SeanF
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You want us to "look up" how gravity works, but not in astronomy books?


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Yes, look up the law of gravity including the law of superposition, and centers of gravity, in physics books.



Kaptain K
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So...
Every astronomer, physicist and mathematician on the planet is wrong, and you (and only you) are right?
Impressive ego you've got there.

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I'm truly sorry that it looks like an ego thing. It is not.
Einstein had to prove Newton wrong and even with his brain power it wasn't easy. I know all to well I'm not in that league.
I'd as soon leave the physicist and mathematician out of this, unless of course they were working not just giving lip service.
If you read a book and then quote that book you could be wrong. I've read quite a bit one book said a cow jumped over the Moon, another said the Moon was made of "blue" cheese. And yet another said the cheese was "green".

Let me make 2 statements.

1) Gravity pulls or holds the Earth and Moon together, gravity causes the tides, the tides push the Earth and Moon apart.
2) The Moon is going slow so it keeps falling farther and farther behind.

Now you tell me, which made the most sense.




Silas

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We give you our word of honor; it is true!


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Very good I except that. Here is all that I ask.
You say the Moon is slowing the Earth, and the Earth is causing the Moon to move away. What I would like is for someone perhaps yourself to make the correct calculations to prove the length of time the Moon could have been in orbit above the Roche limit.
Carl Sagan and others said the number should be around 400,000,000 years. The last time I tried it I came up with about 402 million. I'll not argue over a couple million.
I believe I said this before you must use the present vectored speed of assent, which I 'think' is about 4.888 meters per second, not the 3.8 cm per year.

If you find that Carl Sagan was, and that I am correct, wouldn't it seem rather strange that the Moon was somewhere else for over 4.5 billion years? Perhaps that is where it got the craters. Wouldn't you have a slight question as to how it got up there? And just maybe wonder when within those 400 million years it happened?
I know I sure did!


Gary

What is the vectored speed of assent? How are you using it?

GrapesOfWrath

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What is the vectored speed of assent? How are you using it?

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Thank you again for the best questions.

Let me start with an example.
Let's say that you and I are on the freeway bumper to bumper at 55 mph in the slow lane. Then which ever one of us is in the lead accelerates to 65 mph. The person now doing 65 starts to slowly move ahead even though they are no longer accelerating. They are just going faster.
I think you can follow that. Now then just suppose the one doing 65 moves over into the second lane to pass someone else doing 55. They vectored over one lane. Again just suppose this to be a 6 lane mega highway and suppose that every minute they move over one more lane until they are in the commuter lane.
At this point the distance separating the two cars will be along the vector but still growing because of the speed difference. Even though the commuter lane isn't that far from the slow lane the distance between the cars is great. Just suppose for a second that the lead car had not accelerated to 65, but had made the lane changes at 55 or even started slowing down to 50. What then?

Each year the Moon moves away from the Earth one more lane width about 3.8 cm, but the speed at which it moves along the vector is the critical factor. I hope that makes a little sense.

I posted a drawing earlier trying to showing the Moons 3-D vector.

Recession of nodes .gif

Cars on the freeway are only 2-D not 3-D. The Moon's recession of nodes motion is the other component of the 3-D vector. I tried to show that more thoroughly in my recession of nodes.htm page.


Gary






<font size=-1>[ This Message was edited by: Gary Redmond on 2002-04-19 11:59 ]</font>

Neither link mentions vectored speed of assent--and neither seems to deal with the increase in the distance between the moon and the Earth.

What'd I miss? How are you using it, and why does it make a difference in your calculations?

GrapesOfWrath
Quote:
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Neither link mentions vectored speed of assent--and neither seems to
deal with the increase in the distance between the moon and the Earth.
What'd I miss? How are you using it, and why does it make a difference
in your calculations?

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Sorry.
To me if the Moon moves away from the Earth, or the distance between their centers increases then they are ascending. If the pull of gravity is down, then up would be ascending.
I have re-posted Recession of nodes .gif with a few changes. In it I show the Moon's non-closed loop trajectory in (red). If the Moon were in what might be called a pure orbit it would follow a closed loop (fuchsia dots). As the Moon goes around the Earth it (recedes or ascends) at the same time it recesses (opposite of precesses). So the vectored rate of assent is an imaginary (dark blue) dotted line connecting the ends of the red line. The outward motion is the 3.8 cm per year or 3 mm per month, which is negligible. The recession is about 1 revolution every 18.6 years.

Math time:
If the Moon is 240,000 miles away. Then C = pi * D
1 revolution = aprox. 1,507,964 million miles.
1 / 18.6 of a rev = aprox. -81,073 miles per year or -6,200 mile per month.
I would say that 81 thousand miles per year is important wouldn't you.

Note:
If the Moon were in a perfect orbit it would gyroscopically precess and the -81,073 miles would be a slightly positive number.

Also in the recession of nodes.htm (10 th and last drawing) the green arrow -6200 miles per month of recession motion is far more important then the 3 millimeters of outward motion.

In my cars on the freeway example, it's as if the expanding distance between the cars (65 vs. 55) would be over three trips around the Earth every year.
Yet not only did the lead car not change lanes it didn't even get across the white line.

Using your (the) astronomy books and the "R cross F" theory,
the closer the Earth and Moon were to each other, the stronger would have been the gravitational pull, the faster they should have separated. In other words the vectored assent over the past eons should have followed a reversed ballistic parabola.
"R cross F" vectored trajectory.gif

We know the Moon to be about 240,000 mile away, to divide that by 81,073 and then say the moon has only been in orbit 3 years would be a little weird. What you need do is calculate the length of the parabolic spiral from the Moon's present location back down to the Roche limit. Then divide that distance by the 81,073 miles per year to find the number of years.

Did this help?


Gary

Help? It made my day (and this has been a good day).

"assent" means to agree to something. An "ascent" is a rise. "to ascend" is the verb form.

"vector" is a noun, not a verb. Nor can it be used as an adjective (said adjective seems to be just using the nonexistent verb in this case). You can "express something as a vector" but you cannot "vector", nor can you "vector something". (At least in physics. I gather in aviation you can "vector an airplane", but even they don't use it as an adjective, and we're talking physics here, anyway.) Hence, you cannot have a "vectored something". Speed is a scalar, the absolute value of the vector, velocity.

"recession" is not the opposite of "precession". "Recession" is the noun, "recede" is the verb. "To recess" does not mean to move away from. It means "to put into a recess", "to make a recess in" or "to interrupt for a recess (to take a recess)".

Your "center of gravity" concept is useless except in the case of spherical symmetry, where it's the same as the center of mass, and hence redundant. It will not do what you want it to do. Non-sperically symmetric bodies can and do exert gravitational torque on each other. The math is easy to prove, and we have shown it to you several times.
You have several axioms that are just flat-out wrong, and hence you are developing faulty conclusions. I will quote them here in isolation, in the hope that you will recognize that they don't work. I don't see any point arguing about the conclusions when the premises are flawed.

Flawed premise number 1: "The resultant forces are as if those huge bodies of particles were singular point sources of that force and it is as if that force acts on a singular point within all other bodies." This is simply not true, except in the spherically symmetric case. It can't be generalized.

Flawed premise number 2: Your "torque needs two forces" scenario. What you say may be true on earth, from a mechanist's point of view, but it is purely a selection effect, not intrinsic to the definition of torque. Yes, on Earth, there is always some kind of holding force against which you, the mechanist, are applying torque. This is simply a side effect of being on earth. Perhaps we are using different definitions of torque, as Jeff suggests, but you are using the *effect* of a physicist's torque (a change in angluar momentum), while claiming your machinist's definition (there have to be two forces). Therein lies the problem. We are claiming a change in angular momentum of the earth by the torque from the moon, so you have to use the physicist's definition. An example: you take a wrench (or stick), and throw it up in the air, but just before you release it, you push down. It will start to spin like crazy. You've thus torqued it without any secondary force. And if you think gravity is the secondary force in this case, let's go back to the two by four on the ice: grab it by one end and start pushing it, and then pull back on your end. Again, it will spin (hence there was a torque) and here there is only one force in the problem: you pushing it.

If you correct for these problems and still don't understand how the earth and the moon are torquing each other, we can try again, but I'm not going to try to argue the finer points of the evolution of the earth/moon system as long as we've got these basic problems in the very definitions and premises.

Yours,

Don

<font size=-1>[ This Message was edited by: DoctorDon on 2002-04-21 13:16 ]</font>

GrapesOfWrath

Please forgive me. I have as much trouble with my drawing in 4 dimensions as with my English. ("assent" "ascent")
I have reposted my "R cross F" vectored trajectory.gif


Again:
Using your (the) astronomy books and the "R cross F" theory.
The closer the Earth and Moon were to each other, the stronger would have been the gravitational pull. The stronger the pull the greater would have been the slowing. The greater the slowing the faster should have been the separation. In other words the vectored "ascent" over the past eons should have followed a reversed ballistic like parabola.
The speed at which the Moon is ascending that parabola can be measured as the Moon crosses the Earth's elliptic plane.



Gary


PS Thanks Don, does the above sound better.

Geo3gh

I still haven't found the answers to your original questions, but I've found more.

The Earth spins on its axis once a day. That rotation causes the Earth's diameter at the equator to swell by 27 miles in relation to the pole to pole diameter. When the Earth gravitationally locks to the Sun (or rotates once a year) will we have one continent again, and will the Arctic Ocean be 30 miles deep?


Gary

Continents come and go, come and go, but before the time the Earth has locked to the Sun, the Sun will have probably destroyed the Earth. It would take that long.

Also, that 27 miles is in diameter--that's only an extra 13 miles at each pole. And when the spinning of the Earth slows down, not only will the water recede from the equator, but the solid part of the Earth will relax also. The Earth will become more spherical, and there will no longer be a 13 mile "depression" at the poles. So, no deep arctic ocean.

Sorry I've been away. My job got a little crazy, so I was back to lurk mode. But I'm back.

I agree with Grapes. Plate tectonics works separately from the Moon's orbital meachanics. The configuration of the continents will effect the amount of drag on the ocean's tidal bulge. And the fact that the Moon and Earth are doing their merry dance, with all the tidal deforming and all, should help keep the Earth's interior hot and molten, so we should keep seeing the tectonis playing shuffleboard with the plates for quite a while.

But it will take so long for the Earth to become tidally locked to the Sun that the Sun will be a red giant by that time. I'd say the depth of the oceans will be zero, since they'll have boiled away. [img]/phpBB/images/smiles/icon_smile.gif[/img]

But seeing as all I've done here is echo Grapes' post, I'll shut up now.

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!    --Invader ZIM

On page 41 of my physics book I find the following.

Quote:

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Newton's Second Law

F = ma

"Newton's Second Law is among the most important scientific laws ever discovered. It says that if we know the force acting on any particle we can calculate its acceleration. Thus, if we know the acceleration at any time we can assume that the acceleration will remain constant for an instant, and so calculate the velocity and position of the particle at the end of that instant. This can be repeated for the next instant, the next, and so on. Thus, if we know the position and velocity of a particle at any time, and know what the force will be for all time, we can calculate the position and velocity of the particle at any future time.

Therefore, if we know the laws that govern the forces, and the locations and velocities of all the particles in the universe at any given time, we should be able to predict the positions of all the particles in the universe for any time in the future."

================================================== =====================


My point is this:

If the future can be predicted then the past should be reconstructable.

I feel confident that "we know the laws that govern the forces," with a reasonable degree of accuracy. While we don't know "the locations and velocities of all the particles in the universe at any given time.", I believe that we do know "the locations and velocities of" the Earth, Moon, and Sun, quite well.


I know and understand that as time marches on accuracy decreases, as was mentioned earlier.

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The Bad Astronomer Posted: 2002-04-01 00:32   

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You cannot extrapolate the Moon's current recession speed (4 cm/yr) back in time very far. The current rate of recession is known to be high because of a peculiar resonance with the ocean; the friction caused by the moving tidal bulge is very high right now. In the past, although the Moon was closer, the actual amount of friction it caused by tides was smaller. The efficiency with which the Earth's spin energy was transferred to the Moon was lower, and so the Moon's recession rate was slower.

Extrapolating the Moon's recession rate backwards using a constant velocity is therefore an incorrect method.

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Even though the rate of change to the speed maybe variable and unpredictable, it's all that we have.

The big question is not about the slowing or the rate of slowing, but by how much has the Moon slowed? Or, what was the original speed minus the present day speed?

As I just said all the accelerations have been in the negative or a slowing direction. Therefore the Moon must run slower than it did, WHICH IT DOES.


The Moon goes around the Earth once every 27.32166 days, the sidereal month.

The Moon crosses Earth's elliptic plane every 27.21220 days, the nodical month.

The difference of 0.10946 days per month is the amount of time by which the Moon falls back (recesses) every month.


recession of nodes.htm

trajectory.gif

The recession of nodes can be expressed in many different ways, as time or distance or a speed. It is not a slowing it is simply the amount the Moon falls behind every month. This running slow could be due to the slowing, which is or maybe variable. For lack of anything better I suggest this speed be used to ingress the Moon's position. The direction of that speed is the 3.8 cm/yr rate. There are of course several options as to how to do that.

I leave the choices to you.

Forty years ago many texts said the Moon could only have been above the Roche limit for about 400 million years, could they have been off by 4.5 billion years?

I don't think so!

Those texts also said the dinosaurs lived with a 22 day month. A 4 cm/yr radius change would = a 2600 km change in 65 million years. That would be 5200 km to the diameter or 16 336 km to the circumference. That means less than 8 hours difference per month, that would mean a 27 day month rather than the 27.32 we have today. Could the scientist of yesteryear have been wrong by 5 whole days?

I don't think so!

I've said before that my calculations are within 5 million years of those put forth 40 or more years ago. I also believe the old texts were correct. Yes, using the 4 cm per year for the extrapolation is wrong, other than to show as a part of the vector. I would like verification that those early texts were somewhere near close, or some proof that they were wrong.



Gary

[ This Message was edited by: Gary Redmond on 2002-04-29 15:11 ]

<font size=-1>[ This Message was edited by: Gary Redmond on 2002-04-29 15:15 ]</font>

Which texts are they?

GrapesOfWrath

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Which texts are they?

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I've said it before you ask the best questions. I knew when I said that it would come back and bite me. I left all my books in Alaska, paying shipping once was quite enough, so I'm trying to find something here.

I think it might be better to not dredge up and rehash; the redo is not that complicated.

Could you please try to work with me on this. AND PLEASE if I say something that doesn't make sense or offends, don't hesitate to interrupt or make corrections.

When I posted my
vectored trajectory.gif and asked:


Did this help?


You said:
Quote:
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Help? It made my day (and this has been a good day).

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That really didn't (HELP!) me much. Are my drawings really that bad?



Gary



<font size=-1>[ This Message was edited by: Gary Redmond on 2002-05-01 17:50 ]</font>

Well, part of it was your comment that "We know the Moon to be about 240,000 mile away, to divide that by 81,073 and then say the moon has only been in orbit 3 years would be a little weird." That was a good chuckle. My mind is slipping a little, but I'm pretty sure I would have remembered something like that.

On the other hand, I just noticed that your drawings are not coming up now, for me. Are they still there?

GrapesOfWrath

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On the other hand, I just noticed that your drawings are not coming up now, for me. Are they still there?

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I checked from here and it seemed AOK.


GrapesOfWrath
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Well, part of it was that, "We know the Moon to be about 240,000 mile away, to divide that by 81,073 and then say the moon has only been in orbit 3 years would be a little weird." That was a good chuckle. My mind is slipping a little, but I'm pretty sure I would have remembered something like that.


================================================== ======================

The reason I said that, is it's almost the same idea as using the 3.8 cm distance to make the calculation it's a simple distance and just plain wrong.

Again you seem to evade my questions, but egg me on at the same time. I really would like to answer the questions posed by Geo3gh at the start of this thread. But, I'm not sure that I can do it satisfactorily without some help. It looks as if we need to go back to the first pages of the physics book.

Page 12 Chapter 2 "Motion is relative"
Page 13 "Position, displacement, and distance"

"In a mathematical description of motion the basic step is a description of the position of an object at a particular instant of time. This is generally done by giving the coordinates of the object; but before the coordinates can be given it is necessary to choose a coordinate system and a time axis."

(Bla bla bla)

"When an object moves from position A to position B it is said to have been displaced. If we follow the motion of the object along the path that it traverses from A to B, we can measure the distance it traveled. We can, however, ignore the actual path and fix our attention on the change in position, which is called the displacement from A to B."

(Bla bla bla)

"The most significant difference between distance and displacement is that distance is expressed as a number in certain units while displacement is a number in those units plus a direction."

(Bla bla bla)

================================================== ======================

So far so good?


Here we go; "before the coordinates can be given it is necessary to choose a coordinate system and a time axis."

Time axis first. The Earth's elliptic plane is nearly perpendicular to the direction, which the sun is traveling around the Milky Way, and because this might be a long time (more than once around the Galaxy), I would think that each time the Earth passes between the Sun and the Galactic center would make a good 1 yr. incremented time axis.

Let's use the sidereal year is that OK?

Now to choose the origin. As the Earth and Moon move apart, it is along the gravitational link that ties them together. The point around which their system mass rotates is their barycenter, and it is this point that moves around the sun.

So do you agree the Earth - Moon barycenter should be the origin of the coordinate system?

And, the elliptic plane formed by that point as it goes around the sun should be one of the coordinates?

Let me take a break here. That's 4 questions more than enough for one sitting.



Gary

What they mean for a time axis is we need to choose a start time. A "T=0" if you will. So we all can agree: I declare 2002-01-01 00:00 (Midnight on January 1st of this year)

As for units on that axis, sideral years is good enough for me.

I can work with x=0, y=0, z=0 of our spatial coordinate system as being the center of mass of the Earth-Moon system.

But the "elliptic plane" will not be a coordinate. It can be represented as two axes though. Let's call +x towards the center of mass of the sun, -x away from it. +y in the direction of the Earth's orbit (right angles from the x axis). +z can then be at right angle to the x and y axes, the best North you can do.

<rl>
At t₁=0, the Moon's orbit has a (measured) radius of R₁.
At t₂=t₁+1 year, the Moon's orbit has a (measured) radius of R₁ + 3.8 cm

We're not talking about why yet, just agreeing on observed facts.



_________________
Jeff Schwarz
__________________________________________________
Oh freddled gruntbuggly thy micturations are to me
As plurdled gabbleblotchits on a lurgid bee.

<font size=-1>
Edited some code for subscripts.

[ This Message was edited by: Geo3gh on 2002-05-03 12:20 ]</font>

<font size=-1>[ This Message was edited by: Geo3gh on 2002-05-03 12:21 ]</font>

Thanks Jeff,

It is easy to tell which of us went to school and which didn't, but please don't give up on me. I know you wanted to go forward with your time line, but I think it more important to find where it is coming from first.


You said
Quote:
___________________________________________

But the "elliptic plane" will not be a coordinate. It can be represented as two axes though. Let's call +x towards the center of mass of the sun, -x away from it. +y in the direction of the Earth's orbit (right angles from the x axis). +z can then be at right angle to the x and y axes, the best North you can do.

At t1=0, the Moon's orbit has a (measured) radius of R1.
At t2=t1+1 year, the Moon's orbit has a (measured) radius of R1 + 3.8 cm
We're not talking about why yet, just agreeing on observed facts.
___________________________________________

Let me suggest rather than radius we opt for distance between.
Sorry, I'm still having trouble with super and sub shifts.

I think I got it, I'm not as good as I was 12 years ago. I'm still not sure this coordinate system is correct, but it's close. From here on I babble a lot, but I hope you can straighten it out.

In the perfect world of Kepler, celestial bodies traveled in closed ellipses called orbits.
So if the Moon-Earth system with its 5+ degree tilt, were to make one rotation it would start at the intersection of the X=0 Z=0 and at near +Y= +240,000 mile.
Then moving up the +Z to a point over Y=0 and -X= near -238,000 mile.
Down +Z to Z=0 X=0 and -Y= near -240,000 again.
Etc., etc.
Always crossing at X=0 Z=0 +Y and again at the Z=0 X=0 -Y intersection and then back to X=0 Z=0 +Y.
In other words the speed required to make one full orbit is exactly correct for the distance covered.

Then with a slight stretch we learn about gyroscopes and that spinning things precess. So the Z=0 X=0 are lost. The Earth's equinox precesses, the Moon's perigee precesses, and Einstein's famous proof of relativity relied on the precession of Mercury's perihelion.
Most low Earth "orbit?" satellites precess at a prodigious rate. It is this inward directed speed, which brings them down.

Note: To keep satellites up they should be launched from the Southern hemisphere into the Northern.

Using the above coordinate system, and precession: If you start the Moon as before at the X=0 Z=0 and +Y and move to the +Z point over Y=0 -X, you will find that you have already started the decent. As you arrive at the X axis you will still be at a -X something. As the precession progresses eventually a time will come when both X values will be positive or both will be negative not having crossed the Y axis. Over 1/2 the ellipse would be South of the systems elliptic plane or in the -Z range.

Recession is the opposite of precession.

Using the coordinate system, and recession: If you start the Moon as before at the X=0 Z=0 and +Y and move to the +Z point over Y=0 -X, you will find that you are still ascending. Upon arrival at the +Z axis you'll still be above the X axis. This time over 1/2 the system's ellipse will be in the Northern or +Z range.

I'm trying to figure the best way to explain this. Hmmmmmmmmmmmmmmmmmmmmm
Doodle Doodle.gif

OK, the Earth and Moon are relatively close together so the Moon's speed of fall around the Earth is controlled mostly by the distance between them. But, there is more to it than that; the Earth-Moon system is moving at another speed at 90 degrees to the plane of rotation (Galactic speed).
So, recession is dependent on the relative speed difference between the larger and the smaller mass in this other direction.
Because over half the Moon's trajectory is in the Northern hemisphere, and the Earth-Moon are going South (tee hee) the Moon recesses. That same trajectory however is equally divided as it passes through the Y and Z axes. Hence the difference between the two types of month.


Nodical 27.21220 days.


Sidereal 27.32166 days.




This speed difference of 0.10946 days is what you must use to regress (in time) the Moon down to the Roche limit or Earth's surface.

At some time in the past, the Earth passed the Moon.
When did that happen?




Gary




BTW here is a chart showing the time difference between high tide peak and the Moon's zenith for Honolulu in 1997. The location and depth of the tide, is the result of all the accelerations on the ocean.
On February 28 and September 7 the difference was 2.316 hours,on January 17 it was 4.933 hours.

Tide Chart.gif






<font size=-1>[ This Message was edited by: Gary Redmond on 2002-05-04 12:29 ]</font>

Just a quick reply here Gary. I'm still working thru your last post, but I have a busy social evening in front of me. I will probably be able to digest it and respond fully tomorrow.

But I do have a quick philosophical notion to put forth before I dart off.

What I'd like to do first is work on the facts. That is, the observed data. Then I would say that whatever mathematical models we work with must agree with the facts. For instance, if we agree that the paleontological evidence indicates that the Moon did indeed exist 3 billion years ago, then any model that shows its origin signifigantly later than that can be discarded. Or at least we can agree that the model in question needs a lot of work. In a similar vein, if the data indicates that the length of the day in the Devonian Era was 22 hours, a successful model will have to explain that.

I think that arguing the math and working out when the Moon was below the Roche limit based on the assumption static rates of change is putting the cart before the horse.

So, before we jump to precession of equinoxes, etc., let's deliniate the observed data, and agree on what definitions we are going to use. OK?

__________________
Jeff Schwarz
__________________________________________________
Argh!! They booby-trapped their sun!!    --Invader ZIM

Geo3gh
Quote:
================================================== ====================

I'm still working thru your last post,

So, before we jump to precession of equinoxes, etc., let's deliniate the
observed data, and agree on what definitions we are going to use. OK?

================================================== ====================

I'm sorry I get so carried away; please forgive me.

OK!

There are essentially 4 types of month.

1) Nodical . . 27.21220 .days
2) Sidereal . . 27.32166 .days
3) Anonalistic. 27.55455 .days
4) Synodic. . . 29.530588 days

I say that #2 the Sidereal month of 27.32166 days is one complete rotation of the Earth-Moon (E-M) system. This is using one of our axes. Therefore, if the (E-M) starts on a line in the elliptical (X) plane and makes one rotation back to a line in that plane in 27.21220 days, then the (E-M) system hasn't made a complete rotation.
So therefore what are the options???????????
Is this what you mean?

Gary