PDA

View Full Version : Why is there a tidal bulge on the far side of the Earth?



Eroica
2003-Sep-22, 04:12 PM
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?

Pinemarten
2003-Sep-23, 08:52 AM
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.

kilopi
2003-Sep-23, 04:29 PM
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.
How do you explain the tidal bulge on the near side to the moon?

Pinemarten
2003-Sep-24, 07:21 AM
Oops. #-o
I guess my 'accelerometer' won't work as an analogy in this instance. Sorry for the confusion.

Tobin Dax
2003-Sep-25, 05:16 AM
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

Eroica
2003-Sep-25, 02:42 PM
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.


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.

Tobin Dax
2003-Sep-25, 04:31 PM
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. :)

Eroica
2003-Sep-26, 07:24 AM
Yes, I see what you mean.

JohnD
2003-Oct-12, 10:33 PM
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

kilopi
2003-Oct-13, 01:28 AM
This way of thinking about tides is more useful than 'gravity less further away' as it may be extended to other situations
How would you use it to explain the tidal depressions at ninety degrees from the tidal bulges?

Eroica
2003-Oct-13, 10:23 AM
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 ... and as the Earth is flexible, it bulges, towards AND away from the Moon.
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.

JohnD
2003-Oct-13, 10:27 PM
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

kilopi
2003-Oct-14, 04:12 AM
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.
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.

It WILL tend to move away away from the CoM, forming a bulge towards the Moon.
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?

JohnD
2003-Oct-14, 10:24 PM
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

Eroica
2003-Oct-15, 07:24 AM
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:

The lunar tidal effect is calculated in an analogous way. Again, one has to realize that Earth is in a free fall towards the Earth-Moon center of mass.
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!

kilopi
2003-Oct-15, 03:59 PM
I'm beginning to clutch at straws to answer your point about the equal bulges
Yes, and that is why I would disagree with your comment (http://www.badastronomy.com/phpBB/viewtopic.php?p=153128#153128) that "This way of thinking about tides is more useful". Rather, I think it adds unnecessary complication.


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.
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.

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.
If you are doing things right, there should be no real difference between the solar case and the lunar case.


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.
Interesting! Sawicki posted his comments about gravity (http://www.badastronomy.com/phpBB/viewtopic.php?p=68449#68449) in March, but I don't remember seeing him referenced in the book before. Perhaps it was added after?


Sawicki's explanation of solar tides, you'll be happy to know, is exactly the one you gave.
Sawicki's post (http://www.badastronomy.com/phpBB/viewtopic.php?p=68449#68449) has a link to an online version of that paper. JohnD characterizes his (JohnD's) view (http://www.badastronomy.com/phpBB/viewtopic.php?p=153639#153639) as the "different orbit" explanation--but I don't see any aspect of that in Sawicki's paper.


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
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.


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:

That's the problem with that sort of analysis.


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.

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.

Eroica
2003-Oct-15, 06:31 PM
Interesting! Sawicki posted his comments about gravity (http://www.badastronomy.com/phpBB/viewtopic.php?p=68449#68449) in March, but I don't remember seeing him referenced in the book before. Perhaps it was added after?

The reference was in the Recommended Reading section on page 261.

Mikolaj Sawicki is a physicist at John A. Logan College in Illinois. his website on tides ... cleared up some of my own tidal misconceptions. It has a very clear and interesting explanation of tides, and is one of the very few that not only is correct but carries out the idea to its logical conclusion.
Perhaps I read too much into that statement. If so, I apologise for misleading you.

Eroica
2003-Oct-15, 07:01 PM
This is what Sawicki says about solar tides:

Consider the point C on Earth closest to the Sun and the point F on a far side of Earth. The Sun pulls harder on a unit mass at the point C, not as hard on a unit mass at Earth center O, and weaker yet on a unit mass at point F. The acceleration "a" of Earth as a whole in free fall towards the Sun is determined by the gravitational pull of the Sun on Earthís center. Hence the unit mass at C has a tendency to accelerate towards the Sun with acceleration a + Delta a, i.e. more than the center of Earth, while a mass at the far side F has a tendency to accelerate towards the Sun with acceleration a - Delta a, i.e. to lag behind the center of Earth.
I thought this amounted to what JohnD was saying in his "different orbit" theory. If not, I obviously misinterpreted it. Sorry again.


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

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.

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.

kilopi
2003-Oct-16, 05:07 AM
The reference was in the Recommended Reading section on page 261.
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 this amounted to what JohnD was saying in his "different orbit" theory. If not, I obviously misinterpreted it. Sorry again.

I thought that the BA's version agreed with Sawicki's--and that JohnD's didn't. JohnD?


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?
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.


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.
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.

JohnD
2003-Oct-16, 07:13 AM
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

Eroica
2003-Oct-16, 11:36 AM
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:

Matter tells space how to bend.
Space tells matter how to move.
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.

Eroica
2003-Oct-16, 11:57 AM
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?

kilopi
2003-Oct-16, 05:02 PM
All,
There is a lot of confused thinking about this (mine and the Bad Astronomer!)
I disagree that the BA's explanation is confused

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.
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.

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.
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.

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

JohnD
2003-Oct-16, 09:00 PM
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

kilopi
2003-Oct-17, 09:42 AM
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.
The equatorial bulge is 20,000 meters, whereas the tidal bulge is less than a meter.


Anyway, that flattening is at right angles to the tidal effect, which will be superimposed on the equatorial bulge.
Yes, that is why Sawicki states that the rotation doesn't affect the conclusion about tides.


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
Earth in orbit is in "free-fall," so that comment is wrong.


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.

Changing reference frames is valid--but ignoring an effect is not.



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".

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.

Eroica
2003-Oct-17, 02:19 PM
I've been doing a bit of googling and I came across this explanation of the tidal bulges on the MadSci Network. (http://www.madsci.org/posts/archives/dec96/848737537.As.r.html) It sounds a bit like your theory, JohnD:

Indeed, one tide gets raised because the part of the Earth closest to the
Sun feels a stronger gravitational attraction than the rest of
the Earth. Sea water is gathered up and pulled towards the direction of the
Sun. On the other hand, the "backside" of the Earth, away from
the direction of the Sun feels a reduced gravity. As a result
sea water has tendency to escape the attraction of the Sun and
therefore gathers up in a tide on the backside of the Earth.

Although the use of the words "centrifugal force" tends to make physicists
uneasy, it can be a useful concept to figure out what is going on. In effect,
the Earth has a circular orbital velocity around the Sun which leads to
a centrifugal force which exactly balances the gravitational attraction
from the Sun at the center of the Earth. Regions of the Earth closer to the
Sun feel more gravitational attraction which is insufficiently compensated by
the "centrifugal force" from the orbit. Regions of the Earth further from the
Sun feel less gravitational attraction which is overcompensated by the orbital
"centrifugal force".

Marc Herant
Astronomy Mad Scientist Moderator


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.

kilopi
2003-Oct-17, 02:45 PM
If JohnD's theory is wrong, I wish someone would tell us why, not that, it is wrong.
It isn't wrong (at least, if done correctly). I have no problem with using centrifictional force (http://www.badastronomy.com/phpBB/viewtopic.php?p=29784#29784).

However, it is not simpler. It is unnecessarily more complicated. Errors tend to creep in (http://www.badastronomy.com/phpBB/viewtopic.php?p=154728#154728). That's why I disagree with your contention in the OP (http://www.badastronomy.com/phpBB/viewtopic.php?p=144333#144333) when you said that the BA's explanation was "unnecessarily complicated." As I said before (http://www.badastronomy.com/phpBB/viewtopic.php?p=155633#155633), changing reference frames is valid--but ignoring an effect is not.

PS: Part of the reason (http://www.badastronomy.com/phpBB/viewtopic.php?p=155284#155284) 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.

The Bad Astronomer
2003-Oct-17, 07:48 PM
My head is starting to swim reading through all this. I guess my basic question is: was I right or wrong in the book? :D

kilopi
2003-Oct-17, 08:05 PM
I guess my basic question is: was I right or wrong in the book?
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. :)

Eroica
2003-Oct-17, 08:11 PM
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.

kilopi
2003-Oct-17, 09:23 PM
Whereas when I read JohnD's explanation, I had a moment of epiphany.
Are you saying that you now feel that you didn't understand the tide effect when you wrote the OP (http://www.badastronomy.com/phpBB/viewtopic.php?p=144333#144333)? I had thought it summed up the BA's position pretty well--other than the other questions you had about Mercury.

JohnD
2003-Oct-17, 11:17 PM
Dr.Plait! I wondered if the conversation had got to a point when we needed to appeal to a Higher Power, but that of course would be against everything you stand for. Nevertheless, please take part (no one else is!).

Eroica quotes an explanation that invokes centrifugal force - that Great Chimera - but, Yes, that is what I mean. Swing a weight around on a string, and you feel a pull. Swing it faster and the pull gets stronger. The weight is going faster than it needs to stay in "orbit" about your finger and so tends to seek a higher orbit where at that speed it will be stable. The converse is difficult to simulate in this way, and the whole idea can only be an analog, as a string doesn't work like gravity.

I am flattered by Eroica's epiphany, and I can share her feeling. I've had the same moment, for instance when trigonometry became 'true' for me - it involved a 'vision' of a right angled triangle that changed its size, angles and sides, but always in proportion! But away from solipsism.

I think we have evolved a test. Sowicki used his model to calculate a value for the force that raises the tides. Can that force be calculated from the "different orbits"(DO) model? If the DO value agrees with Sawicki's value, then the two models are probably correct but different views of the same reality.

I regret that I would have to go back to school to learn how to do so. Eroica, if you are able, please will you give it a try? Dr.Plait? Kilopi? All? What I tell you three times is true!
John

kilopi
2003-Oct-17, 11:37 PM
then the two models are probably correct but different views of the same reality.
I've never said your model was not correct--but there are some gotchas along the way.

The main one, as near as I can tell (the one that seems to trip up even the pros), is using a rotating Earth to calculate the tide. You don't have to ignore reality, but if you do the calculation with the Earth rotating, then you have to account for the effect of the Earth rotating. That's why it appears as a complication-- but that doesn't make the theory wrong.

The second one, which seems to surprise a lot of people, is that the centrifugal force caused by the revolution is exactly the same at every point on the Earth. Any difference is caused by the rotation of the Earth.

There are a lot of ways to explain the concept--but that doesn't mean the BA's way is wrong, or worse than any other.

JohnD
2003-Oct-17, 11:38 PM
All,
I was reviewing this thread, and saw what I had missed before - Kilopi referring to a posting by Dr.Sawicki. Following that thread to its origin, I found a posting by JohnWitt, expressing precisely the "different orbits" point of view, though he spoils his pitch by mentioning centrifugal force!

John, are you out there? Or even Dr.Sawicki? Come and join us! It is good to know that there are others as well as Eroica and myself who see the Universe this way, and I would welcome more who can test the two points of view.
John

kilopi
2003-Oct-18, 06:26 AM
It is good to know that there are others as well as Eroica and myself who see the Universe this way
You can add me (http://www.badastronomy.com/phpBB/viewtopic.php?p=12031#12031) to the list if you like.

Eroica
2003-Oct-18, 07:23 AM
Are you saying that you now feel that you didn't understand the tide effect when you wrote the OP (http://www.badastronomy.com/phpBB/viewtopic.php?p=144333#144333)? I had thought it summed up the BA's position pretty well
Funny. I thought my OP expressed a completely different and completely wrong view (and one which I now retract)! Originally I thought that the tidal bulge on the far side was being "left behind" because the rest of the Earth was being pulled towards the Moon more strongly. But I now realize that if this were literally happening, the Earth would soon collide with the Moon! The Earth is not "lurching towards" the Moon. The CoM of a planet in a circular orbit about a star remains at the same distance from the barycentre. (In fact, over time it moves away; but let's not complicate things!)

But if there is something in the "different orbits" model, I still don't understand why the rotation of the Earth is not a factor. After all, we launch rockets from as close to the equator as possible because their initial velocity due to the Earth's rotation plays a very real part in helping them achieve escape velocity. Why should the same not be true for a mass of water on the far side of the Earth as it tries to 'climb' to a higher orbit (http://www.badastronomy.com/phpBB/viewtopic.php?p=10965#10965)? How does its velocity due to the Earth's revolution differ from its velocity due to the Earth's rotation (other than in magnitude)?

kilopi
2003-Oct-18, 07:45 AM
The Earth is not "lurching towards" the Moon.
I'm not sure if the word "lurching" was used before, but it is true that the Earth can be thought of as accelerating towards the Moon. If you want to call that lurching, then I don't see a problem with that point of view either.


But if there is something in the "different orbits" model, I still don't understand why the rotation of the Earth is not a factor. After all, we launch rockets from as close to the equator as possible because their initial velocity due to the Earth's rotation plays a very real part in helping them achieve escape velocity. Why should the same not be true for a mass of water on the far side of the Earth as it tries to 'climb' to a higher orbit (http://www.badastronomy.com/phpBB/viewtopic.php?p=10965#10965)? How does its velocity due to the Earth's revolution differ from its velocity due to the Earth's rotation (other than in magnitude)?
As the Earth rotates, its equator bulges. If you only look at the near side and the far side (from the moon), you'd notice that they both had a bulge because of the rotation. It is tempting to attribute those bulges to the tides.

But, if you look at other places along the equator, they bulge also! There is no tidal depression. And the rotational bulge is more or less permanent. It doesn't change with time.

Plus, the tidal bulges are near the ecliptic, whereas the bulges due to rotation are aligned along the equator. (The ecliptic--the path of the Sun from our point of view-- is tilted over twenty degrees with respect to the equator, and the plane of the moon's orbit is much closer to the ecliptic than to the equator)

So the rotation of the Earth does cause a bulge, but not what we'd normally call a tidal bulge. The rotation of the Earth has produced a bulge about the equator that is twenty kilometers farther from the center of the Earth than the poles.

PS: I just checked, and "lurch" was used in the OP.

Eroica
2003-Oct-18, 08:58 AM
As the Earth rotates, its equator bulges ... It is tempting to attribute those bulges to the tides.
Not to me it ain't. I'm not talking about or thinking of the rotational bulge, since it has nothing to do with the Sun or the Moon. That's why I introduced the idealized model of a planet in locked rotation, where the rotation is so slow that it can be discounted.

Eroica
2003-Oct-18, 09:30 AM
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
Sawicki's figure is the difference between the acceleration due to the Sun's gravity at the centre and at the near (or far) point on the Earth's surface. My own calculations led to a figure of 1.01e-6 metres per second-squared for the difference between the near and far points. That's actually the same as Sawicki's figure (if you double it and replace g with 9.81 metres per second-squared). But all that this means is that I can still do arithmetic.

To reconcile your theory with Sawicki's you need to calculate the sizes of the bulges that would be raised, and at the moment I don't know how to do that. I fear you are overestimating my abilities. I think we need reinforcements.

(PS: Eroica is a guy.)

Eroica
2003-Oct-18, 11:44 AM
Boy, these tides are starting to do my head in!

JohnD, I have been thinking about your first post (http://www.badastronomy.com/phpBB/viewtopic.php?p=153128#153128) to this thread. At the time I criticised it because you said that the Earth's CoM was in orbit about the Moon, whereas it is actually in orbit about the barycentre.

Now, a question occurs to me: why are we talking about the Earth-Moon barycentre at all, when we are trying to explain lunar tides? The barycentre is the point where the Moon and the Earth's gravity are centred. But the Earth's gravity plays no role in creating the lunar tides (other than preventing the lunar tides from tearing the Earth apart).

Maybe JohnD was right. We should be considering the Earth's motion relative to the Moon's CoM! I'll try crunching some numbers when I get a chance.

kilopi
2003-Oct-18, 02:17 PM
As the Earth rotates, its equator bulges ... It is tempting to attribute those bulges to the tides.
Not to me it ain't. I'm not talking about or thinking of the rotational bulge, since it has nothing to do with the Sun or the Moon. That's why I introduced the idealized model of a planet in locked rotation, where the rotation is so slow that it can be discounted.
But why not go all the way, and get rid of the rotation entirely? That way, you can be sure it can be discounted.


At the time I criticised it because you said that the Earth's CoM was in orbit about the Moon, whereas it is actually in orbit about the barycentre.

Now, a question occurs to me: why are we talking about the Earth-Moon barycentre at all, when we are trying to explain lunar tides? The barycentre is the point where the Moon and the Earth's gravity are centred. But the Earth's gravity plays no role in creating the lunar tides (other than preventing the lunar tides from tearing the Earth apart).

Maybe JohnD was right. We should be considering the Earth's motion relative to the Moon's CoM! I'll try crunching some numbers when I get a chance.
What's hard for me to understand is why you would say the BA's approach is hard to understand, yet we don't have a calculation for the alternative theory. My goal is to convince you that that approach is not as simple as it may seem.

Eroica
2003-Oct-18, 03:01 PM
I've been thinking about the barycentre again, and I now think that Sawicki and the BA are wrong when they claim that a person at the Earth's CoM doesn't feel the force of the Moon's gravity because they are in free fall about the barycentre. They are in free fall about the barycentre, but that's not why they don't feel the Moon's gravity. The correct reason is that they are also in free fall about the Moon's CoM!

I haven't done (and perhaps can't do) the maths, but Starry Night Pro clearly demonstrates that if you take the Moon's CoM as your fixed frame of reference, then the Earth's CoM does indeed revolve about it in a neat elliptic path, and is therefore in free fall (like all objects travelling through a gravitational field along a path that is a conic section).

If this analysis is correct, then my objection (http://www.badastronomy.com/phpBB/viewtopic.php?p=153320#153320) to JohnD's first post is no longer valid. The fact that the barycentre is between the tidal bulges is irrelevant to lunar tides. What is relevant is that they are both on the same side of the Moon's CoM.

JohnD
2003-Oct-18, 04:51 PM
Eroica,
Apologies for mistaking your gender - Eroica is a fine 'handle' but would be a splendid and original first name for a girl! Though I suppose adding a 't' could cause all sorts of embarassment.

I was delighted to read that you had calculated the same value for delta-a(?) as Sawicki, but then realised you were doing it from his model; Yes? So proves nothing, except as you say your arithmetic is reliable. But why is it necessary to calculate the size of the bulge? That follows from the value of the force.

I'm happy to be considering the barycentre rather than the CoM of the Moon. My original mind-picture was of a small satellite around a large primary, Earth-space probe or Sun-Earth, so that the barycentre virtually coincided with the primary's CoM, and I had not visualised it. You corrected that. the model still works and it is the different orbits of parts of the satellite around the barycentre that lift the tides, not orbits around the CoM of the other partner in the binary

Kilopi,
Are you getting a bit hung up on the equatorial bulge? That is real, but static, caused as you say by the Earth's rotation. And I'm happy to accept that it is of the order of 20 kilometers - I can't dispute it! But we are discussing tides that nowhere on Earth vary by more than 50ft/16m, an ocillation of less than 0.1% compared to the bulge you quote. Surely such an ocillation can be superimposed on the equatorial bulge?

Which raises again my own ignorance - Eroica (did I see Dublin associated with your name) and I live in mid latitudes, where tides are significant. Do tides occur around the Equator? I had always assumed that they did, but are you implying, Kilopi, that the bulge somehow suppresses them, so they are much less high, or absent?
And then what about the Poles. Apart from axial tilt, the Poles have the same orbit around the primary as the CoM, so no tidal effects there - Sawicki points this out from his model, and the DO model does the same. (Another parallel!) Following this point back to the start, BOTH models predict that tidal forces will be maximal at the Equator (axial tilt excepted) Unless you can tell me that there are no, or reduced tides on the Equator, I'll insist on wobbling the bulge!

John

kilopi
2003-Oct-18, 05:44 PM
I haven't done (and perhaps can't do) the maths, but Starry Night Pro clearly demonstrates that if you take the Moon's CoM as your fixed frame of reference, then the Earth's CoM does indeed revolve about it in a neat elliptic path, and is therefore in free fall (like all objects travelling through a gravitational field along a path that is a conic section).
Any point of view can be used, not just a luna-centric one. The math does get complicated sometimes though.

Kilopi,
Are you getting a bit hung up on the equatorial bulge?

Hung up? Just the opposite. My warning is that one should make sure that the calculations don't accidentally mistake a rotational component as a tidal component. If your model includes a rotation of the Earth, you have to somehow subtract that out to calculate the tidal effect.


Surely such an ocillation can be superimposed on the equatorial bulge?

That's my whole point.


Do tides occur around the Equator? I had always assumed that they did, but are you implying, Kilopi, that the bulge somehow suppresses them, so they are much less high, or absent?

?? No, to the last question, yes to the first.

The Bad Astronomer
2003-Oct-18, 08:41 PM
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.

I used the idea of subtracting vectors because it's easier to swallow, for me, than one using centrifugal force. I have no problems with centrifugal force, but it implies a rotating system. You get tides even if the system is not rotating. Let the Earth fall into the Sun, and you still get differential tides across the Earth, even when the net angular momentum is zero. So using centrifugal force won't apply there.

I also wanted to use a different explanation than is usually seen, because why repeat something in every textbook? Actually, I have gone through many textbooks, and very few explain tides in any detail at all, a fact I find astonishing. Tides sculpt many objects in the Universe, from our shorelines to clusters of galaxies. It's worth spending a few hundred words explaining carefully!

George
2003-Oct-19, 05:18 AM
I sure hope you do, BA, as it needs someone to handle it properly. Too bad it's not simple, however.

For instance, the gravitational force from F=G*M1*M2/r^2 between me (90kg) and the moon says I am only .0007 lbs. lighter when the moon is overhead. Surprisingly, this same force equation says I am .12 lbs lighter when the sun is overhead. So the Sun's gravitational force is 170x's greater than the moon.

The websites show that tidal forces, however, vary as the inverse cube of the distance not the square. I wish I could find a clear explanation for this.

Also, it appears the horizontal component not the radial component is the main player in an ocean tide.

Here is an animated vector tidal force site that is pretty neat....

>>> Round n Round We Go (http://www.ifmo.ru/butikov/Projects/Tides3.html) <<<

For fun, and to use exaggeration to emphasize gravity forces, I used the gravitational equation for a blackhole. If I get within 1/2 million miles of a million sun BH, the force on me becomes 4 billion lbf. The only difference in force for 1 meter closer is only about 10 lbf. However, at 100,000 miles this difference in 1 meter is about 6,000 lbf. Ouch!

Eroica
2003-Oct-19, 08:03 AM
I used the idea of subtracting vectors because it's easier to swallow, for me, than one using centrifugal force. I have no problems with centrifugal force, but it implies a rotating system. You get tides even if the system is not rotating. Let the Earth fall into the Sun, and you still get differential tides across the Earth, even when the net angular momentum is zero. So using centrifugal force won't apply there.
I wish I had never clouded the issue by talking about the Earth's rotation or by linking to that MadSci article. So let me clarify one thing: the DO (Different Orbits) theory does not - NOT involve centrifugal forces. If anything, the BA's "negative force" is closer to the notion of centrifugal forces than anything in the DO theory. The only force in the DO theory is gravity, which is always attractive. The DO theory tries to account for the tidal bulge's movement away from the Moon in terms of its velocity through the Moon's gravitational field.

So let's simplify things and ignore completely the Earth's rotation. If necessary I'll start a new thread some day to address that issue. kilopi challenged me to describe the DO with a non-rotating Earth. I'll try.

Eroica
2003-Oct-19, 08:28 AM
You get tides even if the system is not rotating. Let the Earth fall into the Sun, and you still get differential tides across the Earth, even when the net angular momentum is zero.
As far as I see there is no real difference at all between the DO theory and the DF (Differential Forces) theory in the case where a non-rotating Earth is simply falling into the Sun. In that case, the tidal bulge on the far side really is being left behind by the rest of the Earth, while the bulge on the near side is outstripping the rest of the planet. It's the way I envisaged the tides in my OP (http://www.badastronomy.com/phpBB/viewtopic.php?p=144333#144333).

But imagine, for a moment, what would happen in such a case if the Earth were prevented from falling. For example, imagine there is a spike driven through the centre of the Earth and protruding from the poles; now imagine that a giant being took hold of the ends of that spike and held the Earth above the Sun. What sort of tidal bulges would you get? There would still be a bulge on the near side as the Earth tried to slip off the spike and into the Sun, but there would be no bulge on the far side. The Earth would be pear-shaped. It would be like an ice-cream lolly that is being held horizontally to the ground, and is starting to melt and slide off the stick!

In this (admittedly fantastic) scenario, the far side of the Earth does not bulge away from the Sun because it is not moving through the Sun's gravitational field with the necessary velocity to climb up the gravitational gradient. In fact, it's not moving at all. It's like a rocket sitting on the launch-pad: it ain't going nowhere!

But those same differential forces that produced the normal tidal bulges still apply. In this case they don't produce the normal tides because the Earth's CoM is no longer in free fall. A person at the centre of the Earth would feel the Sun's gravity. But someone on the far side of the Earth would still feel a weaker force than someone at the CoM!

SimonCB
2003-Oct-19, 12:39 PM
What goes around comes around, this seems to me a rather more
eloquent version of the debate I had here http://www.badastronomy.com/phpBB/viewtopic.php?t=887.

I am still not altogether happy with the books explanation, but I did gain a much better understanding of tides largely thanks to GrapesofWrath.

If I might chip in a few points.

George wrote:

For instance, the gravitational force from F=G*M1*M2/r^2 between me (90kg) and the moon says I am only .0007 lbs. lighter when the moon is overhead. Surprisingly, this same force equation says I am .12 lbs lighter when the sun is overhead. So the Sun's gravitational force is 170x's greater than the moon.

The websites show that tidal forces, however, vary as the inverse cube of the distance not the square. I wish I could find a clear explanation for this.

The key to tides is not the absolute force, but the relative force. Try redoing the calculation for the force exerted on you at midday and midnight.

Kilopi said

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.

The cetrifictional force (a name a hadn't heard before :D) is constant, but it seems to me something of a stretch to say it can now be ignored.

I think another wild analogy is called for. Imagine the earth were made of some material whose stiffness were controlled by an electric current. So if we start in the high stiffness mode, we have small tidal bulges. If we flip a switch and suddenly the Earth is less stiff, the tidal bulges increase. Part of the Earth moves away from the moon. I think the best way to explain this is the cetrifictional force.


Simon

P.S. I feel the need to point out that I don't just come here to argue about tides. I haven't visited in a while, and this was at the top of the board.

kilopi
2003-Oct-19, 02:14 PM
The websites show that tidal forces, however, vary as the inverse cube of the distance not the square. I wish I could find a clear explanation for this.
The one that I found helpful was based on the idea in BA's book. The tidal force is a result in the difference between the gravitational forces at two different places. In other words, it's more or less the derivative of the gravitational force. Take a derivative of the inverse square, and you have an inverse cube.


Also, it appears the horizontal component not the radial component is the main player in an ocean tide.

I'm not sure exactly what you are referring to here. I've seen some similar discussion, but I'm not clear on what you mean by "horizontal." Could you go into more detail?

George
2003-Oct-19, 09:30 PM
The key to tides is not the absolute force, but the relative force. Try redoing the calculation for the force exerted on you at midday and midnight.

Good idea. My body will experience a difference between these two forces of 0.000067 lbf. [Lunar attraction only]

So, the simple gravitational attraction of the moon seems far too weak to explain tides.

I stumbled into this site that brings up a very interesting idea.

>>> here (http://www.es.flinders.edu.au/~mattom/ShelfCoast/notes/chapter05.html) <<< .

It says that tides are more of a standing wave. The bodies of waters are like oversize tanks of water that if you shake at a resonant frequency you get bigger and bigger waves (tides) in the tank.

Possibly my only outstanding contribution to mankind, as a youth at least, may apply here. I can get into almost any swimming pool at about 4 feet depth and oscillate my body in a certain way and at a certain frequency, and in little time, create white caps on the enlarging waves generated. 8) =D> :D *chest swelling* (no autographs please).

I suppose the "shaking" force would be the Lunar gravity force vector component that acts along the east-west plane of the sea. The resonant frequency, of course, is critical to the ocean depth so this helps explain the significant differences in local tides.

So ya'll's thoughts on this would be appreciated (ocean resonance, that is not my pool ability).

BTW - my wife can tie a cherry stem into a knot with just her tongue. 8) =D> :) [No, it's not why I married her] :lol:

George
2003-Oct-19, 09:47 PM
The websites show that tidal forces, however, vary as the inverse cube of the distance not the square. I wish I could find a clear explanation for this.
The one that I found helpful was based on the idea in BA's book. The tidal force is a result in the difference between the gravitational forces at two different places. In other words, it's more or less the derivative of the gravitational force. Take a derivative of the inverse square, and you have an inverse cube.


Yep. It just didn't seem intuitive at first but it does with a little thought as there can be little difference my body feels as a result of the Sun's extensive distance.




Also, it appears the horizontal component not the radial component is the main player in an ocean tide.

I'm not sure exactly what you are referring to here. I've seen some similar discussion, but I'm not clear on what you mean by "horizontal." Could you go into more detail?

Good question. "Horizontal" is a bit akward term for a vector component

The "pull" of the Moon is vertical at only one point on the Earth's surface at any given time. All other points will be "pulled" at an angle. This force at an angle (vector) can be broken into two components - vertical and horizontal to the Earth's surface. The further from that vertical point, the greater the horizontal force. However, it won't be as large as the wimpy vertical component as the angle is never that great. (Of course, you knew all this but I figure someone might not).

So how can it help make a tide? It may be very important if the tide is a standing wave and this component keeps building the wave up along with other forces like the sun.

JohnD
2003-Oct-19, 10:39 PM
All,
As I said at the start, I honestly believed that the different orbits and the gravity vector models were compatible and described the same situation in different ways. However, I've been thinking hard about the Earth free-falling into the heart of the Sun, and in this situation I can't explain tides using the DO model. So by Occam's Razor, the DO model fails, the GV wins, and I retire from the discussion. Sadly, as I still find the GV model difficult to understand, Thankfully, as the discussion has been most enjoyable!
John

kilopi
2003-Oct-19, 11:17 PM
The bodies of waters are like oversize tanks of water that if you shake at a resonant frequency you get bigger and bigger waves (tides) in the tank.
The actual tides vary all over the place. As has been mentioned before, the force which produces the tides would result in less than a meter difference from high tide to low tide, but places like the Bay of Fundy (http://www.valleyweb.com/fundytides/), where the tidal range is on the order of a dozen meters, are infamous.

I still find the GV model difficult to understand,
We all do. :)


Thankfully, as the discussion has been most enjoyable!

Cheers!

Eroica
2003-Oct-20, 11:12 AM
I've been thinking hard about the Earth free-falling into the heart of the Sun, and in this situation I can't explain tides using the DO model.
I can. And I think I did in my last post (http://www.badastronomy.com/phpBB/viewtopic.php?p=156483#156483).
Of course, in this situation, every point on (and in) the Earth is orbiting too slowly to stay in orbit and so they all fall into lower orbits. The difference is that their orbital speeds are not just too slow, they're zero.

George
2003-Oct-20, 01:18 PM
The actual tides vary all over the place. As has been mentioned before, the force which produces the tides would result in less than a meter difference from high tide to low tide, but places like the Bay of Fundy (http://www.valleyweb.com/fundytides/), where the tidal range is on the order of a dozen meters, are infamous.

Thanks for that site. It verifys my point that resonance plays a key role.
As per your site...


The primary cause of the immense tides of Fundy is a resonance of the Bay of Fundy-Gulf of Maine system. The system is effectively bounded at this outer end by the edge of the continental shelf with its approximately 40:1 increase in depth. The system has a natural period of approximately 13 hours, which is close to the 12h25m period of the dominant lunar tide of the Atlantic Ocean

I do not wish to diminish the DO - GV arguments, as they are stimulating, but the resonant factor explains much.

kilopi
2003-Oct-20, 03:00 PM
I've been thinking hard about the Earth free-falling into the heart of the Sun, and in this situation I can't explain tides using the DO model.
I can. And I think I did in my last post (http://www.badastronomy.com/phpBB/viewtopic.php?p=156483#156483).
There is nothing wrong with the DO model per se--but it should be noted that the velocity of every point on the Earth is essentially identical, in magnitude and direction, disregarding rotation. Thus, the velocities do not induce a differential from farside to center to nearside.

Still, the model may be instructive for some. That doesn't make the BA's not as good.

George
2003-Oct-20, 11:09 PM
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?

[You probably are wore out already, but you did answer my threads and I appreciate it].

I'm sure the simple answer is that it's ratio of it's rotation (58.65 days) to it's revolution (88 days) is a nice 3/2 harmonic motion. It is "comfortable" there like a vibrating string.

Now, if you want to know why it is not 1/1 (like a vibrating string prefers) well, this is for someone wiser. Most moons are 1/1, I believe. Maybe planetary surface characteristics such as bulges, mountains and water play a stronger tidal roll. It wouldn't surprise me to see our friend "General Relativity" step in here in the case of Mercury.

BTW - If you don't mind, how do you say your name? Mine is obvious but I have not heard yours before. Is it Eh-Roy-Ka, or what?

Eroica
2003-Oct-21, 11:26 AM
Since posting that first message, I have done a bit of googling and discovered that the primary reason that Mercury is locked into a 3:2 resonance, rather than a 1:1 like our Moon, is because its orbit is much more eccentric (0.2056) than the Moon's (0.0549). If it were locked into a 1:1 resonance, it would not be able to keep the same side facing the Sun because its orbital velocity would fluctuate a good deal between aphelion and perihelion, whereas its rotational velocity would be more or less unchanging.

Since it can't easily rotate at different speeds, there's no way for Mercury to keep the same face, and its tidal bulge, pointed at the sun at all times.

Instead, it does the best it can: it makes sure that it always has its tidal bulges lined up with the sun when it is closest to the Sun and the force of gravity is strongest. The best way to do that while rotating at a constant speed is to complete 1.5 rotations every year. That way the tidal bulges always line up with the sun at perihelion.

Our Moon's orbit is also elliptic, so its orbital speed fluctuates, but not as much. Nevertheless, these fluctuations do cause longitudinal librations, allowing us to see a little more than 50% of the Moon's face. (There are other causes for the various other forms of libration.)

PS: Eroica is the Italian word for "Heroic." I borrowed it from my favourite piece of music, Beethoven's third symphony, the Sinfonia Eroica. I pronounce it er-ro-ik-ka. I've been using it as a nickname since long before the Internet.

kilopi
2003-Oct-21, 02:28 PM
Since it can't easily rotate at different speeds, there's no way for Mercury to keep the same face, and its tidal bulge, pointed at the sun at all times.
The tidal bulge is always pointed at the Sun, except for the slight offset that results in tidal slowing. If Mercury is in a resonance, the slowing on one part of the orbit must be offset by speeding up on another part.


Instead, it does the best it can: it makes sure that it always has its tidal bulges lined up with the sun when it is closest to the Sun and the force of gravity is strongest. The best way to do that while rotating at a constant speed is to complete 1.5 rotations every year. That way the tidal bulges always line up with the sun at perihelion.
There may be density differences that were emplaced a long time ago that have nothing to do with the tides that tend to orient the planet. Perhaps they were a result of the impact that produced the Caloris basin.

Eroica
2003-Oct-21, 03:02 PM
The tidal bulge is always pointed at the Sun, except for the slight offset that results in tidal slowing. If Mercury is in a resonance, the slowing on one part of the orbit must be offset by speeding up on another part.
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?

kilopi
2003-Oct-21, 03:31 PM
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).
Yes, that's what I meant.


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?
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.

Eroica
2003-Oct-21, 04:45 PM
Thanks for that, kilopi. I found this interesting PDF (http://www.sp.ph.ic.ac.uk/~giacomo/Papers/mercuryrev.pdf) which explains the whole process in mathematical detail.

They do confirm my suspicions about the tidal bulge:

the bulge is carried forward with respect to the subsolar point, except at perihelion, where the converse happens (since during this short phase the Sunís apparent motion is retrograde....
However, they conclude that:

the torque acting on the permanent deformations dominates [over the tidal torque], and causes the preferred orientation of Mercury.
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]

kilopi
2003-Oct-21, 09:33 PM
the bulge is carried forward with respect to the subsolar point, except at perihelion, where the converse happens (since during this short phase the Sunís apparent motion is retrograde....
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.

Eroica
2003-Oct-22, 10:50 AM
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. (http://www.sp.ph.ic.ac.uk/~giacomo/publications.html)

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

... at perihelion the orbital angular velocity becomes larger than the spin angular velocity, causing the Sun to move on a retrograde orbit for a short interval of time.

kilopi
2003-Oct-23, 08:11 AM
The retrograde motion of the Sun only lasts about 6 or 7 days:
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?

Eroica
2003-Oct-23, 04:16 PM
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:

(i) the 3:2 resonance is due to permanent deformations on the equatorial plane [of the planet], and to the non-zero eccentricity [of its orbit]
(ii) tidal torque by itself is not able to produce a non-synchronous resonance. In contrast, it could prevent reaching or maintaining it if the permanent deformations were not large enough

In an earlier post I wrote:

Instead, [Mercury] does the best it can: it makes sure that it always has its tidal bulges lined up with the Sun when it is closest to the Sun and the force of gravity is strongest.
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.

kilopi
2003-Oct-23, 08:56 PM
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.
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!

George
2003-Oct-25, 02:57 AM
(i) the 3:2 resonance is due to permanent deformations on the equatorial plane [of the planet], and to the non-zero eccentricity [of its orbit]
(ii) tidal torque by itself is not able to produce a non-synchronous resonance. In contrast, it could prevent reaching or maintaining it if the permanent deformations were not large enough



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?

Ian R
2004-Jan-14, 07:31 PM
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

milli360
2004-Jan-14, 09:07 PM
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 (http://www.badastronomy.com/phpBB/viewtopic.php?p=124513#124513). 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.

RichardMB
2004-Feb-29, 04:40 PM
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?

Ian R
2004-Feb-29, 04:55 PM
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.


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.

Eroica
2004-Feb-29, 06:15 PM
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.

RichardMB
2004-Mar-01, 06:54 PM
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.

milli360
2004-Mar-01, 08:46 PM
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.
No, not really.


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?
The tidal bulge is also proportional to the radius of the body. A tennis ball is a lot smaller than the Earth.

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.
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.

Eroica
2004-Mar-02, 11:58 AM
At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.
Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?

Eroica
2004-Mar-02, 12:01 PM
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' ... [snip]
My definition of "free-fall" is "subject only to gravitational forces." What's yours?

milli360
2004-Mar-02, 02:01 PM
At point F centripetal acceleration exceeds gravitational and at point C gravitational exceeds centripetal. This is what 'raises' the tides not differential gravity.
Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?
I'm pretty sure he is talking about the centrifictional force (http://www.badastronomy.com/phpBB/viewtopic.php?p=29784#29784) due to the body's revolution around the central body. Gravity is the centripetal force.

One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body. That's why most explanations use the differential gravity, because it doesn't matter whether the body is in the free fall of an orbit, or falling "straight down."

I don't have the trouble that many have in recognizing centrifugal effects as a force, but in this case, it is not necessary.

RichardMB
2004-Mar-03, 03:27 PM
Eroica wrote:

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

Oh, cheers, mine's a Boddingtons. Your definition looks fine to me, but are you absolutely certain this is what Sawicki means? because if he means something else you'll not be understanding his model. My definition isn't important, I gave it a stab and milli360 jumped on me so I've lost interest - I've really no idea what it means.

Eroica wrote:

Are you suggesting that there would be no tides if the Earth were not rotating on its axis? Or have I misunderstood you?
You've misunderstood me, I didn't mention earth's axis, I was referring to earth's annual rotation about the solar barycentre same as Sawicki was - I'll come back to it in a moment. What I was saying is that without centripetal acceleration (annual solar) the sun's gravity would produce a single tide, much the same as on the tennis ball and for the exactly same reason.

[b]milli360 wrote

One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body.
I once found a site once that explained this idea in detail. It's wrong. Look again at Sawicki's fig.1. The mistake is to think of points F, O and C as points on earth, this is wrong, they're points in space. Point O is 1AU from the sun so we can plug this into a(g) = GM/D^2 to determine the sun's gravitational acceleration at point O. By the same token we can use a(c) = w^2r to find it's centripetal acceleration. If we do this we'll find that they are exactly the same. None of this requires the presence of earth. You can remove earth altogether if you wish. You can replace it with jupiter. You can spin it backwards or forwards, fast or slow, pole over pole. It won't make the slightest bit of difference - at point O solar gravitation will equal centripetal acceleration. Point F is farther out, at 1AU+1 earth radius, gravitational acceleration (=GM/D^2) is therefore lower and centripetal acceleration (= w^2r) is greater. Again, I must stress, it has nothing whatever to do with earth. Once we've done our calculations and we know what a(g) and a(c) is at all three points, we can stick the earth back and ask what affect these accelerations might have - tides for instance.

Don't be gulled by smart graphics. Get hold of a decent textbook.

milli360
2004-Mar-03, 04:00 PM
What I was saying is that without centripetal acceleration (annual solar) the sun's gravity would produce a single tide, much the same as on the tennis ball and for the exactly same reason.
No, that is definitely not true, unless by "centripetal acceleration" you are actually referring to gravity itself. As I pointed out, if the body is falling straight down, a tidal bulge would be induced. No centrifugal force necessary.


milli360 wrote

One aspect of this issue that is often overlooked is that on a rigid body in revolution, the centrificitional force is constant, in magnitude and direction--and so does not induce a difference across the body.
I once found a site once that explained this idea in detail. It's wrong.
The site was wrong?


Look again at Sawicki's fig.1. The mistake is to think of points F, O and C as points on earth, this is wrong, they're points in space. Point O is 1AU from the sun so we can plug this into a(g) = GM/D^2 to determine the sun's gravitational acceleration at point O. By the same token we can use a(c) = w^2r to find it's centripetal acceleration.
You mean centrifugal. Gravity is centripetal. However, F, O, and C are points on Earth--we are trying to calculate the tide on Earth, not in space.

If we do this we'll find that they are exactly the same. None of this requires the presence of earth. You can remove earth altogether if you wish. You can replace it with jupiter. You can spin it backwards or forwards, fast or slow, pole over pole. It won't make the slightest bit of difference - at point O solar gravitation will equal centripetal acceleration. Point F is farther out, at 1AU+1 earth radius, gravitational acceleration (=GM/D^2) is therefore lower and centripetal acceleration (= w^2r) is greater.
If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth. The reason for that is you have to remove all effects of the rotation of the Earth. Otherwise, the situation would be akin to the moon's, where there is one complete rotation in every revolution.

RichardMB
2004-Mar-05, 12:27 PM
milli360 wrote:

'The site was wrong?


Yes. Where it says that centripetal acceleration, on a body in orbit, is the same at all points on the surface of the body (or some thing like that), the site is wrong.


'If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth.'


If you say so - but since I never mentioned either 'centrifugal' or 'force' I'd rather not comment. But I do agree that gravity is centripetal - by definition.

Let's get back to Sawicki's model though, again referring to his fig.1.

Point O is 1AU from the sun and it takes 365.3 days to complete an orbit so we can work out a(c) (centripetal acceleration), at O, using a(c) = w^2 r - it comes to 5.9303e-03 ms^-2. We also know the solar mass (1.989e30 kg) so we can work out a(g) (gravitational acceleration), at O, using a(g) = GM/d^2 - it also comes to 5.9303e-03 ms^-2. At point O therefore centripetal and gravitational acceleration are the same.

Point F is farther from the sun at 1AU+ 1 earth radius but it's period is the same at 365.3 days. Repeating the above calculations, using this increased distance, gives a(c) = 5.9306e-03 ms^-2 and a(g) = 5.9298e-03 ms^-2. Solar gravity is less than the centripetal acceleration, which means that (at point F) the sun in not able to supply all the acceleration required - something else is making up the difference.

Stop and think about this, we started looking for the force (acceleration) that pulls the tidal bulge at F up, but after doing 3 simple calculations we find that we really need a force (acceleration) to pull it down. Not an outward force, an inward force - the same as the sun's gravity.

Good isn't it? and it makes a lot more sense - the earth itself can provide this, no trouble at all. The tiddly little 2.59e-05 ms^-2 needed comes from earths mighty 9.81 ms^-2, as a consequence the weight force at F is reduced by the same amount - and a reduction in the weight force will result in a bulge - as required.

This explanation does away with the troublesome outward force, it does away with doubtful vector manipulations, it does away with asymmetric tidal forces.

Still happy with 'differential gravity' are we ?

milli360
2004-Mar-05, 07:22 PM
Where it says that centripetal acceleration, on a body in orbit, is the same at all points on the surface of the body (or some thing like that), the site is wrong.
I'm not sure why you keep using "centripetal" where you should be using centrifugal, but with that in mind, that it's the same at all points is not wrong. It is correct, assuming a sphere, once you completely subtract the effect of the rotation of the Earth.



'If done correctly, you'll find that the centrifugal force is the same, in magnitude and direction, across the Earth.'

If you say so - but since I never mentioned either 'centrifugal' or 'force' I'd rather not comment. But I do agree that gravity is centripetal - by definition.
If gravity is centripetal, why do you differentiate between gravitational and centripetal acceleration, below?


Let's get back to Sawicki's model though, again referring to his fig.1.

Point O is 1AU from the sun and it takes 365.3 days to complete an orbit so we can work out a(c) (centripetal acceleration), at O, using a(c) = w^2 r - it comes to 5.9303e-03 ms^-2. We also know the solar mass (1.989e30 kg) so we can work out a(g) (gravitational acceleration), at O, using a(g) = GM/d^2 - it also comes to 5.9303e-03 ms^-2. At point O therefore centripetal and gravitational acceleration are the same.

Point F is farther from the sun at 1AU+ 1 earth radius but it's period is the same at 365.3 days. Repeating the above calculations, using this increased distance, gives a(c) = 5.9306e-03 ms^-2 and a(g) = 5.9298e-03 ms^-2. Solar gravity is less than the centripetal acceleration, which means that (at point F) the sun in not able to supply all the acceleration required - something else is making up the difference.

Stop and think about this, we started looking for the force (acceleration) that pulls the tidal bulge at F up, but after doing 3 simple calculations we find that we really need a force (acceleration) to pull it down. Not an outward force, an inward force - the same as the sun's gravity.

Good isn't it? and it makes a lot more sense
It makes an intuitive sense, but it's wrong. Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force.


This explanation does away with the troublesome outward force, it does away with doubtful vector manipulations, it does away with asymmetric tidal forces.
All that troublesome math? :)


Still happy with 'differential gravity' are we ?
More than ever. I'm understanding better where the alternative leads.

RichardMB
2004-Mar-06, 05:14 PM
milli360 wrote:
'I'm not sure why you keep using "centripetal" where you should be using centrifugal'

I use centripetal because it means towards a central point (inward) and that is precisely what I mean. Centrifugal means away from a central point (outward), my previous mail explained that there is no need for an outward acceleration - which is just as well because none exists. If you came across a model that so much as hints at an outward force be assured that it is wrong.

'If gravity is centripetal, why do you differentiate between gravitational and centripetal acceleration .... ?'

A good question: Gravitational acceleration a(g) is an acceleration produced by a mass. Although gravitation is centripetal, the term ‘centripetal acceleration’, a(c), is generally used to mean the acceleration required by a mass to cause it to move in a circular (elliptical) path. So in our present discussion a(c) is the centripetal acceleration required to move earth in a 365.3 day circle and a(g) is the centripetal acceleration provided by the sun (as a consequence of its mass). This is similar to the relationship between a 12v battery and a 12v bulb, both are 12v devices but one supplies it whilst the other requires it. Yes?


'Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force. '

Point L in Sawicki's fig 1 is at 90 degrees is this what you mean? Point L is 1AU from the sun so a(g) = a(c) (same as point O), there is no need for a contribution from earths gravity, which is just as well because there is no component of earths gravity in the direction of the sun anyway. The weight force is at a maximum at L so there is no bulge - this is low tide. That's what we want isn't it?


‘All that troublesome math?’
The maths is really very simple (just + - ^ / and *) and anyway I haven’t actually given you any maths – I’ve given you the equations and their results. This is the minimum I think you need to test my model – and I hope you are testing it.
I don’t think you’ve tested Sawicki’s maths though, have you? Take a look at his equation 3. It’s o.k. for Da(s) at point F but then he says it the same as Da(s) at point C – and that’s wrong. I suspect Mik knows it’s wrong too, but if they’re not the same it undermines the whole model.

Still happy with 'differential gravity' are we ?

milli360
2004-Mar-06, 07:15 PM
A good question: Gravitational acceleration a(g) is an acceleration produced by a mass. Although gravitation is centripetal, the term Ďcentripetal accelerationí, a(c), is generally used to mean the acceleration required by a mass to cause it to move in a circular (elliptical) path.
Ah.


'Try to use the same technique to calculate the tidal force ninety degrees or so away from the bulges. At those points where your calculations of the centrifugal acceleration and the gravitational acceleration would still be the same, you won't be able to use that explanation for the tidal force. '
Point L in Sawicki's fig 1 is at 90 degrees is this what you mean? Point L is 1AU from the sun so a(g) = a(c) (same as point O), there is no need for a contribution from earths gravity, which is just as well because there is no component of earths gravity in the direction of the sun anyway. The weight force is at a maximum at L so there is no bulge - this is low tide. That's what we want isn't it?
Approximately at 90 degrees, but at the many points where a(g_=a(c), is what I mean. Low tide is a depression, the opposite of a bulge. That's my point, your theory shows no effect. Whereas, the "differential gravity" theory is in accord with observation.

I'm partial to another treatment, which involves a calculation of the potential. To me, it's more comprehensive, and easier to understand--and I can use it to judge the other explanations, too.

This is the minimum I think you need to test my model Ė and I hope you are testing it.
It seems to fail.


I donít think youíve tested Sawickiís maths though, have you? Take a look at his equation 3. Itís o.k. for Da(s) at point F but then he says it the same as Da(s) at point C Ė and thatís wrong. I suspect Mik knows itís wrong too, but if theyíre not the same it undermines the whole model.

Not sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?


Still happy with 'differential gravity' are we ?

Yep, except as noted.

PS: Link to Sawicki's paper (http://www.jal.cc.il.us/~mikolajsawicki/gravity_and_tides.html)

RichardMB
2004-Mar-07, 09:35 PM
milli306 wrote:
‘I'm partial to another treatment, which involves a calculation of the potential’
Excellent – whatever you find helpful, I was discussing Sawicki’s paper.


‘Low tide is a depression, the opposite of a bulge. That's my point, your theory shows no effect. Whereas, the "differential gravity" theory is in accord with observation.’
There is no need to postulate an acceleration at point L. Earth is elastic so think of Sawicki’s fig.1 as a balloon, pulling at point F and O will raise bulges at these points. Point(s) L will naturally depress as the balloon accommodates the changed internal stresses. Where’s the problem?

But let’s have a look at what Sawicki says. According to Sawicki there is a sin alpha component of the sun’s pull at point L operating towards O (downwards), and he more or less just states that this amounts to D(a)s/2. By my calculations sin alpha is 4.25e-05 that’s 0.00000425 not a half so, even if we use the biggest D(a)s, the additional acceleration at L is 2.15e-11 ms^-2. That’s in accord with observation is it?

(I must confess I’ve only given this a quick look - because you raised it - so if I’ve misunderstood it in someway do please explain it to me.)


‘Not sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?’
Sawicki uses equ.3 to calculate Da(s) on the far side, that is point F at a distance d(s)+R (1AU+radius of earth). Using my data it comes to 5.0476e-07 ms^-2. On the near side, point C, the distance is d(s)-R (1AU-earth radius) and I make this –5.0483e-07 ms^-2. According to Sawicki these are the same. Well, what do you reckon, they don’t look the same to me.

So what? Well, Sawicki says the acceleration at point C is a(s) + D(a)s, but since D(a)s is negative the acceleration at C is less than at O, so like F it lags behind the centre – the maths is telling us that there is a low tide at point C!
That’s not all either. You now have a choice of positive or negative D(s)a so at point L you can have a bulge or a depression as you wish. You must watch these ‘higher order terms’ – they’re a bit like the ‘the curl of e’.

You’re still not convinced are you?

milli360
2004-Mar-08, 08:48 AM
There is no need to postulate an acceleration at point L.
It's not a postulate, it's pretty much a fact. And, as you admit, your theory ignores it completely. It is about half the effect at the points of the tidal bulge.


According to Sawicki there is a sin alpha component of the sunís pull at point L operating towards O (downwards), and he more or less just states that this amounts to D(a)s/2. By my calculations sin alpha is 4.25e-05 thatís 0.00000425 not a half so, even if we use the biggest D(a)s, the additional acceleration at L is 2.15e-11 ms^-2. Thatís in accord with observation is it?
Alpha is the angle at the Earth's center, so at L, alpha is 90 degrees and the sine of alpha is one, not .00000425.



ĎNot sure what you are referring to here. He does mention, in the text, that he has ignored higher order terms. Is that what you mean?í
Sawicki uses equ.3 to calculate Da(s) on the far side, that is point F at a distance d(s)+R (1AU+radius of earth). Using my data it comes to 5.0476e-07 ms^-2. On the near side, point C, the distance is d(s)-R (1AU-earth radius) and I make this Ė5.0483e-07 ms^-2. According to Sawicki these are the same. Well, what do you reckon, they donít look the same to me.
As I said, he has ignored the higher order terms. They're a lot smaller than the main semidiurnal tide terms, but they contribute to the diurnal, and even sidereal tides.

Youíre still not convinced are you?
I'm never convinced. :)

RichardMB
2004-Mar-09, 04:55 PM
milli360 wrote:



RichardMB wrote:
There is no need to postulate an acceleration at point L.

’It's not a postulate, it's pretty much a fact. And, as you admit, your theory ignores it completely’.


Gosh, a theory. No I don’t have a theory. For a simple problem like this I find Newton’s theory does admirably.

The origin of tides is an orbital mechanics problem and there are two basic, and complementary, equations in orbital mechanics: GM/D^2 characterising the gravitational component and w^2r characterising the (elliptical) motion. Anybody reading these pages should know this by now.

Where Sawicki goes wrong is to ignore the w^2r (centripetal acceleration) component and, by replacing it with ‘free-fall’ (whatever that means), he fatally cripples his model, and all that follows is corrupted.

Whether acceleration at L is needed or not isn’t something I could get exited about (I suspect it isn’t), the point is it’s that Sawicki’s explanation of it has to be wrong.

If you want a more plausible explanation for this acceleration (or depression) tell me where I can find out more about it and I’ll get onto it. Is it mentioned in Encarta, for instance or Webster’s, or Britannica? Where have you heard about this ‘fact’– other than in Sawicki’s paper?

milli360
2004-Mar-09, 07:50 PM
Gosh, a theory. No I donít have a theory. For a simple problem like this I find Newtonís theory does admirably.
Your explanation, then.


Where Sawicki goes wrong is to ignore the w^2r (centripetal acceleration) component and, by replacing it with Ďfree-fallí (whatever that means), he fatally cripples his model, and all that follows is corrupted.
That's wrong. As you point out above, the only centripetal acceleration component is due to gravity, and Sawicki definitely does not ignore gravity.


Whether acceleration at L is needed or not isnít something I could get exited about (I suspect it isnít), the point is itís that Sawickiís explanation of it has to be wrong.

I'm not so sure, and certainly not for the reasons that you've given.


If you want a more plausible explanation for this acceleration (or depression) tell me where I can find out more about it and Iíll get onto it. Is it mentioned in Encarta, for instance or Websterís, or Britannica? Where have you heard about this ĎfactíĖ other than in Sawickiís paper?
Are you asking how I know that there is a low tide??

RichardMB
2004-Mar-11, 11:33 AM
milli360 wrote:


Are you asking how I know that there is a low tide??


No, I didn't mention low tide. What I said was:



If you want a more plausible explanation for this acceleration .... Where have you heard about (it) other than in Sawicki’s paper?


The question is perfectly clear - where have you heard about this acceleration (D(a)s/2 at point L) - other than in Sawicki's paper.


That's wrong. As you point out above, the only centripetal acceleration component is due to gravity, and Sawicki definitely does not ignore gravity.


No, I did not say that. What I said was that gravity is a centripetal acceleration. In fact there are at least 5 centripetal accelerations to consider with this problem.
Yes, I do agree that 'Sawicki definitely does not ignore gravity' - the trouble is that's allhe considers.

Now, I do appreciate your contribution to this discussion but in future you're going to have to try harder. I'm not going to bother to answer your questions if it's obvious that you haven't read my contribution.

Please, is there anybody else out there with views or questions on this subject? See if we can't get it going again.

milli360
2004-Mar-11, 05:08 PM
milli360 wrote:


Are you asking how I know that there is a low tide??


No, I didn't mention low tide. What I said was:



If you want a more plausible explanation for this acceleration .... Where have you heard about (it) other than in Sawickiís paper?


The question is perfectly clear - where have you heard about this acceleration (D(a)s/2 at point L) - other than in Sawicki's paper.
Your ellipsis covers the parenthetical "(or depression)" that apparently refers to that same acceleration. That depression is the low tide. The tidal acceleration is recorded by accelerometers around the world. It's in every book on the theory of tides that I've ever seen.


Now, I do appreciate your contribution to this discussion but in future you're going to have to try harder. I'm not going to bother to answer your questions if it's obvious that you haven't read my contribution.
I have.

Eroica
2004-Mar-11, 09:58 PM
Point O is 1AU from the sun so we can plug this into a(g) = GM/D^2 to determine the sun's gravitational acceleration at point O. By the same token we can use a(c) = w^2r to find it's centripetal acceleration. If we do this we'll find that they are exactly the same ... Point F is farther out, at 1AU+1 earth radius, gravitational acceleration (=GM/D^2) is therefore lower and centripetal acceleration (= w^2r) is greater.
This sounds to me like another way of expressing the Different-Orbits theory that has already been mooted in this thread. What do you think? Do both explanations (yours and the DO theory) amount to the same thing?

RichardMB
2004-Mar-12, 10:55 PM
milli360 wrote:

The tidal acceleration is recorded by accelerometers around the world. It's in every book on the theory of tides that I've ever seen.

That would explain why I’ve never heard of it, but I can confirm that I’ve recently found it from another source. Having said that, my position remains that, whilst such an acceleration may exist, it is not a pre-requisite. If it does exist then yes, a complete theory of tides should address it – mine included (if ever I should produce one).


This sounds to me like another way of expressing the Different-Orbits theory that has already been mooted in this thread. What do you think? Do both explanations (yours and the DO theory) amount to the same thing?

The DO theory is correct as far as it goes and I have no problem with it. But I’m sure the author could have taken it further.

I’ve given more detail in particular why no outward force is necessary. I’ve shown how centripetal acceleration, solar gravity and terrestrial gravity interact, and I’ve given you the tools to check the results for yourself.

Does it help any or does it raise more problems?

milli360
2004-Mar-15, 06:50 AM
Iíve given more detail in particular why no outward force is necessary. Iíve shown how centripetal acceleration, solar gravity and terrestrial gravity interact, and Iíve given you the tools to check the results for yourself.
Which outward force? I think we've discussed before, neither your explanation nor the BA's uses anything but gravity, right? As a centripetal force?

RichardMB
2004-Mar-16, 08:51 PM
milli360 wrote:


Which outward force? I think we've discussed before, neither your explanation nor the BA's uses anything but gravity, right? As a centripetal force?

I was making a general point that any explanation should show that no outward force is required, to be credible it must show this explicitly.

However, since you ask, here a snip from Sawicki:

'... there appears to be another force, the tidal force, pulling the mass m at points C and F up...' (Sawicki's emphasis),

and one from the BA:

'It is mathematically correct to then subtract the force of the Moon on the center of the Earth from the force felt on the near and far sides. This is called vector addition. If we do that, our diagram will look like this:

<- X ->
far center near
side of Earth side '

Not what I’d call explicit (that didn't print properly did it? - take a look at the BA home page).

But I’m not much bothered about this, what concerns me most is that Sawicki’s paper totally ignores the centripetal acceleration of earths orbit. I’m frankly baffled that he should be so familiar with the suns gravitational gradient (D(a)s) yet (apparently) so ill informed about the complementary centripetal gradient.

Replacing centripetal acceleration with ‘The Earth is simply in free fall towards the Sun’ is staggeringly naÔve – it produces in a wrong explanation and it produces incorrect results.

milli360
2004-Mar-16, 10:48 PM
I was making a general point that any explanation should show that no outward force is required, to be credible it must show this explicitly.

::snip::

Not what Iíd call explicit (that didn't print properly did it? - take a look at the BA home page).
I assume you got that from the webpage. There may be more there, but the book seems explicit enough: "It seems paradoxical that gravity can act in such a way as to make something feel a force away from an object, but in this case it's because we are measuring that force relative to the center of the Earth."

That's fairly explicit, saying that gravity is the only force required.


But Iím not much bothered about this, what concerns me most is that Sawickiís paper totally ignores the centripetal acceleration of earths orbit. Iím frankly baffled that he should be so familiar with the suns gravitational gradient (D(a)s) yet (apparently) so ill informed about the complementary centripetal gradient.
As I've mentioned before, if you calculate the centripetal force required for each piece of a non-rotating Earth, the gradient is zero. In other words, the force is constant, in direction and magnitude, across the body of the Earth.


Replacing centripetal acceleration with ĎThe Earth is simply in free fall towards the Suní is staggeringly naÔve Ė it produces in a wrong explanation and it produces incorrect results.
I disagree. The Earth is in free fall, after all, and the calculations match the data. The alternative you've suggested doesn't seem even to explain the low tide.

As someone said Einstein said, you want to make the theory as simple as possible, but no simpler.

Lunatik
2004-Mar-17, 02:45 AM
OPPOSITE FACE LUNAR TIDAL BULGES ARE ELECTROSTATIC?

"Why is there a tidal bulge on the opposite side of the Earth?" Excellent question! But in reading the posts, I'm still not satisfied... :o

In fact I've never found a truly satisfactory reason in all my readings why on the opposite side of the globe facing away from the moon or sun, are nearly the same as the tidal bulge facing them. The idea that the gravitational pull on the opposite side of the planet is canceled by the planet does not satisfy me. 'Push' gravity does not explain it effectively either, nor does the "three point pull". So I would like to present an idea which I suspect is original, if not crazy, though another may have thought of it first: the risen tide on the planet's opposite face away from the moon or sun is due to water's electrostatic charge.

As outlandish as this may appear at first blush, consider the following suggestions:

1. There is a magnetic field from pole to pole, if off axis, that permeates the globe. This magnetic field gives a negative charge value to the northern hemisphere and a counter positive charge to the southern hemisphere, with a more neutral reading around the equator, where the positive and negative meet.

2. Tides appear to be greater towards the poles and lesser near the equator, which cannot be accounted by for planetary spin, since that would yield the opposite effect. So another cause must be found, which leads to the idea that like charges repel, even in the planet's ocean mass.

3. If like charges for large water mass, such as the oceans, repel, then the waters on the northern hemisphere, as they rise in response to the moon's or sun's gravity, will create a bulge towards the gravitational attraction and one opposite to it. This bulge creates a large negatively charged mass which, being of like charge, will be repelled on the opposite side of the planet, so that there too the water will bulge outwards. So due to like charge repelling like charge, the opposite side of the planet's mass is now counter-bulging in response to the mass bulge on the side facing the moon or sun. On the equator, where the positive-negative charge more or less cancels, there is less tidal opposite face response, so that there is less bulge than where the electrostatic charge is greater. On the southern hemisphere, the positively charged water mass likewise creates its own opposite bulge, same as in the northern.

4. Why does the equatorial water not bulge as much as towards the poles? My best guess is that it is already bulging, because of the planet's spin, so that the additional gravitational pull of the moon or sun does not displace this water as much as in either hemisphere because the centrifugal force already displaces it. Thus, equatorial waters should experience lower tides. I would expect, by this reasoning, that the waters near the artic or antartic circles should rise the most, and then taper off gradually towards the poles, or flatten out towards the equator. Of course, this basic principle is further modified by ocean currents and land mass interferences.

So by this reasoning, the tidal bulges on the opposite side of the planet are actually electrostatic phenomena, where like charge repels like charge, even in the ocean waters. The same would be expected in the land mass, though to a much lower degree. In effect, the planet is pulsating with a gravitationally activated bulge which then triggers a like response on the opposite side of the planet through like electrostatic force repulsion. If this did not happen, the planet would wobble chaotically, which it does not.

Pardon my ignorance, but I am not familiar with the above mentioned electrostatic theory of tides from any known sources, but would love to find out this it is not so. To my thinking, this much better explains why the tides create a like bulge on the opposite side of the planet, than any theory now found in existing textbooks. Blame it on the hot sun, but I got this idea while hiking up and down the rocky peaks near Pena Springs, Anza Borrego desert, last weekend, watching out for snakes and other prickly things. New barbs won't hurt me, so all challenges are welcome. 8)

* * * * * (edited 3/18/04) * * *
... Having said all that, I must confess that this idea is really crazy, so ignore it... thanks. #-o

RichardMB
2004-Mar-17, 10:24 PM
… the book seems explicit enough: "… gravity can act in such a way as to make something feel a force away from an object..."

That's fairly explicit, saying that gravity is the only force required.
No, it isn’t.

‘Gravity cannot act in such a way as to make something feel a force away from an object.’

That would be explicit.



As I've mentioned before, if you calculate the centripetal force required for each piece of a non-rotating Earth, the gradient is zero. In other words, the force is constant, in direction and magnitude, across the body of the Earth.

O.k. Let me see if I understand you correctly here. You reckon the centripetal acceleration, of a 365.24 day orbit at Sawicki’s 3 points O, F and C, is exactly the same – there will be no difference in centripetal acceleration at F, C and O.
Is this what you are saying?
I use w^2r to calculate centripetal acceleration and I don’t actually get that result, so (bearing in mind your quote of Einstein’s) could you maybe show your calculations, just for points O and F say. That would be most useful.


I disagree. The Earth is in free fall, after all, and the calculations match the data.

Ah, yes. Free fall! When we’ve sorted out centripetal acceleration I’d like to get back to this.

milli360
2004-Mar-18, 06:25 PM
Ö the book seems explicit enough: "Ö gravity can act in such a way as to make something feel a force away from an object..."

That's fairly explicit, saying that gravity is the only force required.
No, it isnít.
But this is what you asked for: "I was making a general point that any explanation should show that no outward force is required, to be credible it must show this explicitly." And no "outward" force is necessary, gravity suffices.


ĎGravity cannot act in such a way as to make something feel a force away from an object.í

That would be explicit.
And, as we've shown, wrong. Gravity is not causing the Earth to feel a force away from the moon, sure, but the difference in gravity causes the tides.


O.k. Let me see if I understand you correctly here. You reckon the centripetal acceleration, of a 365.24 day orbit at Sawickiís 3 points O, F and C, is exactly the same Ė there will be no difference in centripetal acceleration at F, C and O.
Is this what you are saying?
I use w^2r to calculate centripetal acceleration and I donít actually get that result, so (bearing in mind your quote of Einsteinís) could you maybe show your calculations, just for points O and F say. That would be most useful.
No problem. If the points F, C and O are not rotating (and we have to do the calculations with this in mind, otherwise we are factoring in an equitorial bulge that has nothing to do with tides), then as they orbit, all three describe a circle with the same radius. Halfway around the orbit, the point F is oriented like O was, so that the radius of both their orbits is the same as the radius of the orbit of O. That's why the w^2r computes to the same value--because they have the same r. And if you follow it through, even the direction is the same.

RichardMB
2004-Mar-19, 05:04 PM
No problem. If the points F, C and O are not rotating (and we have to do the calculations with this in mind, otherwise we are factoring in an equitorial (sic) bulge that has nothing to do with tides), then as they orbit, all three describe a circle with the same radius. Halfway around the orbit, the point F is oriented like O was, so that the radius of both their orbits is the same as the radius of the orbit of O. That's why the w^2r computes to the same value--because they have the same r. And if you follow it through, even the direction is the same.

I am not impressed. This is a very silly idea and I’ve already explained it to you once. Until you can demonstrate a much firmer grip on centripetal acceleration there is really no point in my discussing it with you.

But it’s an interesting subject and I’m keen to explore it further – anybody out there up for some sensible discussion?

milli360
2004-Mar-19, 05:48 PM
I am not impressed. This is a very silly idea and Iíve already explained it to you once. Until you can demonstrate a much firmer grip on centripetal acceleration there is really no point in my discussing it with you.
Here is an illustration (http://mensware.home.mindspring.com/dan/centrif2.jpg) that I drew a couple years ago. This (http://mensware.home.mindspring.com/dan/centrif.jpg) is a larger version of the same.

The red dot is the center of the circle, and both orbit the red X. The circle does not rotate as it orbits, as can be seen from the orientation of the black dot at each step. The center of the black dot's orbit is the black X. The radius of the red dot's orbit is equal to the radius of the black dot's orbit.

Eroica
2004-Mar-21, 11:21 AM
If the points F, C and O are not rotating ... then as they orbit, all three describe a circle with the same radius ... That's why the w^2r computes to the same value--because they have the same r. And if you follow it through, even the direction is the same.
Yes, they are equidistant from the points about which they are revolving, but they are not always equidistant from the Sun's centre of mass. So while their centrifugal accelerations are constant, the gravitational accelerations which the Sun imparts to them vary.

Take point F, for example. Sometimes it is closer to the centre of the Sun, so gravitational acceleration is greater than centrifugal acceleration - hence there is a tidal bulge towards the Sun. At other times it is further from the Sun, so centrifugal acceleration is the greater - hence the tidal bulge away from the Sun.

SeanF
2004-Mar-21, 02:56 PM
If the points F, C and O are not rotating ... then as they orbit, all three describe a circle with the same radius ... That's why the w^2r computes to the same value--because they have the same r. And if you follow it through, even the direction is the same.
Yes, they are equidistant from the points about which they are revolving, but they are not always equidistant from the Sun's centre of mass. So while their centrifugal accelerations are constant, the gravitational accelerations which the Sun imparts to them vary.

Take point F, for example. Sometimes it is closer to the centre of the Sun, so gravitational acceleration is greater than centrifugal acceleration - hence there is a tidal bulge towards the Sun. At other times it is further from the Sun, so centrifugal acceleration is the greater - hence the tidal bulge away from the Sun.

That's true, but if at any single point in time the centrifugal acceleration is the same at all those points and gravitational acceleration is different, then it is still the difference in gravitational acceleration between the points that causes the bulge, ain't it?

And so even if there were zero centrifugal acceleration (not orbiting, just falling), there would still be differences in gravitational acceleration, and the net difference would be the same as when orbiting, so the tidal forces - and thus, the tidal bulges - would be the same.

Eroica
2004-Mar-21, 03:40 PM
And so even if there were zero centrifugal acceleration (not orbiting, just falling), there would still be differences in gravitational acceleration, and the net difference would be the same as when orbiting, so the tidal forces - and thus, the tidal bulges - would be the same.
A point I made back on Oct 19. (http://www.badastronomy.com/phpBB/viewtopic.php?p=156483#156483) 8)

milli360
2004-Mar-21, 11:54 PM
I take it that you are still advocating for the differential orbits explanation, and you think it is similar to RichardMB's.

How does the DO explanation deal with the low tide on the side of the Earth? That question came up earlier (http://www.badastronomy.com/phpBB/viewtopic.php?p=219054#219054).

RichardMB
2004-Mar-22, 10:59 AM
Yes, they are equidistant from the points about which they are revolving, but they are not always equidistant from the Sun's centre of mass. So while their centrifugal accelerations are constant, the gravitational accelerations which the Sun imparts to them vary.

Take point F, for example. Sometimes it is closer to the centre of the Sun, so gravitational acceleration is greater than centrifugal acceleration - hence there is a tidal bulge towards the Sun. At other times it is further from the Sun, so centrifugal acceleration is the greater - hence the tidal bulge away from the Sun.
You’ve taken your eye off the ball. Milli360’s drawing is just a distraction that’s best ignored and a simple experiment will show why.

Remember the tennis ball? Tie a string to it, dunk it in water again and whirl it round your head – what do you see happens to the water? It goes round to the far side, isn’t that right? This is another tidal bulge – a centripetal acceleration tidal bulge.
Now ask yourself what possible difference will it make to this bulge if the ball were to rotate on its axis. Even if you’re unsure about centripetal acceleration you must surely realise that the balls own rotation doesn’t matter squat. The whole idea is just plain daft. So don’t analyse it – put it in the recycle bin.
And don’t wait 2 years.
----------------
In astronomy, things are only as complicated as you care to make them.

milli360
2004-Mar-22, 11:23 AM
Now ask yourself what possible difference will it make to this bulge if the ball were to rotate on its axis. Even if youíre unsure about centripetal acceleration you must surely realise that the balls own rotation doesnít matter squat. The whole idea is just plain daft. So donít analyse it Ė put it in the recycle bin.
The Earth's own rotation causes a bulge, not of just a meter or two like the tidal bulge, but of twenty kilometers. That is how much difference there is between the equatorial radius and the polar radius. The flattening of Jupiter and Saturn is even more pronounced. But that flattening is definitely not a tidal bulge.

In order to study the tidal bulge, you have to ignore that bulge, which is due to the rotation of the planet. As far as the tide is concerned, that flattening bulge is "squat", and you have to remove it from the calculations somehow. You have to treat the case of a non-rotating planet.

Eroica
2004-Mar-22, 12:07 PM
Now ask yourself what possible difference will it make to this bulge if the ball were to rotate on its axis ...
I thought we had all agreed that the rotation of the Earth on its axis is irrelevant to tides? milli360's diagram is of an Earth that is not rotating! (By the way, I was defending you in the post you're now criticizing!)

Eroica
2004-Mar-22, 12:10 PM
I take it that you are still advocating for the differential orbits explanation, and you think it is similar to RichardMB's.
Yes.

How does the DO explanation deal with the low tide on the side of the Earth? That question came up earlier (http://www.badastronomy.com/phpBB/viewtopic.php?p=219054#219054).
Obviously the water that makes the tidal bulges on the near and far side's of the planet must come from somewhere. The tidal depressions are simply left behind when the water is displaced.

milli360
2004-Mar-22, 01:40 PM
How does the DO explanation deal with the low tide on the side of the Earth? That question came up earlier (http://www.badastronomy.com/phpBB/viewtopic.php?p=219054#219054).
Obviously the water that makes the tidal bulges on the near and far side's of the planet must come from somewhere. The tidal depressions are simply left behind when the water is displaced.
That is the intuitive idea, but it's not enough. The data on the accelerometers worldwide, and the more involved theory, show an increase in acceleration, depressing the equipotential surface at the areas of low tide. It doesn't seem that either of you account for that in the explanation.

I am not arguing that there are not alternatives to ways of explaining the tides--different reference frames can be effective. However, it's not as simple as it seems, and the BA's approach is not incorrect.

PS:
Here's a simplified calculation.

At the center of the Earth, the acceleration due to the Sun's mass M is GM/R^2, where R is the distance to the Sun. At the far side of the Earth, the acceleration is GM/(R+r)^2, where r is the radius of the Earth. Subtracting the two gives GM(-2Rr-r^2)/(R^2 (R+r)^2), which is approximately -2GMr/R^3. That's in accord with the usual description of the tide-rasing force acceleration.

At the points of low tide, the distance to the Sun is very nearly the same as at the center of the Earth, GM/R^2. That's the problem I've mentioned. However, the force of the Sun's gravity is not directed parallel to that at the center of the Earth. There are two components, one directed parallel, and one perpendicular, towards the center of the Earth. The right triangle has sides r and R. When you subtract the two, as vectors, then, the difference is r/R times GM/R^2, or GM/R^3. That's half of the tide-raising accleration calculated above--and that is what is found in the accelerometer data.

The equipotential surface at low tide is depressed half as much as it is raised at high tide.

RichardMB
2004-Mar-22, 10:28 PM
I thought we had all agreed that the rotation of the Earth on its axis is irrelevant to tides? milli360's diagram is of an Earth that is not rotating! (By the way, I was defending you in the post you're now criticizing!)

Please accept my apology’s, there was no criticism intended, none at all.
I just didn’t want you distracted by milli360’s diagram because it doesn’t correctly represent centripetal acceleration as it applies to tidal bulges.

You can easily change the diagram so that it does though. Simply draw two circles (centred on the red cross) a big one of radius 1AU+R (R=earth radius) and a small one of radius 1AU-R. Label the big circle F and the small circle C.

Circle F is the circle through which all the outer high tides pass, whilst circle C is the one through which all the inner high tides pass – they are, of course, coaxial.

For a given angular velocity, the centripetal acceleration for path F is greater than for path C. From which it then follows that the centripetal acceleration at the outer bulge is constant and larger than the small constant centripetal acceleration at the inner bulge.

To me this is so obvious but I’ve learned that it won’t be obvious to everyone.

Does it make sense to you or do you have problems with it?

milli360
2004-Mar-23, 12:54 PM
For a given angular velocity, the centripetal acceleration for path F is greater than for path C. From which it then follows that the centripetal acceleration at the outer bulge is constant and larger than the small constant centripetal acceleration at the inner bulge.

To me this is so obvious but Iíve learned that it wonít be obvious to everyone.
I don't think anyone is arguing against the centripetal force calculation, but the question is whether it is relevant to actually computing the magnitude of the tides is another question. For instance, it doesn't seem to help much with the low tide--without introducing additional complications. The BA's explanation is a bit simpler.

RichardMB
2004-Mar-24, 10:10 PM
I don't think anyone is arguing against the centripetal force calculation, but the question is whether it is relevant to actually computing the magnitude of the tides is another question. For instance, it doesn't seem to help much with the low tide--without introducing additional complications. The BA's explanation is a bit simpler.

You can ignore centripetal acceleration, as Sawicki & the BA have done, this gives you the simple, but incorrect, explanation that seems to be your preferred option.
For a correct explanation however, you need to include centripetal acceleration, and for my money it’s also the simplest.

milli360
2004-Mar-25, 07:52 AM
You can ignore centripetal acceleration, as Sawicki & the BA have done, this gives you the simple, but incorrect, explanation that seems to be your preferred option.
No, it's correct.


For a correct explanation however, you need to include centripetal acceleration, and for my money itís also the simplest.
That's just a backdoor way of including centrifictional force, and unnecessary.

milli360
2004-Mar-25, 04:41 PM
Just to follow up one of my previous posts (http://www.badastronomy.com/phpBB/viewtopic.php?p=219054#219054). There is a derivation of the tidal potential that is presented in Stacey's Physics of the Earth (p115). It includes a once-per-revolution (once per year) rotation of the Earth, so it should satisfy your objections. After making the calculations and simplifying the algebra, Stacey comes up with a formula for the potential that includes three separate terms. The third term is - 1/2 w^2 r^2 sin^2 theta, where w is the rotation, r is the radius, and theta is the colatitude. It's the rotational potential due to the rotation, and it would contribute to the rotational potential if the rotation were much larger (once per day). If the problem were set up without that rotation, the algebra might be more difficult, but the answer would be the same, except this term would disappear. The first term is a constant, independent of position on the Earth, and, as Stacey says, "has no tidal effect." It's the gravitational potential at the center of the Earth due to the moon, with a small correction. The second term is the tidal potential, and is -Gmr^2/R^3 times (3/2cos^2 xi - 1/2), where xi is the angle off the line between the moon and Earth. That last factor is just a second order zonal harmonic, a prolate ellipsoid pointed at the moon.

The gradient of that tidal potential is the tidal force, and it is the same as that computed by the method of differential gravity that we've used above. It shows the depression of the low tide that seems to be missing from the differential orbit method.

So, as I've said, it is possible to include all the terms in a system which includes a rotating earth and come up with the right answer, but you have to be careful--for instance, you have to identify, and then ignore, the bulge induced by that rotation.

RichardMB
2004-Mar-26, 06:12 PM
There is a derivation of the tidal potential that is presented in Stacey's Physics of the Earth (p115). It includes a once-per-revolution (once per year) rotation of the Earth, so it should satisfy your objections...
...So, as I've said, it is possible to include all the terms in a system which includes a rotating earth and come up with the right answer, but you have to be careful--for instance, you have to identify, and then ignore, the bulge induced by that rotation.

Exactly – and haven’t I been telling you that for weeks? Earth’s axial rotation is entirely irrelevant (I actually said it doesn’t matter squat”) so Stacey factors it out – brilliant. Now do you believe me.

My problem however isn’t with Stacey’s Physics of the Earth, it’s with Phil Plait’s Bad Astronomy, specifically this idea that the far tidal bulge is caused by differential gravity (attributed to Mik Sawicki). It isn’t – that’s Bad Physics.
The far bulge would exist even if you removed the sun and its gravity completely, so long as earth’s orbital path was maintained. In fact the far bulge would then be an awful lot bigger. Putting the sun and its gravity back would actually depress the far bulge – exactly the opposite of what Sawicki suggests.
Differential gravity can produce planet rupturing stresses, no doubt about that, but what differential gravity cannot do is induce any kind if push force. On it’s own solar gravity would pull all the oceans, and even the atmosphere, to the near side. To raise the far tide you need (and it is not optional) the centripetal acceleration of earth’s solar orbit.

SeanF
2004-Mar-26, 07:18 PM
The far bulge would exist even if you removed the sun and its gravity completely, so long as earthís orbital path was maintained.

That would depend on how the orbital path is being maintained in the absence of gravity, wouldn't it? Shooting a ball around on the inside of a circular track would cause it to flatten against the track, not stretch out . . .


To raise the far tide you need (and it is not optional) the centripetal acceleration of earthís solar orbit.

Do you contend, then, that an object falling straight into a black hole would not be torn apart by tidal forces?

milli360
2004-Mar-26, 08:23 PM
Exactly Ė and havenít I been telling you that for weeks? Earthís axial rotation is entirely irrelevant (I actually said it doesnít matter squatĒ) so Stacey factors it out Ė brilliant. Now do you believe me.
Not the issue. Back in October (http://www.badastronomy.com/phpBB/viewtopic.php?p=156844#156844), I said there is nothing wrong with the DO model per se, you have to be careful with it and apply it properly. And that doesn't make the BA's model not as good. Two weeks ago (http://www.badastronomy.com/phpBB/viewtopic.php?p=219054#219054), I mentioned a treatment that I was partial to, and that's the one of Stacey's. Even it is still not complete though.

And the interesting thing about it is that the centripetal component is also factored out--it's irrelevant as well.


The far bulge would exist even if you removed the sun and its gravity completely, so long as earthís orbital path was maintained. In fact the far bulge would then be an awful lot bigger. Putting the sun and its gravity back would actually depress the far bulge Ė exactly the opposite of what Sawicki suggests.
But then there would be no near bulge, right? It'd even be a "low", right? That's why it can be said to have nothing to do with the tides.


Differential gravity can produce planet rupturing stresses, no doubt about that, but what differential gravity cannot do is induce any kind if push force. On itís own solar gravity would pull all the oceans, and even the atmosphere, to the near side. To raise the far tide you need (and it is not optional) the centripetal acceleration of earthís solar orbit.
No, as SeanF points out, if the Earth is in freefall (and it is), there will be tides. Whether it is falling straight into the Sun or not.

RichardMB
2004-Mar-27, 10:47 AM
Do you contend, then, that an object falling straight into a black hole would not be torn apart by tidal forces?
That is not my contention. This is what I actually said:


Differential gravity can produce planet rupturing stresses, no doubt about that…

So, yes, an object falling straight into a black hole would be torn apart. This is a valid application of differential gravity, which is not a problem. The earth however is not falling straight – it is following a curved path which requires centripetal acceleration, this is provided by solar gravitation and must be properly accounted for. I am not saying that differential gravity does not exist I am saying that it does not raise the far tide (nor does it raise the near tide side, that is primarily due to direct solar gravity).


That would depend on how the orbital path is being maintained in the absence of gravity, wouldn't it? Shooting a ball around on the inside of a circular track would cause it to flatten against the track, not stretch out . . .

Centripetal acceleration, w^2r, depends only on w (essentially the period) and r (the radius). If your scheme includes some kind of mechanical constraint then, yes, you can defeat the consequential bulge, but it will then manifest itself as some kind of distortion - flattening as you correctly surmise. Think of a 1AU tether instead or a (truly) massive bank of rocket motors.


And the interesting thing about it is that the centripetal component is also factored out

Which centripetal component?


But then there would be no near bulge, right?

Right. The near bulge is produced by direct solar gravity (see answer to SeanF, above), so if we remove the sun and its gravity we loose the near bulge. This is the point of the two tennis ball experiments.



No, as SeanF points out, if the Earth is in freefall (and it is), there will be tides. Whether it is falling straight into the Sun or not.

SeanF has not pointed this out to me – but I’m sure he will if he wants to. As for ‘freefall’ I really don’t know what it means. So far as I’m concerned earth is in orbit.

What I have not yet done is combine these two bulges. I don’t want to give the idea that you can simply stick these two bulges on the earth and get the high tides. It’s more complicated than that – but only a bit.

milli360
2004-Mar-27, 02:05 PM
What I have not yet done is combine these two bulges. I donít want to give the idea that you can simply stick these two bulges on the earth and get the high tides. Itís more complicated than that Ė but only a bit.
And what do you do for the low tides?

The point is, once you're done, you'll get the same answer that the BA gets--and his method is valid. It is not incorrect, as you claim.

SeanF
2004-Mar-27, 03:38 PM
That would depend on how the orbital path is being maintained in the absence of gravity, wouldn't it? Shooting a ball around on the inside of a circular track would cause it to flatten against the track, not stretch out . . .

Centripetal acceleration, w^2r, depends only on w (essentially the period) and r (the radius). If your scheme includes some kind of mechanical constraint then, yes, you can defeat the consequential bulge, but it will then manifest itself as some kind of distortion - flattening as you correctly surmise. Think of a 1AU tether instead or a (truly) massive bank of rocket motors.

A bank of rocket motors would have to go on the outside of the body pushing in, and would be the same as the track - no farside bulge. A tether only exerts its centripetel force on a single point on the body, which means it's not comparable to a gravitational orbit - the effects on the body would be different.



No, as SeanF points out, if the Earth is in freefall (and it is), there will be tides. Whether it is falling straight into the Sun or not.

SeanF has not pointed this out to me Ė but Iím sure he will if he wants to. As for Ďfreefallí I really donít know what it means. So far as Iím concerned earth is in orbit.

Orbit is freefall, and the tides are the same either way.

RichardMB
2004-Mar-28, 09:06 PM
And what do you do for the low tides?
I will maybe get to the low tides in the fullness of time – but first things first. I’m not proposing a theory, I’m critiquing Mik Sawicki ’s.


The point is, once you're done, you'll get the same answer that the BA gets--and his method is valid. It is not incorrect, as you claim.

I’ve already pointed out some of the Bad Mathematics in the Sawicki paper so it’s unlikely I’d get the same answers - and any claims to validity are just a tad premature.


A bank of rocket motors would have to go on the outside of the body pushing in, and would be the same as the track - no farside bulge. A tether only exerts its centripetel (sic) force on a single point on the body, which means it's not comparable to a gravitational orbit - the effects on the body would be different.
Orbit is freefall, and the tides are the same either way.

How truly bizarre.

Anyway, if anyone else out there has genuine concerns about asymmetric tides, push gravity and suchlike and you’d appreciate a nudge in the right direction I’d be glad to help – for a while anyway.

SeanF
2004-Mar-29, 02:57 PM
A bank of rocket motors would have to go on the outside of the body pushing in, and would be the same as the track - no farside bulge. A tether only exerts its centripetel (sic) force on a single point on the body, which means it's not comparable to a gravitational orbit - the effects on the body would be different.
Orbit is freefall, and the tides are the same either way.
How truly bizarre.

Oh, please. Let's not be nitpicking spelling errors now, shall we?

From the NASA website (http://quest.arc.nasa.gov/space/teachers/liftoff/basics.html):


The Space Shuttle orbiter falls in a circular path about Earth. Because the orbiter, astronauts, and all the contents of the orbiter (food, tools, cameras, etc.) are falling together, they seem to float in relation to each other. This is comparable to the imaginary situation that would take place if the cables supporting a very high elevator would break, causing the car and its passengers to fall to the ground. (In such an example, we have to discount the effects of air friction on the falling car.) Since the motion of the falling car and the passengers are relative to each other, the people inside seem to float.
...
The floating effect of Space Shuttles and astronauts in orbit is called by many names. It is referred to as "freefall," "weightlessness," "zero-G" (zero-gravity), or "microgravity." Space researchers prefer to use the term "microgravity" because it better represents the actual conditions of Earth orbit. Thus, even though freefall simulates the absence of gravity, very small (micro) gravitational forces are still detectable.

"Orbit" and "freefall" are the same thing.


Anyway, if anyone else out there has genuine concerns about asymmetric tides, push gravity and suchlike and youíd appreciate a nudge in the right direction Iíd be glad to help Ė for a while anyway.

In other words, you only want to talk to people who agree with you. Bizarre, indeed.

RichardMB
2004-Mar-30, 04:50 PM
In other words, you only want to talk to people who agree with you. Bizarre, indeed.

You mean like milli360??? We’ve been going at it for nearly a month and he doesn’t agree with a thing I say. Don’t talk such nonsense.
Certainly I’d prefer to talk to people with a genuine concern about the Sawicki paper, but you don’t have to agree with me. If you don’t agree let’s have a well-tempered debate, but one thing I’m not up for is surly argument – as you discovered.

If you’re interested in my viewpoint by all means question me – you are assured of a polite response. Meantime let me ask you a question.



NASA wrote:
The Space Shuttle orbiter falls in a circular path about Earth. Because the orbiter, astronauts, and all the contents of the orbiter (food, tools, cameras, etc.) are falling together, they seem to float in relation to each other. This is comparable to the imaginary situation that would take place if the cables supporting a very high elevator would break, causing the car and its passengers to fall to the ground. (In such an example, we have to discount the effects of air friction on the falling car.) Since the motion of the falling car and the passengers are relative to each other, the people inside seem to float.


"Orbit" and "freefall" are the same thing.

The implication of what you say here is that NASA’s elevator passengers, falling towards the ground, are actually in orbit - they’re in ‘freefall’, orbit and freefall are the same thing, ipso facto they’re in ‘orbit’.
Is this the meaning that you intended to convey?

SeanF
2004-Mar-30, 05:14 PM
You mean like milli360??? Weíve been going at it for nearly a month and he doesnít agree with a thing I say.

Milli360 doesn't agree with anything anybody says. Watch, he'll probably disagree with this! ;)


If you donít agree letís have a well-tempered debate, but one thing Iím not up for is surly argument Ė as you discovered.

If youíre interested in my viewpoint by all means question me Ė you are assured of a polite response.

And you feel that "(sic)"ing a spelling error and making statements like "How truly bizarre" are a good way to do this?


Meantime let me ask you a question.


"Orbit" and "freefall" are the same thing.

The implication of what you say here is that NASAís elevator passengers, falling towards the ground, are actually in orbit - theyíre in Ďfreefallí, orbit and freefall are the same thing, ipso facto theyíre in Ďorbití.
Is this the meaning that you intended to convey?

Okay, my statement was too general when taken out of context - but are you deliberately being obtuse? No, of course not all freefalls are orbits*, but still . . . orbit is freefall. You're arguing that a square is not a rectangle - don't take a disagreement to imply that all rectangles are squares.

*One of the definitions of orbit (http://dictionary.reference.com/search?q=orbit) is "The path of a body in a field of force surrounding another body." If I wanted to nitpick, I could point to this definition and say that, yes, the falling elevator does qualify - but that would be just as irrelevant to the subject, anyway.

milli360
2004-Mar-30, 05:49 PM
The point is, once you're done, you'll get the same answer that the BA gets--and his method is valid. It is not incorrect, as you claim.

Iíve already pointed out some of the Bad Mathematics in the Sawicki paper so itís unlikely Iíd get the same answers - and any claims to validity are just a tad premature.
I went back through all your posts to this thread, and I could only find two where you do actual calculations, this one (http://www.badastronomy.com/phpBB/viewtopic.php?p=214954#214954) and this too (http://www.badastronomy.com/phpBB/viewtopic.php?p=219435#219435). Were there others that I missed? In the first one, you mistakenly try to apply the tidal calculations to a dripping tennis ball (the drip is not a tidal bulge)--and even then, you make a correction later (http://www.badastronomy.com/phpBB/viewtopic.php?p=215489#215489). The second one was also a mistake on your part, you misunderstood what angle (http://www.badastronomy.com/phpBB/viewtopic.php?p=219570#219570) was being used in the sine calculation.

Milli360 doesn't agree with anything anybody says. Watch, he'll probably disagree with this!
Hah, I once agreed with Beskeptical (http://www.badastronomy.com/phpBB/viewtopic.php?p=40586#40586), even. And there were others.


*One of the definitions of orbit (http://dictionary.reference.com/search?q=orbit) is "The path of a body in a field of force surrounding another body." If I wanted to nitpick, I could point to this definition and say that, yes, the falling elevator does qualify - but that would be just as irrelevant to the subject, anyway.
No, you're on solid ground here. Excepting atmospheric and lithospheric friction, it will become an orbit. Even falling straight down, if you could drill a path through the Earth, you'd end up in an oscillating motion not unlike an orbit--in fact, the period of the orbit would match that of the ellipse of the same semimajor axis.

SeanF
2004-Mar-30, 09:31 PM
Milli360 doesn't agree with anything anybody says. Watch, he'll probably disagree with this!
Hah, I once agreed with Beskeptical (http://www.badastronomy.com/phpBB/viewtopic.php?p=40586#40586), even. And there were others.
Told ya! ;)


*One of the definitions of orbit (http://dictionary.reference.com/search?q=orbit) is "The path of a body in a field of force surrounding another body." If I wanted to nitpick, I could point to this definition and say that, yes, the falling elevator does qualify - but that would be just as irrelevant to the subject, anyway.
No, you're on solid ground here. Excepting atmospheric and lithospheric friction, it will become an orbit. Even falling straight down, if you could drill a path through the Earth, you'd end up in an oscillating motion not unlike an orbit--in fact, the period of the orbit would match that of the ellipse of the same semimajor axis.

"On solid ground." Good one. :)

I never thought about it that way (although we have discussed the "oscillating through the hole" thing on this board before), but you're right - falling really can be considered in orbit. It's just an orbit that happens to intersect the surface at a couple points.

JohnOwens
2004-Mar-31, 07:34 AM
*One of the definitions of orbit (http://dictionary.reference.com/search?q=orbit) is "The path of a body in a field of force surrounding another body." If I wanted to nitpick, I could point to this definition and say that, yes, the falling elevator does qualify - but that would be just as irrelevant to the subject, anyway.
No, you're on solid ground here. Excepting atmospheric and lithospheric friction, it will become an orbit. Even falling straight down, if you could drill a path through the Earth, you'd end up in an oscillating motion not unlike an orbit--in fact, the period of the orbit would match that of the ellipse of the same semimajor axis.
Well, not quite. After all, once you're under the surface of the Earth, the gravitational pull/curvature approaches zero as you approach the center. That's going to complicate things, so it's not going to be as simple as a regular ellipse. Another thing I've wanted to calculate but not gotten around to (as long as you let me assume uniform density!)....
But the period will almost certainly be different, assuming there isn't some cancelled term that stretches the perigee as much as it slows down the period.

freddo
2004-Mar-31, 07:53 AM
as long as you let me assume uniform density
We can't let you do that John, because it's the fact the density of the Earth is not uniform that means gravity actually increases as you approach the core. (http://www.badastronomy.com/phpBB/viewtopic.php?p=72062#72062)

As far as I understand it though, unform density or not, as you approach the centre you will be undergoing acceleration, and leaving the centre you will be decellerating. Calculating the actual rate of these is tricky though.

milli360
2004-Mar-31, 08:34 AM
That's going to complicate things, so it's not going to be as simple as a regular ellipse. Another thing I've wanted to calculate but not gotten around to (as long as you let me assume uniform density!)....
But the period will almost certainly be different, assuming there isn't some cancelled term that stretches the perigee as much as it slows down the period.
You may be familiar with the formula (http://scienceworld.wolfram.com/physics/GeostationaryOrbit.html) for period of an orbit about a mass, which is sqrt( (2pi)^2 r^3 /(GM) ). Using r=6378km, the radius of the Earth, M=5.976x10^24kg, G=6.672x10^-11 N m^2 /kg^2, and pi=3.1415926535, we get T=84.47 minutes for an orbit right at the surface of the Earth.

For the hole drilled through the center of the Earth, using that model of uniform density, the motion becomes simple harmonic motion (http://scienceworld.wolfram.com/physics/SimpleHarmonicOscillator.html), since the acceleration is just gx/r, g is just GM/r^2. k/m is g/r, and omega zero is sqrt( g/r ). The period is 2 pi divided by omega zero: sqrt( (2pi)^2 r^3 /(GM) ), which is the same formula used above.


as long as you let me assume uniform density
We can't let you do that John, because it's the fact the density of the Earth is not uniform that means gravity actually increases as you approach the core. (http://www.badastronomy.com/phpBB/viewtopic.php?p=72062#72062)
Which is another problem with that "orbit". But it's close enough, and the point has been made that freefall has many characteristics of an orbit.

I'd just like to quote a snippet from that link, which was a post of tracer's:

I gotta agree with kilopi on this one.
Notice, we were both disagreeing with JS Princeton--but with good reason!

RichardMB
2004-Apr-01, 11:16 AM
Were there others that I missed?

Yes. You appear to have missed this one (same link):


Sawicki uses equ.3 to calculate Da(s) on the far side, that is point F at a distance d(s)+R (1AU+radius of earth). Using my data it comes to 5.0476e-07 ms^-2. On the near side, point C, the distance is d(s)-R (1AU-earth radius) and I make this –5.0483e-07 ms^-2. According to Sawicki these are the same.

By claiming that they are the same Sawicki
(a) makes Da(s) general which is wrong, it applies only at point F and
(b) incorrectly locates the differential gravity mean at the centre of the earth, which is again wrong.

These errors combine to mask a tidal force asymmetry in Sawicki’s model which reaches 6% in the lunar tide calculation (Da(m) equ.4).

If you’re happy with these asymmetries that’s fine, but if anyone wants to know how to get rid of them give me a shout.

milli360
2004-Apr-01, 02:06 PM
You appear to have missed this one (same link):
I answered it before (http://www.badastronomy.com/phpBB/viewtopic.php?p=219570#219570) at the same time I answered the rest of the post at that link.

According to Sawicki these are the same.
No, that's not true, as I said before. The second sentence after his equation three says "Note that in the second step in Eq.(3) we neglected corrections from higher order R/ds terms."

If that is your last objection to Sawicki's math, then I have answered all of your objections. You can no longer say (http://www.badastronomy.com/phpBB/viewtopic.php?p=232825#232825) you've "already pointed out some of the Bad Mathematics in the Sawicki paper."

if anyone wants to know how to get rid of them give me a shout.
Go ahead and post it. Let's have a look at it.

RichardMB
2004-Apr-03, 02:44 PM
Go ahead and post it. Let's have a look at it.
Certainly.

The net acceleration at the far (F), centre (O) and close (C) points on earth in the Earth – Moon system is:

a(n) = w^2D(b) – GM/D(m)^2

where D(b) is distance to the barycentre, M is mass of the moon, and D(m) is distance to the moon.

Data: mean distance earth-moon = 3.844e08 m
lunar sidereal period = 2360591.5 sec
mass of moon = 7.353e22 kg
radius of earth = 6.378e06 m
earth centre to barycentre = 4.6867e06 m

Using the above data and equation for net acceleration gives:

at F: 4.626035e-05 ms^-2
at O: 0 ms^-2
at C: -4.63163e-05 ms^-2

As can be seen, the net acceleration at the centre of the earth is zero, as required, and at the far and close points the net acceleration is equal to 0.1%, well within the uncertainty in the data.

Using Sawicki’s equ.4 I get 1.0745e-06 and 1.1299e-06 – a difference of 5% (ouch!).

What this shows is that simple mechanics predicts a symmetrical tide to with 0.1% or less.
Sawicki’s ‘differential’ scheme, which ignores centripetal acceleration, predicts a tidal asymmetry of 5%.

You choose.

If anybody would like clarification on any of this I’ll be very glad to help.
If anybody wants to argue with it please do me the courtesy of first checking the maths yourself and telling me your results.

milli360
2004-Apr-03, 05:48 PM
The net acceleration at the far (F), centre (O) and close (C) points on earth in the Earth Ė Moon system is:

a(n) = w^2D(b) Ė GM/D(m)^2
That first term on the right hand side is the centrifictional force (http://www.badastronomy.com/phpBB/viewtopic.php?p=29784#29784) term.

That is similar to how Stacey starts his analysis, except he uses potential instead of force. I was checking out the Wolfram pages on tides (http://scienceworld.wolfram.com/physics/Tide.html), and actually found Stacey's diagram there.


Using the above data and equation for net acceleration gives:

at F: 4.626035e-05 ms^-2
at O: 0 ms^-2
at C: -4.63163e-05 ms^-2

As can be seen, the net acceleration at the centre of the earth is zero, as required, and at the far and close points the net acceleration is equal to 0.1%, well within the uncertainty in the data.

Using Sawickiís equ.4 I get 1.0745e-06 and 1.1299e-06 Ė a difference of 5% (ouch!).
Your first clue should have been the fact that your answers are fifty times bigger than those (which, BTW, don't seem to come from Sawicki's equation four at that link). You still have the once-per-month rotational component in your answers, which gives an extremely large bulge all the way around the Earth equator. That swamps the tidal component--which is why your percentage is fifty times less.

It also misrepresents the actual size of the tide--that centrifictional bulge appears at all points along the equator, and does not act as a tide.

You choose.
I'll choose Stacey over Sawicki over yours. But, as I've said before, if you're careful, any method can arrive at the same values. For your case you have to subtract the rotation of the Earth, w^2D, where w=(2pi)/month and D is the radius of the Earth. That value is ((2*3.14159)/(2360591.5 sec )^2 x 6.378e06 m, or 4.5186e-5 m/s/s. If you subtract that from your two answers, at F: 4.626035e-05 ms^-2, and at C: -4.63163e-05 ms^-2, you get at F: 1.07435e-06 and at C: -1.13030e-06, essentially the same as what you say Sawicki would get--without as much work.

In other words, the difference between your two answers is the same as the difference between the other two values, but you increased yours by a large factor that is not relevant to the calculation of the tide.

RichardMB
2004-Apr-05, 03:30 PM
You still have the once-per-month rotational component in your answers, which gives an extremely large bulge all the way around the Earth equator. That swamps the tidal component--which is why your percentage is fifty times less.

It also misrepresents the actual size of the tide--that centrifictional bulge appears at all points along the equator, and does not act as a tide.

You seem to be saying (and this is not contentious) that the (monthly rotation) centripetal acceleration is uniform at all point around the equator so it should be factored out, whereas I say that this centripetal acceleration, because it is centred on the barycentre is lopsided so it should remain in. That would be fair wouldn’t it?
We're still not in agreement about centripetal acceleration are we?
O.k. then, some time ago you sent this diagram http://mensware.home.mindspring.com/dan/centrif.jpg with this explanation:

The red dot is the center of the circle, and both orbit the red X. The circle does not rotate as it orbits, as can be seen from the orientation of the black dot at each step. The center of the black dot's orbit is the black X. The radius of the red dot's orbit is equal to the radius of the black dot's orbit.
Just so I understand what the diagram means, can you perhaps elaborate a bit, for instance how does it show that centripetal acceleration is the same at all points and maybe how this is affected by an eccentric orbit – such as the earth around the barycentre?

milli360
2004-Apr-05, 03:47 PM
You seem to be saying (and this is not contentious) that the (monthly rotation) centripetal acceleration is uniform at all point around the equator so it should be factored out, whereas I say that this centripetal acceleration, because it is centred on the barycentre is lopsided so it should remain in. That would be fair wouldnít it?
My point is that it doesn't matter how you calculate it, it turns out the same, if you do it right. And the answer agrees with Sawicki. There is nothing fundamentally wrong with his approach.


We're still not in agreement about centripetal acceleration are we?

I'm not sure.


O.k. then, some time ago you sent this diagram http://mensware.home.mindspring.com/dan/centrif.jpg with this explanation:

The red dot is the center of the circle, and both orbit the red X. The circle does not rotate as it orbits, as can be seen from the orientation of the black dot at each step. The center of the black dot's orbit is the black X. The radius of the red dot's orbit is equal to the radius of the black dot's orbit.
Just so I understand what the diagram means, can you perhaps elaborate a bit, for instance how does it show that centripetal acceleration is the same at all points and maybe how this is affected by an eccentric orbit Ė such as the earth around the barycentre?
The radius and the period of the black dot's path is the same as the radius and period of the red dot's path. That means the magnitude is the same. The direction is the same also, since the black dot is offset from the red dot in the same direction as their respective centers. And it doesn't matter which point on the sphere you choose, the black dot could be anywhere.

The same would be true if the diagram had the red center inside the red circle, the orbits would be smaller than the red circle is all.

RichardMB
2004-Apr-06, 04:06 PM
The radius and the period of the black dot's path is the same as the radius and period of the red dot's path. That means the magnitude is the same. The direction is the same also, since the black dot is offset from the red dot in the same direction as their respective centers. And it doesn't matter which point on the sphere you choose, the black dot could be anywhere.

It does seem very persuasive but it still gives me trouble.

See, I can’t decide from your diagram how an ocean in would be affected. Suppose that the (thick) red circle was an ocean, would the acceleration have any affect at all on such an ocean? Would it cause any kind of bulge?
Your diagram shows a red and a black arrow, these aren’t labelled so I’m not sure what they mean but they do suggest a force pointing outwards, wouldn’t this create a bulge on the outside?

I’m sorry if I appear thick but I’m trying to get a firm grip on how you understand centripetal acceleration – I’m sure as hell not there yet.

milli360
2004-Apr-06, 04:50 PM
See, I canít decide from your diagram how an ocean in would be affected. Suppose that the (thick) red circle was an ocean, would the acceleration have any affect at all on such an ocean? Would it cause any kind of bulge?
Since there is no difference from side to side, no. Some other force would have to be involved. It's like pushing on the back of a car--the occupants are pressed against the seats, or they might even tumble out of the car if they weren't restrained. If you could somehow affect the car and occupants equally, they wouldn't feel any pressure against the seats, nor would they tend to tumble out. That'd be akin to dropping the car over a cliff--gravity "pulls" equally on all and they don't feel a pressure from the car, they're in free fall. There would be a tidal force, but on the scale of a car, that force is tiny.


Your diagram shows a red and a black arrow, these arenít labelled so Iím not sure what they mean but they do suggest a force pointing outwards, wouldnít this create a bulge on the outside?
The arrows represent the centrifictional force. Reverse them to represent the centripetal force, if you like. But every point on the ball experiences the same force. Since there is no differential, there is no tidal force.


Iím sorry if I appear thick but Iím trying to get a firm grip on how you understand centripetal acceleration Ė Iím sure as hell not there yet.
I'm still struggling with it myself.

RichardMB
2004-Apr-07, 02:15 PM
RichardMB wrote:
Suppose that the (thick) red circle was an ocean, would the acceleration have any affect at all on such an ocean? Would it cause any kind of bulge?

Since there is no difference from side to side, no.

I’m going to agree with you that the centripetal acceleration at the red dot is the same as at the black dot, and I’ll agree that the centripetal acceleration at all points on, and in, the ball is exactly the same, both in magnitude and in direction.

But this does not mean that there would be no bulge. Let me give you three examples:

1) On the left hand circle an observer, at the black dot, releases a tennis balls from 1m above the ground. He notes the balls path and observes that it falls slowly to the ground. This is because the ball moves in a straight line from the release point (we’re ignoring gravity) whilst the ground moves to the right to intercept it.
At the same time another observer repeats the experiment on the far (outer) side, but he sees the ball slowly rise, gently curving back as it climbs into the sky, this is again because the ball moves off in a straight line as the ground moves to the right, but now ‘to the right’ is away from the ball.
In place of the ball we can imagine water, if sufficiently fluid it would flow round to wherever the ball would rise, to the outside of the diagram – it would form a bulge.

2) at a more parochial level we might consider your diagram to be a nice cup of tea. The red circle is the cup and the black dot is the handle. I’m sat here wondering if I should tell you what’s going to happen to the tea when you move the cup in the manner indicated by your drawing – why don’t you tell me?

3) and let’s not forget the wet tennis ball whirling round on a string – there’s that pesky bulge again.

Even though the centripetal acceleration is everywhere the same, nonetheless all three, quite different, experiments clearly show a bulge.

milli360
2004-Apr-07, 03:44 PM
Even though the centripetal acceleration is everywhere the same, nonetheless all three, quite different, experiments clearly show a bulge.
Calculate the magnitude of the tidal bulge using that method. You'll find that you have the same problems as before, and the answers will end up being close to Sawicki's, just as before.

SeanF
2004-Apr-07, 03:52 PM
Even though the centripetal acceleration is everywhere the same, nonetheless all three, quite different, experiments clearly show a bulge.

They're not all quite as different as you seem to think, but they are certainly different than a gravitational orbit. :)

Numbers 1 and 2, for example, suffer from the same flaw I pointed out before: to wit, the gravity of the Moon (and/or the Sun) affects the entire Earth.

In your "soaked tennis ball" example (analogy 3), the tether holds on to the ball, but not the water. This, allows the water to move "away" from the ball. Gravity would be holding on to the ball and the water.

In the "released ball" example (analogy 1), you have the red circle and the released ball affected differently - if it's gravity (by the X's) that causes the red circle to "move to the right," then that gravity will also cause the released ball to "move to the right," won't it? You have to consider that.

So those two examples would be valid if gravity were keeping the Earth moving in a circular path but not the oceans . . . and that's not right.

I'm still thinking about analogy 2, with the cup of tea. :)

RichardMB
2004-Apr-08, 01:04 PM
I'm still thinking about analogy 2, with the cup of tea. :)

Might I suggest you start by looking more closely at milli360’s diagram.
It purports to show the case of a single rotation, with all irrelevant rotations removed.
What it actually shows is a single translation.
Look at any pair of red circles. You can get from one to the other by a single reflection, a single rotation or a single translation.
Once you add the black dot, however, you can only get there by a single translation. Reflection doesn’t work at all.
You can get there by rotation but it takes two, not one.
Does that surprise you?

The two rotations are:
a) a rotation about the red cross that takes all points on the circle, including the black dot, through the same angle and
b) a rotation about the red dot that takes all points on the circle through the opposite angle, restoring the original angle of all points on the circle, including the black dot.

Both rotations require their own centripetal accelerations (CA).
The CA due to rotation b) is symmetrical about the red dot, so it gives rise to an equatorial bulge which can pretty much be ignored.
The CA due to rotation a) is symmetrical about the red cross, so it gives rise to an outer tidal bulge which cannot be ignored.

It’s the CA due to rotation a) that is causing the tea tide.
Can you now see that it also causes the earth’s outer tidal bulge? – they are exactly the same thing.

You’ll notice that I haven’t mentioned gravity. That’s because it isn’t necessary, Earths outer tide is caused by rotation b) not by gravity – differential or otherwise.

SeanF
2004-Apr-08, 01:37 PM
I'm still thinking about analogy 2, with the cup of tea. :)

Might I suggest you start by looking more closely at milli360ís diagram.
It purports to show the case of a single rotation, with all irrelevant rotations removed.
What it actually shows is a single translation.

No, it shows a single revolution, with no rotation at all. The red circle is moving in a circular path, but it is not rotating - at all.


Youíll notice that I havenít mentioned gravity. Thatís because it isnít necessary, Earths outer tide is caused by rotation b) not by gravity Ė differential or otherwise.

But your "rotation" is itself caused by gravity, so you can't have the rotation if you're ignoring the gravity.

That's why your analogy with the ball on the string doesn't really work. You're creating a circular path using a non-gravitational acceleration, but there's no reason to expect that to give the same results as a circular path resulting from gravitational acceleration. After all, your string-ball results in less water on the "inward" side of the ball, but we actually have a high tide on the "inward" side of the Earth. It's not the same thing.

milli360
2004-Apr-08, 02:49 PM
Might I suggest you start by looking more closely at milli360ís diagram.
It purports to show the case of a single rotation, with all irrelevant rotations removed.
A single revolution, with no rotation.


The two rotations are:
a) a rotation about the red cross that takes all points on the circle, including the black dot, through the same angle and
b) a rotation about the red dot that takes all points on the circle through the opposite angle, restoring the original angle of all points on the circle, including the black dot.
Those two rotations "undo" each other.

Take a block that moves from (-1,0) to (1,0). The step can be accomplished by sliding it along the x-axis through the origin, or it can be turned 180 degrees through a circle with radius 1 about the origin, and then rotated 180 degrees once it is at (1,0). Both result in the same finish, and neither is "wrong." Similarly, your sort of analysis would, if you did it correctly, come up with the same answer as Sawicki.

RichardMB
2004-Apr-09, 01:14 PM
No, it shows a single revolution

Yes, of course it does Sean.
Now, why don’t you explain to me how you think the teacup thing works?


Those two rotations "undo" each other.

No they don’t. They would if they had the same centre – surely you realise this much.


Take a block that moves from (-1,0) to (1,0). The step can be accomplished by sliding it along the x-axis through the origin, or it can be turned 180 degrees through a circle with radius 1 about the origin, and then rotated 180 degrees once it is at (1,0). Both result in the same finish, and neither is "wrong."

Excellent, something we can agree on (who said anything about ‘wrong’?).


Take a block that moves from (-1,0) to (1,0). The step can be accomplished by sliding it along the x-axis through the origin

Yes, that’s what I said.
Notice that this does not avoid the outer tidal bulge.
As you accelerate from (-1,0) you get an outward facing bulge to the left then, as you decelerate towards (1,0) you get an outward facing bulge to the right. That’s because the outer tidal bulges are an intrinsic feature of the transformation, they’re nothing whatsoever to do with gravity, that is a mistake but, hopefully, understanding the transformation should clear it up.
BTW we don’t call it a ‘slide along’ we call it a translation.


or it can be turned 180 degrees through a circle with radius 1 about the origin {about the point (0,0) RMB}

Yes, that’s the first rotation – around the sun, it causes the outer tide and of course, it causes Sawicki’s 5% error.


and then rotated 180 degrees once it is at (1,0). {about the point (1,0) RMB}

Yes, that’s the second rotation – around earths axis, it causes the equatorial bulge so Stacey correctly factors it out.

Good.

Now to summaries:
I’ve shown that your diagram consists of the 2 rotations.
I’ve shown that this causes an outer bulge – both the teacup and earth’s outer tidal bulge.
I’ve shown that the bulge is the result of circular motion not gravitation.
I’ve shown that there is no outward force.
I’ve shown why Stacey factors out the axial rotation.
I’ve shown how to remove Sawicki’s systematic 5% tidal asymmetry.

Unless anybody has any more questions I think this would be a good time to sit back.

Byeee…

SeanF
2004-Apr-09, 01:42 PM
No, it shows a single revolution

Yes, of course it does Sean.
Now, why donít you explain to me how you think the teacup thing works?

As far as I can see, assuming the teacup is moving smoothly (no jiggling or speeding up, slowing down), there'd be no real effect on the tea at all. I certainly don't see how'd you end up with a tea "bulge" that rotates around the rim of the cup.



Iíve shown that this causes an outer bulge Ė both the teacup and earthís outer tidal bulge.

Not to me, you haven't it. All I've seen you do is ask about the teacup - you haven't shown anything regarding it.

milli360
2004-Apr-09, 05:48 PM
Those two rotations "undo" each other.

No they donít. They would if they had the same centre Ė surely you realise this much.
See below



Take a block that moves from (-1,0) to (1,0). The step can be accomplished by sliding it along the x-axis through the origin, or it can be turned 180 degrees through a circle with radius 1 about the origin, and then rotated 180 degrees once it is at (1,0). Both result in the same finish, and neither is "wrong."

Excellent, something we can agree on
Notice that those two rotations do not have the same center.


(who said anything about Ďwrongí?).

My point was that there is often more than one way of doing things. You keep insisting, without basis, that Sawicki is wrong. You've said that his math is wrong in a few cases, but you've been shown why it is not, in every case.


Yes, thatís the second rotation Ė around earths axis, it causes the equatorial bulge so Stacey correctly factors it out.
Actually, Stacey uses the second one to cancel out the effects of the first one.

JohnOwens
2004-Apr-09, 07:01 PM
Iíve shown that this causes an outer bulge Ė both the teacup and earthís outer tidal bulge.
Not to me, you haven't it. All I've seen you do is ask about the teacup - you haven't shown anything regarding it.
You mean besides the tempest in it?
(ducks & runs)

JohnD
2004-Nov-18, 11:00 PM
All,
I revive this thread having retired defeated by my inability to contradict the arguments of the Bad Astronomer, Sawicki and others. 'My' theory, that we dubbed 'different orbits' seemed to fail in certain special circumstances. I still can't answer that as a criticism, but I can put forward (!) on my behalf none other than the late Dr. Robert L.Forward.

I have just read, for the first time, Forward's SF book "Starquake", about a civilisation living ON a neutron star. Suspend your disbelief, as he added an appendix on the physics of placing a human crewed craft in orbit. I quote:
"We wanted Dragon Slayer [the human craft] at a 406 kilometer altitude so it would be in a synchronous orbit........ At this distance from a neutron star, even though the orbital motion cancels the gravity attraction at the centre of the spacecraft, the acceleration due to the tidal effects is 200 Earth gravities per meter outward in the radial direction to theneutron star and 100 gravities per meter inward in a plane tangent to the star."
If that isn't an argument for 'different orbits' I'll eat my neutronium hat.

There is a thread here, too. I acquired my insight into this view of the causation of tides after reading another SF story, Larry Niven's "Neutron Star". Moreover, I think LN's degree was in maths; was he taught by Forward? Anyway, he acknowledged the assistance of Dr.Forward in writing other stories, especially "The Hole Man". So the 'different orbits' view of tides should be Forward's view.

Too bad Dr.Forward isn't here any more, so I apologise to his shade if I am wrong.
John

A Thousand Pardons
2004-Nov-19, 11:46 AM
I have just read, for the first time, Forward's SF book "Starquake", about a civilisation living ON a neutron star. Suspend your disbelief, as he added an appendix on the physics of placing a human crewed craft in orbit. I quote:
"We wanted Dragon Slayer [the human craft] at a 406 kilometer altitude so it would be in a synchronous orbit........ At this distance from a neutron star, even though the orbital motion cancels the gravity attraction at the centre of the spacecraft, the acceleration due to the tidal effects is 200 Earth gravities per meter outward in the radial direction to theneutron star and 100 gravities per meter inward in a plane tangent to the star."
If that isn't an argument for 'different orbits' I'll eat my neutronium hat.
Eat away. :)

RichardMB is still fighting the good fight in the thread what is the cause of tides? (http://www.badastronomy.com/phpBB/viewtopic.php?p=367895#367895) If you look over that thread, you'll see where the freefall tidal explanation does explain that 2:1 ratio that you found in the book. The differential orbit explaination would have 300 Earth gravities outward--and zero in a plane tangent.

The reason is, that the rotation itself produces a 100 Earth gravity differential, in all four directions.

JohnD
2004-Nov-20, 09:48 AM
Please, explain "Tangent to the star". I was puzzled by these words in Forward's explanation.
They give me a picture of a plane that touches the star's surface at one point, like a ball on a table. An orbiting body will pass through such a plane, briefly and at an angle relative to the direction of the star's centre. Is the acceleration applied at an angle?

If this tangent plane always touches the star at a point immediately below the satellite, it will never pass through it. In which case, I can't see how acceleration can be applied in a plane that doesn't pass through the satellite.

If the plane is parallel to a tangent to the star, then this is the same as though the centre of the satellite, at right angles to the direction of the stars centre. In other words sideways. No, I really can't see that happening!

So what is meant by "tangent to the star"?

Thanks for pointing me to the other thread - I'll go and be educated there too.
John

PS Well, I went and I was educated and I have enormous respect for RichardMB as a teacher. His three-tethered-balls-at-the-distance-of -the-Moon thought experiment, with the simple calculations appended, is JUST what I need to get a better picture in my mind of what is happening.
AND it reveals that even if my half baked view of 'different orbits' is wrong it was half right! The tension between the three balls nearly doubles if they are in orbit, rather than merely falling straight in! Does this mean that the faster the orbit, the greater the tide? I'll have to try to follow Richard's calculations better.
But I still don't understand 'tangent to the star"!

Thanks again to all.
John

A Thousand Pardons
2004-Nov-24, 09:24 PM
If the plane is parallel to a tangent to the star, then this is the same as though the centre of the satellite, at right angles to the direction of the stars centre. In other words sideways. No, I really can't see that happening!
Yes, that's it, and it really does happen. We know this from actual measurements.

That effect is pretty much balanced, on the equator, by a centrifictional force produced by a rotation of the satellite that coincides with the time of orbit.

theholycrap
2004-Dec-07, 09:02 PM
If the plane is parallel to a tangent to the star, then this is the same as though the centre of the satellite, at right angles to the direction of the stars centre. In other words sideways. No, I really can't see that happening!
Yes, that's it, and it really does happen. We know this from actual measurements.

That effect is pretty much balanced, on the equator, by a centrifictional force produced by a rotation of the satellite that coincides with the time of orbit.

cool :P