effect is proably compensated for by the effect on a mass concentration that is configured so that it gets more of a "kick" at the close approach of a more elliptic orbit..

Squashed
2006-Sep-01, 05:39 PM
Supplementary to my previous post. If you want to read more about the whole "centrifugal problem" as applied to thinking about tides, there's an exchange between Ken G and myself on page 2 of a previous "Spinning Moon" thread (http://www.bautforum.com/showthread.php?t=43086&page=2) that walks through interpretations with and without centrifugal force.
There's also a website dealing with the same material here (http://www.lhup.edu/~DSIMANEK/scenario/tides.htm); I think it's fair to say that the author isn't that taken with deploying centrifugal forces in the explanation of tides, because of the potential for confusion.

Grant Hutchison

I must be gettin' confused for in other dialogs centrifugal force is equated with gravity, i.e.: artificial gravity centrifuge, whereas in the tides the centrifugal force is totally discarded as a cause and it is gravity alone.

Also in discussing satellite orbits it is pointed out that the centrifugal force of the orbit balances the force of gravity and ergo we have orbit. In this case the centrifugal is anti-gravity.

Now if centrifugal force is the cause of the equatorial bulge in the earth, a rotating system, then the rotating earth/moon system can be viewed in a similar manner, I would think.

hhEb09'1
2006-Sep-01, 05:43 PM
Now if centrifugal force is the cause of the equatorial bulge in the earth, a rotating system, then the rotating earth/moon system can be viewed in a similar manner, I would think.It can, and Ken G is a strong advocate for that (as am I), but once the analysis is done, there is no significant portion of the tidal bulges that can be attributed to centrifictional force.

grant hutchison
2006-Sep-01, 09:06 PM
I had to look the word up to see why you were using it against me! :)"Against" you?
Your post moved me to remark on a general tendency I've observed, in which people make short declarative statements about centrifugal force, usually involving negation: it isn't a real force, it doesn't do this or that.
I do worry that this confuses the hell out of people who are having their first struggle to think about rotating systems.


I was talking about the usual sort of gravity, but I would entertain generalizations--I'm just not sure why that makes a difference in the discussion. Could you elaborate on the question?Imagine the moon orbiting the Earth in a gravity field which does not fall off with distance. We need to make the orbit perfectly circular and the force perfectly central for the sake of stability, but otherwise this corresponds to the situation with the real moon, except there's now no gravity gradient across our imagined moon. Bulges, or no bulges?


... he added a postscript that basically says that centrifugal force can be a convenient device, but plays no essential role.Sure. My point is just that a bald statement ruling out centrifugal force for consideration in this setting can undermine any future attempt to deploy it as a convenient device, and can render the inexperienced reader uncertain about anything they've read in the past which used that convenient device.

Anyway. This has some of the hallmarks of turning into another of our fruitless discussions. Since everything I want to say on the topic has now been said or appears in one of my links, I hope you'll forgive me if I bow out now. :)

Grant Hutchison

hhEb09'1
2006-Sep-02, 01:15 AM
"Against" you?
Your post moved me to remark on a general tendency I've observed, in which people make short declarative statements about centrifugal force, usually involving negation: it isn't a real force, it doesn't do this or that.
I do worry that this confuses the hell out of people who are having their first struggle to think about rotating systems.Yes, that's what I meant. I think confusion doesn't arise from short declarative true statements, though, it mostly comes from not addressing the issue. If my statement is wrong, show me, it would not be the first time that I was wrong.
Imagine the moon orbiting the Earth in a gravity field which does not fall off with distance. We need to make the orbit perfectly circular and the force perfectly central for the sake of stability, but otherwise this corresponds to the situation with the real moon, except there's now no gravity gradient across our imagined moon. Bulges, or no bulges?I wasn't sure, so it took awhile to work up. Assuming similar distances and magnitudes, it seems that there are insignificant bulges, less than ours by a couple orders of magnitude.
Sure. My point is just that a bald statement ruling out centrifugal force for consideration in this setting can undermine any future attempt to deploy it as a convenient device, and can render the inexperienced reader uncertain about anything they've read in the past which used that convenient device.Most of the writeups accessible to the inexperienced reader should be treated with uncertainty, because they're wrong. If they use that device correctly, they're probably not accessible to the inexperienced reader.
Anyway. This has some of the hallmarks of turning into another of our fruitless discussions. Since everything I want to say on the topic has now been said or appears in one of my links, I hope you'll forgive me if I bow out now.I never consider them fruitless. I've learned a lot. Thanks.

But Ken G should be able to address the issues from here on out.

DALeffler
2006-Sep-02, 02:32 AM
If there is no gravitational gradient across a body in a gravitational field, the only "stretching" force available is the vectoral sum (difference?) in the angles from different points on the one body to the gravitating body?

hhEb09'1
2006-Sep-02, 04:59 PM
If there is no gravitational gradient across a body in a gravitational field, the only "stretching" force available is the vectoral sum (difference?) in the angles from different points on the one body to the gravitating body?In general, that is true, though, right? In that case, it might be more of a compressive force, rather than a stretching force, if I understand you correctly.

hhEb09'1
2006-Sep-05, 05:05 AM
I wasn't sure, so it took awhile to work up. Assuming similar distances and magnitudes, it seems that there are insignificant bulges, less than ours by a couple orders of magnitude.OK, I made a mistake. It's only a third. Assuming, you know, I didn't make other mistakes.

What's that all about? Why did you ask?

Ken G
2006-Sep-07, 07:34 AM
I think it is 1/3, as I recall. That's why I always object to the pat answer you sometimes see "tidal bulges are caused by the fact that gravity is weaker at the far side than the near side". As you know from other discussions, tidal bulges will appear whenever the gravity along the center line does not increase as rapidly as the centrifugal force for an object in circular orbit. This statement works because the centrifugal force along the center line is a proxy for the lateral gravitational squeezing effect, which always balances the lateral centrifugal force, which is the same as the line of centers centrifugal force. I imagine the centrifugal effect would only be viewed as a causal player in tides in complete general relativity, and none of us are using that picture, thank heaven.

hhEb09'1
2006-Sep-08, 09:03 PM
I think it is 1/3, as I recall. That's why I always object to the pat answer you sometimes see "tidal bulges are caused by the fact that gravity is weaker at the far side than the near side".I've never seen quite that pat of an answer. :)

Most of the analyses I've seen depend upon a calculation of differences--which would translate well into even the "increasing gravity" scenarios that you are envisioning. Even better, I think, since they would apply even when more esoteric functions would fail your simple two-dimensional linearized test. I don't think we should expect high level mathematics, but I am more than satisfied with the principle.

PS: be that as it may, I just want to say I appreciate the discourse and hope we can continue, I don't mind at all delving into higher math myself

Ken G
2006-Sep-08, 10:04 PM
I've never seen quite that pat of an answer. :)

You haven't looked far. For example, if you look up "Tide" in Wikipedia (often considered the first go-to source for online information around here), you will find:


The reason for this is that the tidal force is related not to the strength of a gravitational field, but to its gradient. The field gradient decreases with distance from the source more rapidly than does the field strength..
So that's just what I'm talking about, in black and white. There is actually no such thing as a "gradient" of a vector field (gradients act on scalar fields), but Wikipedia seems to be saying they are comparing what is happening to the field strength with what is happening to its gradient (and the field strength is indeed a scalar and does have gradient). Problem is, this is precisely what I have objected to as being too "pat". Even a field strength with zero gradient induces tidal bulges. I think I would not have to look far on the web to find more of the same. It is really only in books, with glossy 2D pictures, that you are likely to get all the little arrows you need if you are not using the device of the centrifugal force in your explanation.


Most of the analyses I've seen depend upon a calculation of differences--which would translate well into even the "increasing gravity" scenarios that you are envisioning.
What one means by "differences" requires vector thinking (or the centrifugal force). The ones that try to talk about differences without either of those elements are the faulty ones, like the Wikipedia entry, and are "too pat".

I don't think we should expect high level mathematics, but I am more than satisfied with the principle.
That's exactly why my tidal explanation involves only an analysis of the line of centers-- how high level is it to say that gravity and the centrifugal force are out of balance? "The principle" is that no purely one-dimensional analysis of tidal bulges is correct unless it invokes the centrifugal force for circular orbits (or equivalently, Kepler's law).

PS: be that as it may, I just want to say I appreciate the discourse and hope we can continue, I don't mind at all delving into higher math myself

Yes, the linearized eigenvalue picture of tidal deformation is a useful one, as it applies to anything with the right symmetry and a small planet at a large distance.

hhEb09'1
2006-Sep-09, 03:27 AM
It's called Tidal Locking, and it is pretty common. Other's can explain it far better than I can, but here's the Wiki. (http://en.wikipedia.org/wiki/Tidal_lock)Whoa, that said "In the case of the Moon, both its rotation and orbital period are just over 4 weeks." It should read "just under".
You haven't looked far.Far enough :)
For example, if you look up "Tide" in Wikipedia (often considered the first go-to source for online information around here), you will find:
The reason for this is that the tidal force is related not to the strength of a gravitational field, but to its gradient. The field gradient decreases with distance from the source more rapidly than does the field strength.. So that's just what I'm talking about, in black and white.Actually, those particular sentences are referring to "the reason" that the lunar tide is larger than the solar tide. You might want to look farther. :)
It is really only in books, with glossy 2D pictures, that you are likely to get all the little arrows you need if you are not using the device of the centrifugal force in your explanation.You can actually explain it without centrifugal force and without arrows. One of the explanations that I've mentioned before, Stacey's, uses just potential--no arrows--but also uses centrifugal force. However, it is not necessary to use centrifugal force.
That's exactly why my tidal explanation involves only an analysis of the line of centers-- how high level is it to say that gravity and the centrifugal force are out of balance? As I've said many times, I do not have a disagreement with the use of centrifugal force in the explanation. I just disagree that the others are somehow fatally flawed.
"The principle" is that no purely one-dimensional analysis of tidal bulges is correct unless it invokes the centrifugal force for circular orbits (or equivalently, Kepler's law).I think I would say no purely one-dimensional analysis of tidal bulges is correct.

The question that is usually addressed in all those substandard explanations of the tides is why there are two bulges instead of just one bulge. Your objection seems to be that in some (non-physical) circumstances there is no bulge. I don't think that invalidates the explanations, as far as their intended scope.

Ken G
2006-Sep-09, 02:11 PM
Actually, those particular sentences are referring to "the reason" that the lunar tide is larger than the solar tide.
I was quite aware of that, it refutes my point not in the least. My point is not that it has to be wrong everywhere to be a problem, merely that if it is wrong anywhere authoritative, it will foster the misconception I am attempting to correct. How do you know that this isn't the only part someone will read, as it quite clearly implies the cause of tides along with explaining why one tide is stronger than the other? Is the latter not a topic that must contain the former?


You might want to look farther.
All right, let's do that. Let us google "cause of tidal bulges", fair enough? The first hit is from NOAA, http://tidesandcurrents.noaa.gov/education.html, and contains this explanation of the bulges:


The gravitational attraction between the Earth and the moon is strongest on the side of the Earth that happens to be facing the moon, simply because it is closer. This attraction causes the water on this “near side” of Earth to be pulled toward the moon. As gravitational force acts to draw the water closer to the moon, inertial force attempts to keep the water in place. But the gravitational force exceeds it and the water is pulled toward the moon, causing a “bulge” of water on the near side toward the moon (Ross, D.A., 1995).

On the opposite side of the Earth, or the “far side,” the gravitational attraction of the moon is less because it is farther away. Here, the inertial force exceeds the gravitational force, and the water tries to keep going in a straight line, moving away from the Earth, also forming a bulge (Ross, D.A., 1995).

In this way the combination of gravity and inertia create two bulges of water.
If read carefully, this is just the argument I give, as it is all along the line of centers, and "inertial forces" are the centrifugal force. Note however the words I bolded, which although correct as written, quite clearly foster the misconception I am attempting to clear up. That's the first hit, and it's NOAA!

Shall we go on to the next hit? It's a forum, so let's move on to #3, which is
http://mb-soft.com/public/tides.html, passing itself off as very authoritative, and this is what it has to say:

A molecule of water in an Ocean that is directly below where the Moon is in the sky, is about 4,000 miles closer to the Moon's mass than the center of the Earth is. That closeness (and the inverse square distance part of the gravitational law) means the Moon's gravitation attracts that molecule of water more than the Moon attracts the Earth as a whole, or any other water in any other oceans. The Moon is certainly attracting the body of the Earth, too, but there is just a "differential" because of the stronger pull of the water that is closer to the Moon.
I don't think I can spell out the misconception any clearer than that, people are led to think we would not have tides if the Moon's pull wasn't stronger on the closer side of the Earth. Have I looked far enough yet?


You can actually explain it without centrifugal force and without arrows.
I'm quite skeptical of the truth of this remark, if by "without arrows", you mean "without appealing to the way a force picks up a lateral component as its direction turns". That is precisely the kind of multi-dimensional thinking that I try to avoid because I think it is harder for nonmathematical people to picture.


I think I would say no purely one-dimensional analysis of tidal bulges is correct.
The centrifugal force argument is one-dimensional and correct. It does require one further insight, which is that along the lateral direction you have material orbiting stresslessly just like at the center of the planet. That is the only lateral consideration you need, and it's obvious. After that, it's all entirely 100% one dimensional, this is the point I have struggled to make.

Your objection seems to be that in some (non-physical) circumstances there is no bulge. I don't think that invalidates the explanations, as far as their intended scope.What I object to with the incomplete explanations is that, if someone says "A causes B", the obvious corollary is "the absence of A should generally not cause B". That is what is incorrect in those explanations, therefore they are poor explanations. They do not capture the true essence of the thing they are explaining. It's like saying, "making money in the stock market is caused by buying stocks when they are below their true value". No, that can work, but you can still make money by buying stocks above their true value, if you can sell them for even more. That's the true essence that includes all the possibilities, within reason. Now, sometimes we accept a poor explanation because a correct one is too difficult. This is not the case here, most people have a pretty good "feel" for the centrifugal force.

hhEb09'1
2006-Sep-09, 07:54 PM
I was quite aware of that, it refutes my point not in the least. My point is not that it has to be wrong everywhere to be a problem, merely that if it is wrong anywhere authoritative, it will foster the misconception I am attempting to correct. How do you know that this isn't the only part someone will read, as it quite clearly implies the cause of tides along with explaining why one tide is stronger than the other? Is the latter not a topic that must contain the former?It's just not the pat answer that you were asserting. I think we should judge these things on face value, not the possible shortcomings of the potential reader.
All right, let's do that. Let us google "cause of tidal bulges", fair enough? The first hit is from NOAA, http://tidesandcurrents.noaa.gov/education.html, and contains this explanation of the bulges:That NOAA website is infamously bad. At one time, it said that the two tidal bulges were the sublunar one and the subsolar one!

Here's a page there (http://tidesandcurrents.noaa.gov/restles3.html) that says "The tide raising force of the moon, is, therefore, entirely insufficient to 'lift' the waters of the earth physically against this far greater pull of earth's gravity. Instead, the tides are produced by that component of the tide-raising force of the moon which acts to draw the waters of the earth horizontally over its surface toward the sublunar and antipodal points."

Nevermind that there is such a thing as the solid earth tide! :)

The person who first wrote that webpage for NOAA produced a centrifugal force based argument that was so screwed up that it has taken years to straighten out. You would think that NOAA could do better, especially after two or three tries.

But none of the examples that you give are as "pat" as you claimed--the NOAA website goes on and on for dozens of pages. Taking a single sentence out of context--especially one that is essentially correct--does not prove that the website explanation is erroneous.
I'm quite skeptical of the truth of this remark, if by "without arrows", you mean "without appealing to the way a force picks up a lateral component as its direction turns"."Skeptical of the truth" of my remark? I've referenced Stacey's proof before. As I said, it's done with potentials. No vectors.

The centrifugal force argument is one-dimensional and correct. It does require one further insight, which is that along the lateral direction you have material orbiting stresslessly just like at the center of the planet. That is the only lateral consideration you need, and it's obvious. After that, it's all entirely 100% one dimensional, this is the point I have struggled to make.Then, why call it "purely" one dimensional?
What I object to with the incomplete explanations is that, if someone says "A causes B", the obvious corollary is "the absence of A should generally not cause B". That is what is incorrect in those explanations, therefore they are poor explanations.Let me see if I can paraphrase that: IF A THEN B implies IF NOT A THEN NOT B. Is that correct? :)

Ken G
2006-Sep-10, 03:17 AM
.That NOAA website is infamously bad. At one time, it said that the two tidal bulges were the sublunar one and the subsolar one!
Be that as it may, the site is a confirming instance of my contention that the "pat" answer is quite widespread in authoritative websites.



Here's a page there (http://tidesandcurrents.noaa.gov/restles3.html) that says "The tide raising force of the moon, is, therefore, entirely insufficient to 'lift' the waters of the earth physically against this far greater pull of earth's gravity. Instead, the tides are produced by that component of the tide-raising force of the moon which acts to draw the waters of the earth horizontally over its surface toward the sublunar and antipodal points."
Yes, that is an example of a perfectly fine answer, but note that it also requires two-dimensional thinking and understanding of vector components. For example, how many people do you think could take that answer and be able to tell that a gravity that increases linearly with distance would not produce that effect? You could, because you can do vector subtraction. Most others? This is precisely what I'm avoiding with my approach, though it is certainly a fine alternative. Moreover, the fact that some websites get it right is again no refutation of my claim that the incorrect ones are not uncommon. You are a good enough logician to know that.


But none of the examples that you give are as "pat" as you claimed--the NOAA website goes on and on for dozens of pages.
So? Do you think everyone will read them all? A pat answer is an answer that sounds good enough for people to stop reading. They think they "got it", and are happy with a misconception, as though the illusion of understanding was as good as real understanding. That's practically the definition of "pat", in my book. Now a good explanation is allowed to take some liberties, but it must capture the essence or we can do better. And how you can claim that I have taken an obvious effort to explain the effect as "out of context" is completely beyond me, it's "argument by outrageous claim".


I've referenced Stacey's proof before. As I said, it's done with potentials. No vectors.
Well then it includes the centrifugal force in an effective potential, yes? Proves my point, there's no way to do it with pure gravity and in only one dimension. One either invokes centrifugal forces of a circular orbit as a device, or one considers a 2D gravity field. That's what I'm skeptical of-- unsubstantiated claims to the contrary. (And you don't have to explain the whole proof, any essence of it will suffice for me to understand it.)



Let me see if I can paraphrase that: IF A THEN B implies IF NOT A THEN NOT B. Is that correct? :)I suspected you would try a pedantic response based in formal logic, as if I didn't know that. Let me clarify the argument. As part of a good explanation of the cause of B, one must be able to contrast against situations where B will not appear. It is the difference between saying "smoking cigarettes can cause lung cancer" versus "lung cancer is caused by smoking cigarettes". Which type of causal statement would you say is "tides are caused by the fact that the Moon's gravity is stronger on the near side than the far side of Earth"? Is it just trying to cite one contributing factor in generating tides (out of a wider set of possibilities), or is trying to explain the cause of tides? If you actually believe it is the former, then I guess you would have no objection with the pat answer, obviously. If you believe the latter, then note it is not hard to do better, and also please judge for yourself the relevance of your logic puzzle.

hhEb09'1
2006-Sep-10, 03:33 AM
Be that as it may, the site is a confirming instance of my contention that the "pat" answer is quite widespread in authoritative websites.Not the NOAA website. It goes on for pages.
And how you can claim that I have taken an obvious effort to explain the effect as "out of context" is completely beyond me, it's "argument by outrageous claim".Because it was referring to a different aspect of the effect than the one you claimed it was. It was not "explaining" the tides, it was explaining why the Sun's tide was smaller than the moon's tide.

Well then it includes the centrifugal force in an effective potential, yes? Proves my point, there's no way to do it with pure gravity and in only one dimension.There is a way to do it with pure gravity but not in only one dimension. As you've admitted, there is no way to do it in one dimension.

I suspected you would try a pedantic response based in formal logic, as if I didn't know that. You're the one who said it was obvious. It's not even obvious. It's wrong.

Ken G
2006-Sep-10, 01:02 PM
Not the NOAA website. It goes on for pages. ...Because it was referring to a different aspect of the effect than the one you claimed it was. It was not "explaining" the tides, it was explaining why the Sun's tide was smaller than the moon's tide.I already refuted both those arguments above, in a manner that has not been countered. You're just repeating yourself.


There is a way to do it with pure gravity but not in only one dimension.
Obviously. The effect is entirely caused by pure gravity, as we've always agreed.


As you've admitted, there is no way to do it in one dimension.
You're the one who said it was obvious. It's not even obvious. It's wrong.
"Admitted"? There is a way to do it in one dimension, using the centrifugal force as a device, in the sense that the other dimension is dispensed with by asserting it is all the same orbit (and yes, I do find that obvious). What are you talking about when you say not obvious and wrong? Is there some particular statement of mine that you think is wrong, or are you still talking about that NOAA website (whose explanation is quite correct, but badly worded).