The Earth's tidal bulge is always slightly ahead of the Moon, isn't it? And the Moon pulling on it slows the Earth's rotation (Bad Astronomy pp 72-73).

In the case of Mercury, though, the tidal bulge lags behind the Sun at perihelion (due to the increase in orbital velocity) and runs ahead of it at aphelion (due to the decrease in orbital velocity). So the Sun slows the planet's rotation during the slower half of the orbit, and speeds it up during the faster half. Is this what you mean?

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Yes, that's what I meant.
No, not necessarily. The tidal slowing should in fact be continuous, but if there is a true resonance, then the slowing would be offset by some sort of speed up. For instance, if there were a dipole moment in the density distribution (like the tidal distribution, but permanent) such that it was oriented to speed up the rotation at perihelion then one revolution later (and one and a half rotation later) it would again be oriented at perihelion to speed up the rotation. It's orientation at other parts of the orbit would both speed up and slow down the rotation, but the effects would either cancel or be smaller than that at perihelion.

Thanks for that, kilopi. I found this interesting PDF which explains the whole process in mathematical detail.

They do confirm my suspicions about the tidal bulge:
However, they conclude that:
Which, I guess, is precisely the point you were making about the dipole moment in the density distribution.

I might have known the reason would be far from simple.

[Edited to provide better hyperlink]

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Interesting! I would not have suspected that, without actually computing it. I wonder how long the retrograde motion lasts (I seem to have a problem accessing that link to the pdf file right now).

For a perfectly circular orbit, the apparent motion of the Sun is equal to one rotation, whereas the actual rotation is one and a half. Thus, it has to get close and speed up quite a bit to make the Sun appear retrograde.

I suspect that if there were not a permanent dipole embedded in the material of Mercury, it would have progressed to a real tidal lock.

Sorry about the dodgy link. The article is titled Mercury: The Planet and its Orbit. It's by Andre Balogh and Giacomo Giampieri.

You can also download it from Giacomo Giampieri Publications.

The retrograde motion of the Sun only lasts about 6 or 7 days:

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Which possibly means that the retrograde motion is not great--so the apparent rotation speed is small also. That would tend to reduce the effect of tidal slowing--I mean, speeding.

Did they mention what the apparent maximum retrograde rotation was?

No. I've quoted all they have to say about the retrograde motion of the Sun. Among their conclusions are the following:
In an earlier post I wrote:
It turns out that it is the permanent (non-tidal) bulge or deformation which points towards the Sun at perihelion, not the tidal bulges (which, as you said, point more-or-less towards the Sun at all times). A number of websites I have consulted make this mistake.

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It is subtle, isn't it?

Worse, if the permanent deformation truly did point towards the sun at perihelion, the effect would be more or less symmetric and wouldn't contribute the required speed up--so it probably does not point to towards the sun at perihelion!

Ah ha! Apparently, at least one of my two cents worth of guessing was on the money.

Does this deformation help explain the ecentricity of the orbit as well? Is the Sun "flingining" it around at perihelion with a little extra kick from the deformation tug to hold the eccentricity?

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"The Sun, with all the planets revolving around it, and depending on it, can still ripen a bunch of grapes as though it had nothing else in the universe to do..." Author: Galileo supposedly.

This page is perhaps the best explanation for the tides that I have read so far. While gravity alone CAN produce tidal forces, the apparent negative movement of the far bulge can only really be explained by the centrifugal force of the Earth orbiting the Earth-Moon centre of mass.

http://co-ops.nos.noaa.gov/restles3.html

I haven't looked at it lately, but we've discussed it before. Unless it has been updated, it can't be trusted. It was so full of errors, mistakes, and misconceptions, that I would consider it an embarrassment to NOAA.

PS: The centrifictional force is the same at all points on the Earth--in magnitude and direction. So there is no differential because of it. That webpage was amended to say that years ago, possibly in response to our emails, but other statements were left in place that conflicted with that notion.

I have a problem with the Sawicki (BA) model for tides which goes like this: -

Dunk a tennis ball in water, hold it up and look at it. See the water drain to the underside? That's a tidal bulge.

Now, according to Sawicki's model this tide is due to 'differential' gravity and we can calculate what that is:-

using Sawicki's equation (1) the gravitational pull of the earth a(e) at distance d(e) is: a(e) = G*M(e) / d(e)^2 (where d(e) is the distance to the centre of earth) (1)
= 6.67259e-11 * 5.978e24 / 6.378e06^2
= 9.81 ms^-2 (yes, that sounds familiar)

then, using Sawicki's equation (4) for the difference between the earths pull on the closest/farthest point on the ball:

Da(e) = a(e) * 2R / d(e) (where R is the radius of the ball, 0.05m, say) (4)
= 9.81 * 0.1 / 6.378e06^2
= 2.41e-14 ms^-2 (?????????)

You can see the problem can't you - differential gravity is 2.41e-14 ms^-2 . This is vanishingly small, it isn't even pico-gravity, but according to Sawicki that's what causes the tide.

Please do tell me - where am I going wrong?

I think this is where you are going wrong. The movement of the water in this example is not due to tidal forces - the water moves downwards because of gravity, yes, but the ball is too small for tidal forces to have any appreciable effect upon it.

In Sawicki's paper, the Earth (or Moon, or whatever) is in freefall. If you hold the tennis ball up and look at it, it's not in freefall. You'd have to drop it first and then see how the water moves.

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Ian R: The question is a little more subtle than that Ian, but thanks anyway - you might even be right.

Eroica: I think the term 'free-fall' is extremely dodgy but I accept and will address your comment.

Sawicki states that the earth as a whole is in free-fall about the sun. This is not correct. By any useful definition of 'free-fall', whilst point O (Sawicki's fig.1) is in free fall, but points F and C are not. This is because O is the only point at which centripetal acceleration (w^2*r) is equal to gravitational acceleration (G*M(s) / d(s)^2). At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.

This analysis does not require any mysterious 'reverse' gravity and, if you care to do the sums, you'll see that the imbalances are exactly equal and opposite. In contrast, Sawicki's 'differential' model requires (or at least implies) a gravity reversal and it produces a 0.01% imbalance. This may not seem much but in the earth-moon system it's very significant - I've not calculated it recently but I think it's as much as 5%.

Incidentaly, my original post contains an error:- 'differential' gravity should be 1.54e-07 not 2.41e-14. My mistake - I incorrectly squared d(e). Sorry about that.

No, not really.
The tidal bulge is also proportional to the radius of the body. A tennis ball is a lot smaller than the Earth.
Useful definition? By that definition of free fall, there are no "bodies" in free fall. That's not very useful--a term that applies to no body.

Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?

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My definition of "free-fall" is "subject only to gravitational forces." What's yours?

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