I have just been reading (and enjoying) the chapter of your book that deals with tides. I believe that your explanation of the tidal bulge on the side of the Earth farthest from the Moon is unnecessarily complicated. All this talk about "astronauts in free-fall" and "relative to the centre of the Earth" is a distraction from the essential point: the Moon's gravity acts on all parts of the Earth, and tries to pull them towards the Moon. Because the water on the far side of the Earth is farthest from the Moon, it is pulled towards the Moon less than other parts of the Earth, and so is, as it were, "left behind" by the rest of the planet's lurch Moonwards. This accounts perfectly for there being a high tide on the side of the Earth farthest from the Moon.

One other point. In the same chapter, you write that one day the tidal effects of the Moon will slow the Earth's rotation down so that it will take the Earth one lunar month to rotate on its axis (and thus keep the same face turned towards the Moon). But that isn't what has happened to Mercury. The Sun's tidal effects on that diminutive body have slowed its rotation to just two thirds of its period. Surely the same will happen to the Earth?

Incidentally, I've never really understood why Mercury is locked into such a rotation. I'm guessing that for two thirds of its orbit the Sun's pull on its tidal bulge is slowing it down, while for the next two thirds of its orbit the Sun's pull is speeding it up, thus cancelling out the first effect. Is this correct?

I remember making an 'accelerometer' in school.

It was just a jar with a tight fitting lid, a cork, and a string.
Attach the cork to the string, attach the other end to the inside of the lid at the center. Fill the jar with water, screw the lid on, and flip it upside down.
What you should have is the cork floating at about the center of the jar, and totally under water.
When you move the jar, the cork will swing toward the direction of acceleration. If you put it on the dash of a car, it will move forward when you speed up, backward when you brake, and stay centered when you are going a constant speed.
We used it to prove that centripetal force is to the inside, and centrifugal force is a 'feigned' force.

I assume this is how the earth reacts to the gravitational force of the moon. The water goes away from the direction of the force.

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How do you explain the tidal bulge on the near side to the moon?

Oops. #-o
I guess my 'accelerometer' won't work as an analogy in this instance. Sorry for the confusion.

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Since noone has covered this yet, I'll give a quick explanation. I'm sure the BA covers this somewhere on the site, so you can also go to the front page and look around for it.

Tides involve the gravitational pull on 3 points: the near side, the middle, and the far side. The gravitational pull also differs between any two of these three points. The near side gets pulled more than the center, and the center gets pulled more than the far side. This latter difference seems like the far side getting pulled from the center, which causes the tidal bulge on that side.

I'm not sure about Mercury, and it's too late to look something up.l

Right. That's the point I was trying to make. Some people seem to think that the water on the far side of the Earth moves away from the Moon, which it doesn't.

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Well, that depends on your frame of reference. To us, standing on Earth, it seems that the water moves away from the moon, though it may not to someone on the moon. That was my point.

Yes, I see what you mean.

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All,
May I try another explanation? It parallels Tobin's, but to my mind is more complete:

Any object in orbit around another experiences a force tending to pull it apart in a radial direction. This is because the whole object travels at the speed required to keep in orbit a point of the same mass at the position of its 'centre of gravity', or mass centre. A part of the object further out from the mass centre is travelling at the same speed but that is too fast for the wider orbit. If it was seperate, it would spiral out. An object, or part of the object further in at the same speed is going too slowly and will spiral in.

For a small artificial satellite the force is negligable, but for an object as big as the Earth it is large, and as the Earth is flexible, it bulges, towards AND away from the Moon. The rock bulges are small, but real and measureable, the water bulges are obvious in a large enough body of water. Sun tides are much smaller than Moon tides, because the Sun's gravity gradient is much less - it changes less across the diameter of the Earth.

This way of thinking about tides is more useful than 'gravity less further away' as it may be extended to other situations, for instance a skyhook or space elevator. I'm sure it is well known, but I came to it in a Pauline moment, on reading Larry Niven's story "Neutron Star". Which shows the educational power of Science Fiction!
John

How would you use it to explain the tidal depressions at ninety degrees from the tidal bulges?

You forget that the Earth does not orbit the Moon: it orbits the barycentre of the Earth-Moon system. This is only 1,600 km beneath the Earth's surface. This means that the point on the Earth's surface closest to the Moon is closer to the barycentre than is the Earth's centre of mass; but as it is travelling at the same speed, it's going too slowly for its orbit, and so should (according to your argument) spiral in towards the barycentre away from the Moon. In fact, the opposite happens: it bulges out towards the Moon and away from the barycentre.

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Kilopi,
Simple - hydrostatics. The water to bulge out must come from somewhere! It moves from the "sides" of the Earth, relative to the Moon's direction, into the "bulges", so the water level around the sides goes down.

Eroica,
Thank you! I had not realised that the Earth-Moon CoM was where it is, whereupon my argument falls in ruins. Or does it?

I think you have forgotten that as the Earth orbits an internal, eccentric CoM, the orbital speed of points on the Earth's surface relative to the CoM will vary. A point furthest from the Moon, far from the CoM, will travel much faster than one facing the the Moon, near to the CoM. For the same reason, at that sub-Moon point, a particle slightly further out from the CoM, on the surface of the Ocean for instance, will still travel faster than one deeper, so it will be going faster than it needs to stay in 'orbit'. It WILL tend to move away away from the CoM, forming a bulge towards the Moon.

Please shoot me down again, if I deserve it! The "different-orbit" and "gravity-vector" theories (see the Bad Astronomer's explanation) are compatible I believe - they both depend on the same mechanics. If they don't how is that non-rotating satellite tethers work?
John

I meant, you've explained the bulges in the Earth's equipotential surface using your theory--how do you explain the depressions? The bulges and depressions are there whether there is water or not--hydrostatics has nothing to do with it.
The ocean farthest from the moon is three times farther from the CoM than the ocean nearest the moon--why are the bulges about the same size?

Kilopi,
I assume that you mean the bulges and depressions in the earth itself?

Take a sphere - a balloon, maybe. Attach handles to opposite sides - Glue? Use balloon like a chest expander, to develop your pectorals! As the opposite sides are pulled apart, the sides are drawn together. Try the same with a solid ball of dough - your hands will stick to the opposite sides - and the same thing happens. If an elastic body is made longer, its dimensions at right angles grow smaller - and this is true of the Earth too.

I'm beginning to clutch at straws to answer your point about the equal bulges, but I suspect it is to do with equal radial velocity. Although a point on the Earth's surface away from the Moon is travelling faster than a point on the 'near' side, it is also further away from the CoM. Expressed in degres or radians/second around the CoM, the velocity of each point will be the same, and their acceleration, change of direction, the same too. So the forces pulling up the Ocean will be the same.

John

I've been thinking more about your tidal theory, JohnD, and I think it's largely correct as far as the solar tides are concerned but requires some modification to explain lunar tides. The point I want to make in this post is that the rotation of the Earth about its axis is largely irrelevant to the question of solar tides, but is vitally important in explaining lunar tides.

Solar Tides: Let's ignore the Moon for the moment and start with solar tides. In his book the BA says that he relied heavily on Mikolaj Sawicki's Myths About Gravity and Tides. Sawicki's explanation of solar tides, you'll be happy to know, is exactly the one you gave.

The Earth's centre of mass (CoM) is in free-fall about the barycentre of the Earth-Sun system (which is located near the centre of the Sun), orbiting at 108,000 kph, just the right speed to stay in orbit.

The point on the Earth's surface closest to the Sun is also travelling about the Earth-Sun barycentre at this speed, but being closer it should be travelling faster (Kepler's third law) to be in orbit; so it has a tendency to slip into the Sun's gravity well, and this accounts for the tidal bulge on that side of the Earth.

Meanwhile, the point on the Earth's surface farthest from the Sun is travelling at 108,000 kph, but being farther from the Sun it should be travelling more slowly to be in free-fall; so it has a tendency to climb up out of the Sun's gravity well, and this accounts for the solar tidal bulge on the far side of the Earth.

In all of this, the rotation of the Earth about its own axis is not very relevant. A point on the equator is travelling around the centre of the Earth at a speed of about 2,000 kph, which is only about one fiftieth of the Earth's orbital velocity. This will reduce the orbital speed of the point closest to the Sun to about 106,000 kph, and increase the orbital speed of the point farthest from the Sun to about 110,000 kph. So the rotation of the Earth will enhance the tidal effect, but only slightly.

Lunar Tides: Now let's look at the Moon. After explaining solar tides, Sawicki says:
Unfortunately he does not go into any greater detail than this. I contend that the lunar tidal situation differs from the solar one in two crucial ways:

[1] With solar tides, the Earth's two tidal bulges are always on the same side of the Earth-Sun barycentre. With lunar tides, they're always on opposite sides of the Earth-Moon barycentre (which is located 1,600 km beneath the Earth's surface).

[2] With solar tides, the Earth's speed of axial rotation (2,000 kph) is 50 times smaller than the Earth's orbital velocity (108,000 kph). With lunar tides, it's 50 times greater.

The Earth's CoM orbits the Earth-Moon barycentre in 27.32 days. So its orbital speed about the barycentre is a pedestrian 44 kph (do the math!). This means that the point on the Earth's surface closest to the Moon is actually orbiting the Earth-Moon barycentre at about 2,044 kph. To be in free-fall it should be orbiting at 75 kph (do the math!). Because of the Earth's axial rotation, it's actually travelling at 2,040 kph, so it has a tendency to move away from the barycentre. This accounts for the tidal bulge on the near side of the Earth.

A similar situation obtains on the far side. For free-fall, that point should be travelling around the barycentre at 29 kph (again, I leave you to check these figures). But it's actually travelling round it at 2,044 kph, so it too has a tendency to climb into a higher orbit. This accounts for the tidal bulge on the far side of the Earth.

There are two things about this explanation which worry me, though:

[1] Neither Sawicki nor the BA mention any of this in their explanations. As they're the experts and I'm the plodder, I suspect that I have made some fundamental error in all of this.

[2] As kilopi pointed out, the two tidal bulges are the same size. Can my theory account for that? I don't know. Guess I'll have to crunch some numbers, but I'll leave that for another post.

Phew!

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Yes, and that is why I would disagree with your comment that "This way of thinking about tides is more useful". Rather, I think it adds unnecessary complication.
The points are traveling at the same velocity (velocity, not speed) and their orbit is not centered at the CoM, but offset somewhat. They are at the same distance from the centers of their respective orbits.
If you are doing things right, there should be no real difference between the solar case and the lunar case.
Interesting! Sawicki posted his comments about gravity in March, but I don't remember seeing him referenced in the book before. Perhaps it was added after?
Sawicki's post has a link to an online version of that paper. JohnD characterizes his (JohnD's) view as the "different orbit" explanation--but I don't see any aspect of that in Sawicki's paper.
No, it follows a similar path, but it is not centered on the barycenter. It is centered on a point that is offset from the barycenter by the radius of the Earth. Of course, when the Earth completes a half revolution about the Sun, the point will no longer be the closest to the Sun, but will be the farthest.
That's the problem with that sort of analysis.
They don't mention it because they are using a different approach to the problem. I contend, one that is much simpler to understand and easier to use.

The reference was in the Recommended Reading section on page 261.
Perhaps I read too much into that statement. If so, I apologise for misleading you.

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This is what Sawicki says about solar tides:
I thought this amounted to what JohnD was saying in his "different orbit" theory. If not, I obviously misinterpreted it. Sorry again.

I don't follow. At the moment that the point in question is closest to the Sun, it is travelling on an orbit about the Earth-Sun barycentre whose radius is less than that of the Earth's CoM by the radius of the Earth - not offset by the radius of the Earth.

Incidentally, why wait six months for it to be the farthest point? Isn't twelve hours just as good?

It might be easier if we considered a situation in which the Earth is imagined to be travelling on a circular orbit and to be in a locked rotation, so that the point in question will always be closest to the Sun, and will surely be "orbiting" the Earth-Sun barycentre in a circular path, but at a speed that is too slow for free-fall. The tidal bulges will not move then. Like the Moon, the Earth will be permanently stretched.

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Thanks! I found it in my copy. I must have read it before, but was confused with his BABB posts--I didn't remember seeing it before.
I thought that the BA's version agreed with Sawicki's--and that JohnD's didn't. JohnD?
We have to be careful here--the effect of the rotation of the Earth has nothing to do with the tides, so you have to do the analysis without regard to that rotation.
Of course, that means that the Earth rotates once per every revolution--and the effects you are describing are a result of that rotation, not the tide.

All,
There is a lot of confused thinking about this (mine and the Bad Astronomer!) and all this quoting is making mine worse. Can we debate without constant references please? Put the point in your own words.

I say that the BA is confused/confusing because his explanation starts with the point that the earth is so big that the Moon's gravity diminishes across its the diameter. This is clearly so, but he bases his argument on this effect, when "Tides" are a universal effect. The size of the satellite is irrelevant. Small satellites are influenced, usually by being 'pulled' so that they lie with their long axis radial to their primary;massive stars suffer tides, when they are in a binary group. Moreover, the position of the CoM, inside or outside the primary and the rotation of the earth are also irrelevant.

Any argument for the origin of tides that depends on a special case - eg the Earth-Moon CoM lies within the Earth (Or vice versa for the Earth-Sun) or the earth is rotating - must be weak. Let's discuss the general case.

Eroica - you can do the maths (I can't!) What about the general case of two equally massive planets, in orbit around each other, their rotation locked? (A tidal effect, but one we can ignore to pare away distracting elements) Let's think about the tides on those.

John

JohnD: I will post something about such a case (equally massive binary planets in locked rotation), but first there are a few points I wish to get off my chest.

In Bad Astronomy the BA addresses the paradox that somehow the Moon's gravity causes the far side of the Earth to bulge away from the Moon. Common sense tells us that an attractive force can't repel something. So how can this be? In fact, there is really no paradox. Pioneer 10 is leaving the solar system even though the resultant gravitational force acting on it is in the opposite direction. But Pioneer 10 is travelling at 50,000 kph (or whatever it is) and that's why it's able to climb up out of the Sun's gravity well. If you were God and you reached out and stopped it in its tracks, it would slide back into the inner solar system.

Similarly, every part of the Earth is moving, and so does not necessarily fall towards the Moon. This, I thought, was the basis of your theory.

Sawicki and the BA explain tides using the old Newtonian sense of gravity as a force. This works well as far as the mathematics is concerned, but Newton's forces are fictitious. The Sun does not really pull on the Earth. It bends space, and the moving Earth is deflected by the curvature of spacetime.

You are probably familiar with the old adage:
This means that in explaining a tidal bulge you can't cavalierly ignore the Earth's rotation. The point on the Earth's surface furthest from the Moon (ignoring the Sun for the moment) is moving through the spacetime continuum in a certain direction and at a certain speed. It should make no difference whether this movement is caused by the Earth's motion through space or its rotation on its axis.

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JohnD: Before looking at your idealized situation (equally massive binary planets in locked rotation), I want to describe a simpler one.

Imagine the Earth is orbiting the Sun in a perfectly circular orbit. Imagine the Earth is in locked rotation, so that it keeps the same side facing the Sun at all times (rotating on its axis once a year). And imagine the Earth's axis is at 90 degrees to the plane of its orbit. Finally, imagine these are the only two bodies in the solar system.

The CoM of the Earth is in free-fall about the barycentre of the Earth-Sun system. Its orbital velocity is just right for it to remain permanently in free-fall (Kepler's Third Law) and maintain a constant distant from the barycentre.

Now, it seems clear to me that in such a situation the point on the Earth's surface closest to the Sun is also travelling around the barycentre in a perfectly circular path, but one whose radius is about 6,200 km shorter than that of the CoM's. Because it completes one revolution about the barycentre in one year, its speed is slightly less than that of the CoM. But being closer to the Sun, it is in a stronger gravitational field (or, as Einstein might have put it, the curvature of space is steeper). In order to be in free-fall orbit (and therefore maintain a constant distance from the barycentre), it ought to be going faster (Kepler, again). At its slower speed it begins to slip into the Sun's gravity well, and this produces the tidal bulge.

A similar argument explains the tidal bulge on the far side. There, the point farthest from the Sun is travelling on a circular path about the barycentre that is longer than the CoM's. But as it too completes one revolution about the barycentre in one year, it must be travelling at a greater speed. But to be in a free-fall orbit (and therefore maintain a constant distance from the barycentre), it ought to be going more slowly. At its greater speed it tries to climb up out of the Sun's gravity well (just like Pioneer 10), and this creates another tidal bulge.

What is wrong with this interpretation? Is it not a perfect description of your different orbits theory?

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I disagree that the BA's explanation is confused
Just as the gravitational effect is proportional to GMm/r^2, the tidal effect is proportional to GMd/r^3, where d is the diameter of the small body. So, the size is relevant.
Actually, the rotation of the Earth would produce a bulge all around the Earth that is not related to tides at all. By focussing on the effect of that movement only on the near and far side, you are ignoring the effect on the sides.
I agree that that more or less describes JohnD's argument--but it does not match Sawicki's.

Notice, in that argument though, that the Earth is rotating once per revolution. If you get rid of some of the rotation, you should get rid of all of it. Once you do, however, and you trace the path of each point, you find that every point on the Earth traces a similar path with the same radius. Each point experiences the same velocity and acceleration (not just in magnitude but even as vectors).

The only difference, from side to side, is the strength of gravity, as the BA says. Or, if you insist on using the more modern interpretation, the slope of the curvature of spacetime.

Kilopi,
I accept your point about the effect of the Earth's rotation on its shape, and accurate satellite observations have confirmed that the Earth is 'shorter that it is wide'. But I cannot work out how closely that observation coincides with the predicted flattening from rotation. Anyway, that flattening is at right angles to the tidal effect, which will be superimposed on the equatorial bulge.

I have to confess that I have only today read Sawicki's explanation of tides ("gravity vectors"?). I noted two things, first he makes clear that he neglects the effects of the Earth's rotation "as they do not affect our conclusions" and that his model is of an Earth free-falling towards the Sun, not one in orbit. This is a model some way short of reality and of the model expressed so well by Eroica, although by not relying on an object being in orbit, it is simpler than a "different orbits" model. Sawicki's mathematics are beyond me, so I cannot criticise them.
Maybe that is why I favour "different orbits", a more intuitive way of thinking about tides that does not need mathematics, though as Eroica shows, the maths work out, confirming intuition.


Eroica.
Thank you! You have perfectly expressed my point of view, in a model better than the binary one I suggested. I fear, however, that this is a set-up to a devastating, knock-down, counter argument. No matter, truth will out, and I await your broadside with eager anticipation!

In case you are not plotting my downfall (!), will you please try to reconcile these two views? Can you calculate from this model the size of the force 'lifting' the ocean into the tidal bulges? If this value corresponds with Sawicki's calculated 0.515 x 10^-7g, then either point of view is just that, a point of view, a world-picture, NOT a "New Theory Of Tides".
John

The equatorial bulge is 20,000 meters, whereas the tidal bulge is less than a meter.
Yes, that is why Sawicki states that the rotation doesn't affect the conclusion about tides.
Earth in orbit is in "free-fall," so that comment is wrong.
Changing reference frames is valid--but ignoring an effect is not.

Any point of view is valid--but you do have to be careful that you are not ignoring effects, or attributing to tides an effect that is caused by rotation.

I've been doing a bit of googling and I came across this explanation of the tidal bulges on the MadSci Network. It sounds a bit like your theory, JohnD:
My problem with Sawicki's explanation - correct though it is - is that it can only clear up the paradox of the tidal bulge on the far side of the Earth by speaking of negative forces. But gravity is attractive. If something under the influence of a gravitational force moves away from the source of that force, it can only be because of its velocity (like a rocket being launched, for instance).

If JohnD's theory is wrong, I wish someone would tell us why, not that, it is wrong.

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It isn't wrong (at least, if done correctly). I have no problem with using centrifictional force.

However, it is not simpler. It is unnecessarily more complicated. Errors tend to creep in. That's why I disagree with your contention in the OP when you said that the BA's explanation was "unnecessarily complicated." As I said before, changing reference frames is valid--but ignoring an effect is not.

PS: Part of the reason for ignoring the cetrifictional force, in this case, is that if you actually compute the force for the Earth, the amount and direction of the force is the same for every point on the Earth. There is no difference. Which means that the tidal force--which is a derived force much like centrifictional force is a derived force--has no contribution from centrifictional force. The tide-raising force varies in direction from one side of the Earth to the other.

My head is starting to swim reading through all this. I guess my basic question is: was I right or wrong in the book?

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I think we all agree that you were right*. I further contend that your version was better than any of the other candidates that have been presented in this thread.

*I could be wrong about this.

I don't doubt that you were right. I just think that a lot of people - myself included - feel that explaining the tidal bulge on the far side of the Earth by means of this "negative force," is a mathematical equivalent of pulling a rabbit from a hat. We can't deny the figures, but we are not enlightened by them either. Whereas when I read JohnD's explanation, I had a moment of epiphany.

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