1. Originally Posted by grant hutchison
So there's a gradient of gravity across the moon, which pulls the far side away from the Earth, and the near side towards the Earth: two tidal bulges.
Interestingly, there would still be tidal bulges even if the magnitude of gravity was constant with distance. One way to show this is to note that a constant gravity could not keep pace with the increasing "centrifugal force", in a reference frame where gravity and centrifugal force are the only things going on. Some like to invoke centrifugal forces to help understand tidal bulges, some don't (they aren't required), but the bottom line is that a "top heavy" part of the Moon could end up either on the side facing Earth, or on the opposite side-- either would make that heavy piece "happy".

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Originally Posted by Ken G
Interestingly, there would still be tidal bulges even if the magnitude of gravity was constant with distance.
I've seen you do this one before, and it always makes me feel slightly uneasy, for some reason.

Forgive me for talking myself through a line of reasoning that's certainly familiar to you:
I've always found the tides easiest to understand by first invoking a scenario in which objects revolve around their common barycentre, but do not rotate. So if we watched a model Earth on the computer screen, we'd see it lollop around the Earth-Moon barycentre, but with Africa (say) always pointing to the top of the screen. So the centre of mass of the Earth is moving in a (for the sake of the model) circle around the barycentre, and is subject to a centrifugal pseudoforce which is exactly balanced by the gravitational attraction between moon and Earth. Now, all the other points on the surface and within the volume of the Earth are moving in circles, too: and these circles have exactly the same radius of gyration as the centre of the Earth has, except around different centres. So (in this model on the computer screen) a point on the equator in Africa is circling around a point displaced one Earth-radius towards the top of the screen relative to the barycentre, and so on. So centrifugal force is the same everywhere across the volume of our model Earth, and necessarily cancels with gravity at the centre of mass.
It then becomes an intuitive (and arithmetically simple!) task to look at how the gravity vector varies across the surface of the model Earth, to combine it with the constant centrifugal vector, and derive the residual tidal forces: two tidal bulges, and a circle of compressive forces in a plane at right angles to the Earth-Moon line. Then we can mentally set the Earth rotating on its axis, and imagine an equatorial bulge arising from that rotation, superimposed on the pre-existing tidal forces.

Now I plug your constant-gravity situation into that model, and I initially get only the compressive forces in the plane at right angles to the Earth-Moon line; there is no mismatch at the near and far points, so the tidal force vector is zero where previously we had bulge-forming forces.
Only once I set the Earth spinning do I get a centrifugal bulge at the equator, modified by the gravitational compressive forces, which produces something like the original distribution of forces I saw arising entirely from variations in the gravity force in the original situation.

So the tidal bulge in your constant-gravity situation is achieved by solely compressive gravitational tidal forces (which alone would produce a relative bulge), aided by centrifugal forces (which shift some of the force vectors towards bulge-raising)?

Grant Hutchison

3. Originally Posted by grant hutchison
Originally Posted by grav
That is to say, for instance, if the Earth rotates about 365 times per revolution around the sun (per year), and if one of these is caused by the orbit itself, then it is really rotating 364 or 366 times depending on whether it rotates with or against the revolution.
That's right: the Earth rotates 366 times relative to the stars (what's called a "sidereal day" in the time it takes to rotate 365 times relative to the sun.
Wouldn't this also be true of the Earth-moon system as well? Does this mean that we are really rotating much faster (or slower) than we appear to be when we subtract one rotation per revolution of the moon (about one per month)? If this is the case, then would the sidereal year actually be plus one for the revolution of the Earth around the sun and plus or minus 12 for the revolution with the moon?

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Originally Posted by grav
Wouldn't this also be true of the Earth-moon system as well?
I'm sorry, I don't understand what you're asking.
Everything has a sidereal rotation period, which is fixed relative to the background stars. Stuff that goes around other stuff also has a synodic rotation period relative to the thing it's orbiting: so it takes the Earth a little longer than a sidereal day to turn to face the Sun again (because the Earth has moved through an angle relative to the Sun while it was rotating), and even longer to turn to face the moon again (because it has moved through a larger angle relative to the moon).
Neither of these synodic periods has any influence on the sidereal periods of rotation or revolution: it's just that we're interested in synodic periods because they bring astonomical bodies back to the same position in the sky.
Does that address your question at all, or am I off at a tangent?

Grant Hutchison

5. My thinking is, and I'm probably off-base here, that if the moon completes one orbit per revolution relative to Earth, then the Earth also does the same thing relative to the moon. This would mean that the Earth completes twelve synodic rotations relative to the moon per year in addition to that of once a year relative to the sun. If this is true, then the Earth really rotates upon its own axis even faster (or slower depending on the direction) than we now measure. Am I right on this?

6. Originally Posted by grav
If this is true, then the Earth really rotates upon its own axis even faster (or slower depending on the direction) than we now measure. Am I right on this?
Don't think so. How do you figure that the Earth is rotating on it's axis faster because it is orbiting the Sun or the Moon. You should measure rotation relative to some spot in the sky very far away, such as a quasar. Then you will see that the Earth rotates 366.2425 times per tropical year. The movements with respect to the Sun and Moon have nothing to do with the rotation rate.

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Originally Posted by grav
My thinking is, and I'm probably off-base here, that if the moon completes one orbit per revolution relative to Earth, then the Earth also does the same thing relative to the moon. This would mean that the Earth completes twelve synodic rotations relative to the moon per year in addition to that of once a year relative to the sun. If this is true, then the Earth really rotates upon its own axis even faster (or slower depending on the direction) than we now measure. Am I right on this?
No, there's no extra rotation hidden away inside the orbital motion.
The Earth rotates once in 23 hours 56 minutes or so. It then needs to rotate for another 4 mins to catch up with the position of the sun, because it has moved about one degree in its orbit around the sun during that period (four minutes being a 360th of a day, and therefore the time it takes the Earth to rotate a degree). So we end up with a 24 hour day relative to the sun.
The Earth then has to rotate for a little less than another hour to catch up with the position of the moon, which is moving more quickly in its orbit: so we see successive moonrises coming later each night, because we set our clocks to keep pace with the sun.
And at the end of the lunar month, the moon has to orbit for another couple of days to catch up with the sun, which has shifted in the sky by 30 degrees or so during the course of one lunar orbit. So we see the lunar phases cycling every 29 days or so, instead of with the lunar orbit, which lasts about 27 days.

(All figures are approximate here, but I think that's adequate to convey the general idea.)

Grant Hutchison

8. Originally Posted by grant hutchison
So we end up with a 24 hour day relative to the sun.
I've thought about it and found that it once again depends on the frame of reference. Since we determine a day by the time it takes for the sun to rise and fall until it reaches the same position in the sky (which may or may not be a precise definition) and we also determine a year by the time it takes for the Earth to complete one revolution around the sun, both frames of reference are according to the Earth-sun system, so the orbit of the moon does not matter. A year, then, is about 365 days and a sidereal year is about one extra day because of the revolution. End of story.

But I think I still have a point. If the Earth-moon system revolves about itself about once a month, then it will have about twelve synodic rotations per year. Now, again, I can see what you mean in your post, that this will not effect the length of the year at all, but I guess now the question is, if the moon did not exist, would the length of the year relative to the sun remain the same? That is to say, the synodic relationship between the Earth and moon adds or subtract (depending on the direction) some amount of apparent rotation per month relative to the sun and stars. So if the moon was not present, this should then be a noticeable difference indeed.

I suppose another question relating to this would be if most of the initial angular momentum of the Earth-moon system as the solar system formed was carried by the Earth. Since the Earth is much more massive than the moon, this is probably the case. Therefore, the length of the year if the moon was not present should not differ quite as much as twelve days per year because the energy the Earth initially had wouldn't have been too different than it is now even though it shares it with the moon. In this case, the Earth would have carried about the same rotational energy that the moon now taps into and would include most of the synodic rotation as well (which would then amount to just extra normal rotation) instead of sharing this energy with the moon. If I had to guess, I would say the the energy (or angular momentum) is about proportional to their masses, so the the length of the year with the Earth alone should only differ by about 3.5 hours ([Mmoon/MEarth]*12 days=3.5 hours).

9. Originally Posted by grav
That is to say, the synodic relationship between the Earth and moon adds or subtract (depending on the direction) some amount of apparent rotation per month relative to the sun and stars.
All the Moon does is make the center of the Earth go in a circle, it has no impact on Earth's rotation, other than to create an extremely gradual slowing of said rotation. The point is, the Earth's rotation is what it is, there is no reason to analyze the myriad of reasons why it is what it is. The number of rotations per year are not added to or subtracted from because there is a Moon. Indeed the only significant difference between the synodic year and the sidereal year is simply the need to finish a complete rotation before we can bang our pans and wish everyone a happy new year. That's not like the difference between the sidereal and the synodic month, which has nothing at all to do with the rotation of the Earth (but rather its orbit).

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Originally Posted by grav
I suppose another question relating to this would be if most of the initial angular momentum of the Earth-moon system as the solar system formed was carried by the Earth.
If the Earth had a different angular momentum it would be in a different orbit, so the year would be certainly be different.
But the energy and angular momentum associated with the moon came from outside the Earth: from the big impact that formed the moon in the first place. So I'm not sure you can reason from the current situation back to how a moonless (ie impactless) Earth would behave.

Grant Hutchison

11. Sorry I missed this before:

Originally Posted by grant hutchison
I've seen you do this one before, and it always makes me feel slightly uneasy, for some reason.
I think what makes you uneasy is that you may think I am saying that without centrifugal effects there would be no tides, but that's not what I mean, because an earth/moon system that is simply falling together also experiences tides. I use the centrifugal force as a device to show why there must be tides in a co-orbiting system, i.e., as a device to show that the way gravity behaves is not the right way to avoid tides (the standard one-dimensional explanation is quite misleading in this regard).

Originally Posted by grant hutchison
I've always found the tides easiest to understand by first invoking a scenario in which objects revolve around their common barycentre, but do not rotate.
I agree, rotation is a complete red herring, as all it does is create oblateness, not tidal deformation (of course it does play a role in the ebbing and flowing of ocean tides, but we're talking about the tidal deformation here). But this does not require we use no rotation, it means we may choose whatever rotation rate makes the problem easiest to analyze. If one is using the co-orbiting frame, then tidally locked rotation is the easiest to consider. If using your description, then eliminating rotation is fine.

Originally Posted by grant hutchison
So if we watched a model Earth on the computer screen, we'd see it lollop around the Earth-Moon barycentre, but with Africa (say) always pointing to the top of the screen. So the centre of mass of the Earth is moving in a (for the sake of the model) circle around the barycentre, and is subject to a centrifugal pseudoforce which is exactly balanced by the gravitational attraction between moon and Earth.
Yes, that works too, but note you have to be able to subtract vectors to make this work in general.

Originally Posted by grant hutchison
So centrifugal force is the same everywhere across the volume of our model Earth, and necessarily cancels with gravity at the centre of mass.
Yes, if using that picture, the centrifugal force is not a useful device. But that is an unusual reference frame to apply the centrifugal force anyway. Normally one would call it the centripetal acceleration and stay in the inertial frame (which is of course fine).

Originally Posted by grant hutchison
It then becomes an intuitive (and arithmetically simple!) task
Only if you can subtract vectors. If you want to stick to a one-dimensional analysis, that's where the centrifugal force approach is helpful.
For example-- calculate the type of gravity that yields no tides. It matches the centrifugal force along the line of centers-- that's pretty leading as to why the centrifugal force may be used as a helpful device.

Originally Posted by grant hutchison
Now I plug your constant-gravity situation into that model, and I initially get only the compressive forces in the plane at right angles to the Earth-Moon line; there is no mismatch at the near and far points, so the tidal force vector is zero where previously we had bulge-forming forces.
Yes there is, because it spills over from the lateral squeezing. The Earth must be in a global equilibrium. But you realize this, and your analysis works too.

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OK. I see that my use of "centrifugal force" is, umm, eccentric in the scenario I described: it involves me in an infinitude of infinitesimally separated rotating reference frames, one for every point within the volume of my nonrotating Earth. Staying inertial and thinking "centripetal" is much more parsimonious, though the final destination is the same.

I think my unease is maybe just at watching you travel through centrifugal, as it were, to get to what I've always thought of as a purely gravitational effect. So I rather like the business of setting all the centripetal force vectors equal, and then looking at the local differences in the gravity vector. It seems to automatically sort rotation, revolution and gravity into appropriately different mental "bins".
I'm taking from your responses (for which, thanks) that you don't find the approach I described intrinsically misleading, but perhaps overly complicated, given its initial mix of non-rotation with revolution, and the requirement for some understanding of vectors. Is that a fair summary?

Grant Hutchison

13. Originally Posted by grant hutchison
I think my unease is maybe just at watching you travel through centrifugal, as it were, to get to what I've always thought of as a purely gravitational effect.
And you're right, it is a purely gravitational effect. But so are orbits, yet people perceive orbits as a kind of balance, so they are often described as a balance between gravity and the centrifugal force. It's an analagous use of the latter concept, as a device rather than as a fundamental piece of what is indeed a purely gravitational effect.

Originally Posted by grant hutchison
So I rather like the business of setting all the centripetal force vectors equal, and then looking at the local differences in the gravity vector. It seems to automatically sort rotation, revolution and gravity into appropriately different mental "bins".
This is the most general way to do it, and for mathematically adept people, is the "right" way. The best way to think of tidal forces (which is different from tidal bulges, as we'll see) is to ignore the overall motion of the Earth altogether and just consider the (vector) differential from a constant gravity, like you are saying. If you analyze the force differentials arising from a central force from a distant source, they of course can be expressed as a symmetric linear function of the displacements from the center of the Earth. Thus they decompose into three "stretching" eigenvalues, and any time they are nonzero you have "tidal forces". But only when the center-line eigenvalue is different from the lateral eigenvalues do you have tidal bulges, so the latter is a subset of the former. This latter bit is overlooked in most descriptions of how tidal forces make tidal bulges. What introducing centrifugal forces does is it allows you to extend from the oversimplified line-of-centers explanation (which is inherently incomplete) to the material lying in the orbital plane, at least. Then you explain tidal bulges in that plane as being due to the disconnect between Kepler's law (which is where the centrifugal force comes in) and the solid-body law. You don't need to argue about what is happening out of the orbital plane, because you already have bulges in the orbital plane and it suffices. If students can understand why the cross section in the orbital plane becomes oval, you are done, and Kepler's law is a nice device there because it means you never have to subtract any "arrows".

Originally Posted by grant hutchison
I'm taking from your responses (for which, thanks) that you don't find the approach I described intrinsically misleading, but perhaps overly complicated, given its initial mix of non-rotation with revolution, and the requirement for some understanding of vectors. Is that a fair summary?
Yes, it is basically the vectors piece. I essentially substitute Kepler's laws in the orbital plane, as though the Earth was a pile of dust, for the need to understand vectors. You can at the same time make the point that orbits in the solar system are not like a solid turntable-- so what happens if you try to connect the planets with long rods? The rods get stretched, like the axis through the Earth that points at the Sun (if you were explaining how the Sun could cause tides. Then you say "it's actually the Moon, but for the same reason!").

But for the record, I do also like: "if the Earth were falling toward the Moon, it would stretch due to the gravity gradient, and an orbit is just like that." But note the swindle here-- this stretching would not cause tidal bulges if it were isotropic (which it would never be if gravity is a central attractive force, but you need to understand vectors). Applying Kepler's laws, or the centrifugal force, avoids that swindle by allowing you to treat the whole orbital-plane cross section at once without using vectors.

14. Grant Hutchison & Ken G,

Just in case you're interested (and just for fun), I worked out the forces similar to the way you two seem to be describing them. Let's say we want to know how the forces of gravity and centrifugal force effect the points on the moon that are closest and furthest to the Earth. We will consider the moon to be a perfect sphere (and without tidal bulges). At its center of mass (which is the center of the moon with evenly distributed density), the gravitational and centrifugal forces cancel out and its orbit is in equilibrium at this point. Now we will designate the radius of the moon as a ratio of x amount of the distance from the center of the moon to the center of the Earth. The gravity felt at the two points is equal to the square of the inverse of the distance from the center of the Earth, so for the near side, the gravity is 1/(1-x)2 larger than for the center and the far side is 1/(1+x)2 smaller. For the centrifugal force, it depends on the revolution, where the time for revolution is the same for all points on or in the moon. This means that the angular velocity per radius of orbit is a constant for all points since v/2pir=1/t. Since the centrifugal force is proportional to v2/r, and v is proportional to r, the centrifugal force is proportional to v*(v/r)=v*(constant) is proportional to v, and is therefore also proportional to r as well. But this distance is the distance from the center of the moon to the barycenter, not the center of the Earth, and the distance of each to the barycenter is inversely proportional to their masses. We will call the ratio of the radius of the moon to this distance y, where y=x*[1-Mmoon/(MEarth+Mmoon)]. So since the acceleration of gravity at the center of the moon is equal to the centrifugal force, the acceleration at the point on the far side of the moon becomes equal to [1/(1+x)2-(1+y)] times the acceleration of gravity only as felt at the center of the moon and directed toward the Earth and the near side becomes [1/(1-x)2-(1-y)] of that same acceleration which is also directed toward the Earth.

We notice here that the the tidal bulges that are eventually created are far from being gravitational on one side and centrifugal on the other, but rather the mix of the two whose formulas are otherwise far from equal. In fact, the tidal forces only come out about the same for each side if x is very small, where the radius of the moon is very small as compared to its orbit. x and y are both values between zero and one. If x approaches one, however, the tidal pull on the side closest to Earth becomes tremendously greater. Notice also that the force on the far side of the moon is always negative, so that it is ultimately directed away from Earth while the near side is of course always directed toward it. If we make the acceleration of gravity constant across the gradient, as Ken G suggested, so that it always equals one, we now have [1-(1+y)] for the far side and [1-(1-y)] for the near side, or [-y] and [+y]. In this case, the tidal forces would indeed be exactly equal and opposite. This is the same as for a value of x that approaches or is equal to zero, whereas the radius of the moon would have no radius (or at least extremely small compared to the distance separating the Earth and moon), but for x to equal exactly zero would mean that the moon didn't really exist (except perhaps as a singularity).
Last edited by grav; 2006-Jul-07 at 03:29 AM.

15. Sorry to be so late commenting. I was directed over from the parallel universe
Originally Posted by Ken G
Thus they decompose into three "stretching" eigenvalues, and any time they are nonzero you have "tidal forces". But only when the center-line eigenvalue is different from the lateral eigenvalues do you have tidal bulges, so the latter is a subset of the former.
It would seem to be a misnomer then to call them tidal forces if they don't create tidal bulges. Maybe we should call them "harbor forces".
Originally Posted by grav
Just in case you're interested (and just for fun), I worked out the forces similar to the way you two seem to be describing them.
But you have to be careful. I know Ken G likes to throw centrifugal force into the mix, but the result should have no component derived from it. As he says:
Originally Posted by Ken G
And you're right, it is a purely gravitational effect.
Originally Posted by grav
If we make the acceleration of gravity constant across the gradient, as Ken G suggested, so that it always equals one, we now have [1-(1+y)] for the far side and [1-(1-y)] for the near side, or [-y] and [+y]. In this case, the tidal forces would indeed be exactly equal and opposite.
I think if you follow the same analysis for the lateral points, you would find that the force is also y, directed away from the center. In other words, it represents the equatorial bulge produced by the once-per-revolution rotation of the body.

Originally Posted by hhEb09'1
But you have to be careful. I know Ken G likes to throw centrifugal force into the mix, but the result should have no component derived from it.
Yes, although the centrifugal force itself is not strictly a tidal (or tsunami!) effect (as general relativity is not usually invoked), I like to use it as a device to demonstrate the kind of gravity that induces no tidal bulges (to wit, a hypothetical gravity force that increases linearly with distance, in balance with the centrifugal force). If however you use grav's approach of analyzing the perturbation in the force of gravity, you don't need the centrifugal force, it is enough to show that there is zero force perturbation along the center line (for the hypothetical constant gravity case). That by itself will induce tidal bulges, because there is still a sideways squeezing due to the way the gravity from the Earth points towards the center of the Earth. So you get sideways squeezing even in the absense of stretching along the line of centers, and this will create the usual tidal bulges only weaker.
Originally Posted by hhEb09'1
I think if you follow the same analysis for the lateral points, you would find that the force is also y, directed away from the center. In other words, it represents the equatorial bulge produced by the once-per-revolution rotation of the body.
No, the oblateness of the Moon's rotation is a separate issue superimposed on the tidal effects, and it itself induces no bulges. Indeed, for a corotating system like the Moon, the centrifugal force is the only ficticious force and the lateral forces from it are trying to expand the Moon sideways in exact balance to how the lateral pinching of gravity is trying to squeeze it, leaving no net lateral effect. Put differently, the force that would normally induce oblateness of the rotating Moon is balanced by the pinching effects of gravity. If you analyze it in the corotating frame, the above statement holds whether you use real gravity, a constant-magnitude (but central) gravity, or a gravity that grows linearly with distance. So to avoid tidal bulges, you need the same things to be happening along the central line as in the lateral direction, i.e., you need the third of these examples.

17. Originally Posted by Ken G
No, the oblateness of the Moon's rotation is a separate issue superimposed on the tidal effects, and it itself induces no bulges.
I disagree, it is not a separate issue. It is in fact what grav is calculating (and it does produce a bulge, but not a tidal bulge since it is constant around the equator, as you've mentioned before). The equatorial bulge falls out of the calculations whether you use the center of the object or the barycenter of the orbit--as it should, if you're careful.

18. Well, if you don't like calling it separate, call it extraneous. It does not inform the tidal bulge issue in any way, and did not appear in grav's analysis because he didn't even consider the lateral direction. To summarize all this, there are two ways to determine if you will have tidal bulges. One is to purely consider gravity, but then you must contrast what is happening laterally to what is happening along the line of centers-- you can't do it just along the center line the way many often try to. Or, you can stick to the center line and see if what gravity is doing is in perfect balance to what the centrifugal force is doing. In either case, rotation of the planet is irrelevant, and the answer is the same-- the only central-force gravity that yields no tidal bulges is a gravity that increases linearly with distance.

19. Originally Posted by Ken G
Well, if you don't like calling it separate, call it extraneous. It does not inform the tidal bulge issue in any way, and did not appear in grav's analysis because he didn't even consider the lateral direction.
Yes it did appear. The quantity that he called y is the value of the centrifugal force from the rotation of the moon, it is proportional to its radius x.

20. It is not necessary for the Moon to be rotating to get that y. It appears in the co-orbiting frame, regardless of whether or not the Moon is tidally locked. It is pure coincidence that the Moon is in fact tidally locked, this has no impact on its tidal bulges. Any additional lunar rotation has only axially symmetric effects.

21. Originally Posted by Ken G
It is not necessary for the Moon to be rotating to get that y. It appears in the co-orbiting frame, regardless of whether or not the Moon is tidally locked.
In grav's calculations, the two points of the frame at which he performs the two calculations and computes the "y" for each, are rotating about their common center. The "y" represents the centrifugal force produced by that rotation.
It is pure coincidence that the Moon is in fact tidally locked, this has no impact on its tidal bulges. Any additional lunar rotation has only axially symmetric effects.
That "y" is in fact an axially symmetric effect proportional to the distance from the common center, which is to be expected, since the points of the surface (not the moon) in that frame are rotating about their center with the same angular velocity that the frame is rotating about the barycenter point.

22. Originally Posted by hhEb09'1
In grav's calculations, the two points of the frame at which he performs the two calculations and computes the "y" for each, are rotating about their common center. The "y" represents the centrifugal force produced by that rotation.
I'm not trying to enter grav's mind and say what he intended the y to be, I'm saying that the y does not have anything to do with the Moon's rotation. It comes from the reference frame that rotates with the Moon's orbit-- the Moon is welcome to rotate however it likes and grav's calculation is still applicable, whether he realizes this or not, because the axisymmetric effect of rotation (around the Moon's axis, not the axis of the centrifugal force-- this is the key distinction) is irrelevant to tidal bulges.
Originally Posted by hhEb09'1
That "y" is in fact an axially symmetric effect proportional to the distance from the common center, which is to be expected, since the points of the surface (not the moon) in that frame are rotating about their center.
I hope I've set this issue straight now. I return to the central point: there are two ways to verify the way a central gravity would have to act to avoid tidal bulges. The one that does not require an analysis of lateral pinching forces is the one that compares what gravity is doing along the line of centers to what the centrifugal force is doing (in the frame rotating with the Moon's orbit). The latter issue is the "y" effect, and the rotation of the Moon is irrelevant. This is the sole point I am making, it is about physics, not he said she said.

23. Originally Posted by Ken G
I'm not trying to enter grav's mind and say what he intended the y to be, I'm saying that the y does not have anything to do with the Moon's rotation. It comes from the reference frame that rotates with the Moon's orbit-- the Moon is welcome to rotate however it likes and grav's calculation is still applicable, whether he realizes this or not,
I am not disagreeing with grav's calculations, nor am I trying to infer his intentions. I am only describing what I see as the facts.
because the axisymmetric effect of rotation (around the Moon's axis, not the axis of the centrifugal force-- this is the key distinction) is irrelevant to tidal bulges.
And that is my whole point. The "y" is an axisymmetric effect, around the Moon's axis.
I hope I've set this issue straight now.
Not by 0.4 gm
I return to the central point: there are two ways to verify the way a central gravity would have to act to avoid tidal bulges.
My discussion has not been about that, it has been about grav's calculations. I have only been pointing out that the "y" term is axisymmetric about the point central to grav's two points, "because the axisymmetric effect of rotation (around the Moon's axis, not the axis of the centrifugal force-- this is the key distinction) is irrelevant to tidal bulges."

24. Originally Posted by hhEb09'1
The "y" is an axisymmetric effect, around the Moon's axis.
That's true, which is why centrifugal force itself cannot induce tidal bulges (we agree there). What grav's calculation is capable of doing is using the centrifugal force (due to being in the co-orbiting frame, I think we agree rotation of the Moon itself is irrelevant) as a device to see when you will get tidal bulges. This device need only be considered along the lines of centers, because in the lateral direction the centrifugal force always balances the central gravity no matter what gravity law you assume (I am working within the orbital plane, that is sufficient to test for bulges). So that's the point-- the centrifugal force, call it the "y" analysis, needs only to be considered along the line of centers, this device allows you to ignore what is happening laterally. Without considering the centrifugal force, you need to do a two-dimensional analysis, which is a bit harder, you need arrows and snazzy graphics (many books do it, but you don't need to). That's exactly why I like using the centrifugal force even though I know it doesn't produce tidal bulges (I had a similar discussion with Grant). I think your point is just that the centrifugal force itself does not cause tidal bulges. I completely agree, it's just an elegant way to analyze gravity such that you only need to consider the line of centers. That's all grav did, and is what I think of with his "y" effect. Thus none of the statements you have made are incorrect, what is incorrect is your apparent conclusion from your statements, that one cannot use the "y" effect to show when gravity will induce tidal bulges. One can-- whenever it does not balance gravity along the line of centers.

25. Originally Posted by Ken G
I think your point is just that the centrifugal force itself does not cause tidal bulges. I completely agree,
I'm glad that that issue is straight.

Unfortunately, grav's post refers to the tidal bulges as a combination of gravitational and centrifugal effect--because of the appearance of "y".
it's just an elegant way to analyze gravity such that you only need to consider the line of centers. That's all grav did,
Other than introduce an erroneous conclusion, which I tried to address.
Thus none of the statements you have made are incorrect, what is incorrect is your apparent conclusion from your statements, that one cannot use the "y" effect to show when gravity will induce tidal bulges. One can-- whenever it does not balance gravity along the line of centers.
My apparent conclusion? You're disagreeing with my apparent conclusion?

I've said over and over that one may introduce and use centrifugal effects in the calculation of the tidal bulges, but I've also said that people tend to get confused when they do that. I think this is a good example.

26. Originally Posted by hhEb09'1
Unfortunately, grav's post refers to the tidal bulges as a combination of gravitational and centrifugal effect--because of the appearance of "y".
But this is quite correct, in the sense that this paraphrase is correct: "Tidal deformations in a circularly orbiting Moon appear whenever the gravity (assumed to point toward Earth) does not balance the centrifugal forces along the line of centers." That sure sounds like bulges that appear due to some kind of a combination, even though the centrifugal force does not strictly cause the bulges. There is really no error in grav's analysis, but some of the word choices can foster misconceptions. Here we are talking about the difference between A causes B and A implies B. But my specific objection was to your statement
I think if you follow the same analysis for the lateral points, you would find that the force is also y, directed away from the center. In other words, it represents the equatorial bulge produced by the once-per-revolution rotation of the body.
This statement is true at face value (as I now see what you mean by the once-per-revolution rotation, even when not tidally locked), what is not true is that it is a criticism of grav's analysis. That's what I meant by your "apparent" conclusion. The missing insight is that the special thing about the orbital centrifugal effect is that it always balances the transverse pinching of gravity for any applicable gravity law, so that's why it is useful in the line of centers analysis and why its combination with gravity is the key issue there. However, I agree with you that one must be careful not to imply a strict cause and effect relationship when in fact the relationship is merely a mathematical connection-- the line-of-centers centrifugal effect is a useful proxy for the lateral gravity that is the true cause. But given that it is a proxy, saying that its net resultant with line-of-center gravity is a "tidal force" is actually correct, to the extent that the phrase "tidal force" has meaning as a single entity, even though it should not be mistaken for a true causal connection. As usual, we were never as far apart as we thought!
Last edited by Ken G; 2006-Sep-07 at 07:28 AM.

27. Established Member
Join Date
Dec 2001
Posts
199
Current Gravity Model:

The stretching force on Earth is greater than the pinching force on Earth, resulting in max tides on the Earth where and when the moon, sun, and Earth all align along a straight line connecting each center of mass (the highest tides will be measured on Earth along that connecting line).

Constant Central Gravity Model:

The stretching force on Earth reduces to zero (there's no differential force), while the pinching force on Earth remains equal to that of the Current Gravity model. High tides on Earth will now be measured perpendicular to where and when the moon, sun, and Earth all align along a straight line connecting each center of mass.

Gravity Increases Linearly With Distance:

The stretching force on the Earth is equal to the pinching force on Earth:
No tides... Because there's no differential force. No matter how many bodies orbit Earth or how many bodies Earth orbits, the differential force of gravity across Earth from each individual body is zero.

Right?

28. Originally Posted by DALeffler
Constant Central Gravity Model:

The stretching force on Earth reduces to zero (there's no differential force), while the pinching force on Earth remains equal to that of the Current Gravity model. High tides on Earth will now be measured perpendicular to where and when the moon, sun, and Earth all align along a straight line connecting each center of mass.
No, the high tide will still be as in the first example, the "pinch" still produces a low tide.

29. Originally Posted by DALeffler
Right?
Precisely correct, and a very nice summary at that. All that remains to say is that the second example has tides exactly like the first, only 1/3 as great. Given the symmetry of the problem, there is only one degree of freedom, which is the overall strength of the effect. I think hhEb09'1 is confused about the number of degrees of freedom here, but it is certainly a subtle point that all smale-scale perturbations of the shape of the Earth due to the tidal effect of any central-force gravity will look the same except in magnitude (and sign, in some cases).

30. Originally Posted by Ken G
There is really no error in grav's analysis, but some of the word choices can foster misconceptions.