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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(v_,)' _ )`-. \ ``'`
_.- _..-/ /((.'
((,.-' ((,/

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