No, it's the tidal bulge, not the equatorial bulge.

Gary, wouldn't your version also show an outward spiral for Mercury also, as it orbits? But that doesn't happen.

It's not the "tidal bulge", but "tidal drag". Tidal friction is slowing the rotation of the Earth. Momentum must be conserved. The loss of rotational momentum of the Earth is compensated for by an increase in orbital momentum of the Moon. This will continue until the Earth-Moon system is "tidally locked" some time in the very distant future (IIRC the month and day will stabilize at 55 of our current days each).

__________________
Any day you wake up on "the right side of the dirt" is a good day.

T. Anderson

I stand corrected... Time for me to obey the exhortation in your signature!

Silas

Geo3gh
Quote:
------------------------------------------------------------------------
This is exactly why I've stuck it out myself. Civility like Gary's is to be commended and nurtured.
------------------------------------------------------------------------
Jeff.
I'm glad that I counted to ten, things like "Dude, you're wrong" are a little upsetting. But, comments like yours make life worth while. THANKS.
Before I make everyone angry (again), Let me, tell you how my research is going.
I must be Irish or somehow related to Murphy.
As you recall I said I needed a northern and southern most perigee or apogee. And that those should fall about 4.425 years apart.
I found 6 years of data 1997-2002.
You guessed it on Sept. 8, 1999 the perigee was at 00:00 hour and northern most at 19:00.
On April 24, 2000, the apogee was at 12:00 and southern most at 16:00.
Again on May 22, 2000, the Moon was southern most at 00:00 and apogee was at 04:00.

I can live with a split the difference, but I need more data.

Does anyone know where I could find a couple of more years' data?

Let me ask you this, do you find it interesting, that Silas, the Celestial Mechanic, and now Kaptain K think I'm wrong, but if you look at what they said, they have different ideas as to what causes the regression of nodes or Earth slowing? Both of the theory's are taught, perhaps one of them must also be wrong. I say we have three theory's out here on the table what do you think?

Celestial Mechanic
Quote:
------------------------------------------------------------------------
The answer lies in the interaction with the tides, and I believe also
in the fact that frictional forces are slowing down the Earth's
rotation.
------------------------------------------------------------------------
Silas
Quote:
------------------------------------------------------------------------
Although the increase in semi-major axis is believed to be from
interaction with the earth's equatorial bulge, rather than from the
tides per se.
------------------------------------------------------------------------

Jeff,
If you get your physics book out again, you will find that both the above are wrong.
The explanation is very simple.
The law of conservation of angular momentum. That law is very clear, for the most part it is a quote of Newton's first. If a body is given a spin it will continue to spin at that speed unless it is acted upon by a force, that force must be external to the spinning body.

Please don't tell about ice skaters. Yes, the force is stored within, but the directive to apply the force comes from outside. The computer needs external programing.

When you say the Moon is slowing that means the Earth-Moon system is slowing, the force to cause that slowing must come from outside that system. It cannot be the Earth or the Earth's tides.
If you take that one step farther, you find Newton's law of action reaction comes into play. If the Moon slows the Earth, the Earth would speed the Moon. If the Moon were to speed up it would move downward not upward. So you see, either the astronomy books or the physics books are wrong. I don't believe it's the physics.

I know you didn't like that explanation when I gave it before. I can't help it those are the laws and that is the way it is.

The astronomy books are wrong.

All the Moon data is correct, but the theory's that uses the data are wrong. I sure would like some help to prove that. Just for kicks could you guys help make the calculations I suggested. I really think the Moon is 550 miles north of the Earth. As I said if I'm wrong I'm out of here.

We or I need; apogees and perigees, northern most and southern most, dates and times, for 1996, 1995, and perhaps 1994. And never having used an Ephemeris I could use some help there as well.
????????????????????


Gary


PS. GrapesOfWrath
Quote:
================================================== =====================
Gary, wouldn't your version also show an outward spiral for Mercury also, as it orbits? But that doesn't happen.
================================================== =====================
Quite the contrary.

Precession is of two types. The natural gyroscopic type that is induced by the off set torque, and what I call an unnatural type, which is caused by an excess of speed. The regression of nodes such as the Moon's is of only one type its sole cause is an insufficient speed. Those two statements are one in the same it is just a matter of your reference frame. Mercury precesses at what was thought to be an unnatural rate, until Einstein worked his magic. That motion is due to the warping of space by gravity.
As I said it has to do with the reference frame and which body is the most massive. When the more massive object is the precess-or it continues to move away and provides a stable condition. When the less massive body is the precess-or it is like Silas's balanced broom, as the little guy tries to pull out in front it finds itself crashing into the backside of the larger.

Count to ten again...

Simple. Dude, I was wrong. I was corrected, and I accepted it with grace.

Please don't try to claim that my error justifies your error. That's just shabby. I'm not perfect; do you imagine that this means that you are?

In the meantime, you haven't responded to the problem that your own theory predicts several visible effects that have not been observed. Why is the Moon *not* 100,000 miles out of position in the course of a year?

As I asked before: what is your actual area of professional expertise? What would happen if I came in to a discussion board on that subject and started posting remarkably unconventional ideas?

I *know* I'm a lightweight. Why is it so difficult for you to admit that you are one also?

Silas

Hey, don't forget me!

Wait a minute. You're pulling a bait and switch here. The "directive"??? Besides which, there's no change in torque in a spinning ice skater, anyway. The change in speed is compensated for by a change in the moment of inertia, which is why I said I made a mistake when I used it as an example why back when.

So, by your logic, if you and I are standing on an ice rink, I cannot push you away from me, because any force that could change the you-me system would have to come from outside the two of us. Don't you see a problem with that?

Newton's third law absolutely comes into play, except you are making the same mistake I made with the ice skater. It's not just speed which is changing here. It's momentum. Newton's third law doesn't speak of speed; it speaks of "action", which is forces, which is a change in momentum.

When you say "the moon slows the earth", it would be more accurate to say "the moon decreases the earth's angular momentum". Since the Earth's moment of inertia doesn't change much, this change manifests itself as a change in rotational speed. It slows down.

Now, when you say "the earth speeds up the moon", and if you were to recast that sentence into its more accurate form, "the earth increases the moon's angular momentum", you might be able to see that the moon's angular momentum increases by changing the moment of inertia (like the ice skater), i.e. moving away. But since the moon is in a free-falling orbit, moving away means slowing down. Your intuition is telling you that slowing down is a violation of Newton's third law, because you are thinking in terms of speed, not momentum. But as the ice skater example shows, there are two ways to change momentum. By moving away, the moon increases its angular momentum, despite slowing down.

No, it's your understanding of the physics that's wrong, which is causing you to not understand why the astronomy books say what they do.


Don


<font size=-1>[ This Message was edited by: DoctorDon on 2002-06-25 09:47 ]</font>

Gary, are you aware that the amount of Mercury precession is nearly 600 arcseconds per century, whereas general relativity explains only 43 arcseconds of that?

GrapesOfWrath,

Quote:

<HR>
Gary, are you aware that the amount of Mercury precession is nearly 600
arcseconds per century, whereas general relativity explains only 43
arcseconds of that?

<HR>
Are you talking perahelion or nodes.
No! I wasn't! I thought the difference was only 5 or 6. If the 600 is nodes, then perhaps I've unknowingly made another prediction. This is scary. I truly wish you would help with my calculations.

See the drawing at the end of this post.



<HR>



Silas



This looks like a matter of miscommunication, and I'm truly sorry for the part I play in it.

For instance, you say:

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

"Simple. Dude, I was wrong. I was corrected, and I accepted
it with grace."

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

I'm confused as to what you were corrected about. Was it me? Or was it about your explaination of tidal force cause.



I definitely am not perfect. I just believe the astronomy books are wrong.

You say the effects have not been observed.

I say they have.

Here is the problem as I see it. All the books and the teachers either describe the regression of nodes as a precession, or gloss over it because of a lack of understanding. The other data, which comes in the form of the Ephemeris is huge volumes of numbers. First let me say I've only seen one or two of the printed type and never studied them.

Who has?



Let me try to respond to what should be and are visual effects. The Earth and Moon are gravitationally locked together so the Moon can not run off. I said the Moon was going over 12 mph slower than the Earth actually it is more than 13 mph.



Hum? How can I explain it?

If we go to the hardware store and buy one nut and a 3 foot piece of "all" thread rod, I say buy because I'm going to mess it up. Put the nut on the rod, then bend the rod into a circle and weld the ends together We now have a threaded torus. With modification the nut goes around and around the torus forever.

OK, take a saw cut the nut off and throw the pieces away.

Let's build a ring to fit over the outside of our torus with a "v" tooth on the inside. If the ring is turned around and around on the torus, it stays in the same two teeth of the thread and doesn't go anywhere. That would represent a perfect orbit.

I hope you followed that.

Now lets build two more rings one with a bigger inside diameter and a smaller one that will pass through the torus with the tooth on the outside. These two rings are free to advance a tooth around the torus.
The inner ring has a smaller outside than the inside of the outer ring.
Mechanically speaking they have different surface feet per second speeds.

Yuk, I hope you're not completely confused.

So how can I say it? Because the Moon is traveling slower than the Earth it lags behind, when it does that the diameter of its trajectory is smaller, the speed difference causes the regression of one tooth (6040 miles) each month.



I'm sorry that isn't a very smooth explanation either.



As you might guess I was machinist and electronic tech. Printing, office machines, appliances, and auto.

If it is mechanical, I understand it. I think the solar system is mechanical.



I really believe <BIG>we</BIG> need to work out the orbit (trajectory) and see if Earth is at a focus of the Moon's ellipse or if it's the cones as I describe. I don't want you to take my word for it. I want you to do it. My grandson watches "Magic School Bus" on TV, the saying there is "make mistakes" "get messy".

The last three months here have been "real messy". I don't think I've made the mistake. I refuse to admit defeat until proven wrong.



I previously suggested using a maximum north and south perigee or apogee. I only found these:

Quote:

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

"You guessed it on Sept. 8, 1999 the perigee was at 00:00 hour and
northern most at 19:00.

On April 24, 2000, the apogee was at 12:00 and southern most at 16:00.

Again on May 22, 2000, the Moon was southern most at 00:00 and apogee
was at 04:00."

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

Do you think, that using a perigee of 221,600 miles, and apogee of 252,950 miles we could make the correction?

Besides that the truth should show up as an 8-9 minute difference in angle.

Here is a drawing to make the calculations . This is a two in one drawing Fig 1 northern most on top, and Fig 2 southern most on the bottom.

If my theory is correct then angle "x" should be 8-9 minutes greater than angle "y".





Gary

The latter; I had thought it was due to the equatorial bulge of the earth.

Interesting "hardware wars" explanation... I think I comprehend it... But...

Let me try: I'm in a car, and I'm driving around a big circular racetrack. Now, yes, the outside of my car is going faster than the inside of my car, but they never actually move apart.

So you *could* say that the average speed of the Moon, with respect to the sun, is slightly large than the average speed of the Earth, wrt the sun, since the path the moon takes is longer.

But in the first example, it is *wrong* to say that the right side of my car is going 2 mph faster than the left side of my car *with respect to the center of my car!*

With respect to the car, both sides are going exactly the same speed. If not, the right side would shear off and I'd be in one heck of a mess.

Recession of nodes has nothing to do with the trip around the galaxy, nor with the trip around the sun. What we've been trying to explain to you all this time is the r. of n. would happen even if there were no galaxy, no sun, nothin'! Take an earth that is an oblate spheroid, and a moon, and put them where they are today with respect to one another, in an otherwise completely empty cosmos (except for some very distant "marker" stars so we have a frame of measurement) and the moon's orbital nodes would recess.

The increase of the orbital radius is caused by tidal friction. (This was the thing I was wrong about earlier; I thought it was caused by something else.)

The recession of nodes is caused by torque (there's that word again!) from the earth's equatorial bulge.

If you spin a top, it won't recess. If the top is slightly unbalanced, it *will* recess. (If the top is seriously unbalanced, it won't spin at all...)

(The moon, of course, is doing the same thing to the earth, on a lesser scale since the earth is so much more massive. Newton always wins!)

Silas

DoctorDon,

Welcome back. How could I forget you? By the way Don, did you ever whack that meter stick as I suggested.

I wish to make a suggestion. Rather than telling the world how wrong you think I am, why don't you spend your time wisely and prove that the Earth is at the focus of the Moon's ellipse, rather than the Earth and Moon being at the bases of cones as I suggest. (See previous post)

I've drawn the diagram I've researched the data. All you need to do is prove X = Y rather than X-Y = 8 minutes of angle.


Gary

Both, but it's perihelion

Sure is. Where did you make that prediction?

Did once before. Your turn.

GrapesOfWrath
Quote:
------------------------------------------------------------------------
Both, but it's perihelion
------------------------------------------------------------------------
Blankity blankity proof reader.

Quote:
------------------------------------------------------------------------
Sure is. Where did you make that prediction?
Quote:
------------------------------------------------------------------------
This is a big "IF" somehow my theory is correct, then the precession of nodes by a small less massive body is a very unstable condition such as Silas's "balanced broom stick".


Quote:
================================================== =====================
I truly wish you would help with my calculations.
------------------------------------------------------------------------
Did once before. Your turn.
================================================== =====================
I'm sorry, but I'm sure I'm right.
Here try this on for size.
I keep saying the Moon is going slow correct?
I also say that it's the north south galactic speed correct?
And I just have tried to show the Earth's and Moon's ellipses separated by 550 miles correct?
Big IF the Earth is going faster than the Moon in a southerly direction, then the Earth past the Moon, correct? If the outward motion is 1.5 inch per year, how many years ago did the Earth pass the Moon, and how did it do that?
You see it's not the 240,000 mile you need to work with it's 550.


Gary

If you can tell us why you're sure, you will have answered all my questions.

According to http://www.seds.org/messier/more/mw.html
"Considering the sense of rotation, the Galaxy, at the Sun's position, is rotating toward the direction of Right Ascension 21:12.0, Declination +48:19."

This seems to me to be in a northern direction, rather than a southern. The north ecliptic pole is at 18h +66^o, so the solar system is not moving perpendicular to its plane.

I don't feel I need to repeat your math all the way down to step 35, because I keep finding problems at step two or three. There doesn't seem to me to be any point debating whether or not the moon's orbit is a perpendicular slice of a cone 500 miles north of the Earth's orbit (which, as someone else has pointed out, is not a viable orbital solution, anyway), when you have failed to show that there's any problem with the conventional explanation.

I apologize if you feel attacked by what I'm saying; my intent is not to attack you, but rather your mistakes. I still have seen no evidence that you understand torque in a two-body, free-falling orbital situation. As long as you insist that two such bodies cannot torque each other, I don't see any point to moving further down the chain of logic based on a false premise. Of course, any net torque on the Earth/Moon system would have to come from outside, but the whole point of what we've been arguing is that the torques of the moon and the Earth on each other cancel out, resulting in the Earth slowing down and the moon moving further away. No net torque on the system.

You have not shown why this would not be the case, other than to insist without proof that the moon and the Earth cannot torque each other, by your definition of torque.
I'm going on vacation tomorrow for two weeks. Maybe this thread will have actually died when I get back...

...probably not. [img]/phpBB/images/smiles/icon_smile.gif[/img]

Don

There is no reason for Doctor Don (or anyone else here) to "reinvent the wheel" and "prove" to you accepted math, physics, and astronomy. You are the one with the alternative hypothesis (not even close to a theory yet) and the burden of proof is on you.

__________________
Any day you wake up on "the right side of the dirt" is a good day.

T. Anderson

You want we should do your homework for you? Man, you've got chutzpah! First our textbooks are wrong, and now we've got to do your legwork!

It just doesn't work that way, you know. You made a claim (which, to date, you haven't even fully succeded in making clear to everyone) and thus it's your job to substantiate it.

"And well you may. Yes my word you may well ask what it is, this theory of mine. Well, this theory that I have--that is to say, which is mine-- ...is mine."

Silas

"And well you may. Yes my word you may well ask what it is, this theory of mine. Well, this theory that I have--that is to say, which is mine-- ...is mine."

Anne Elk (Monty Python)

What we need here is a good Gumbyism.


"My brain hurts!"

And the obvious reply:

"It will have to come out!"

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

Just want to correct some erroneous impressions that some may have about the relative contributions of various sources of perturbations.

For planetary satellites there are perturbations from the following sources:

• Other satellites
• The Sun
• Other planets
• Mass distribution of the planet
• Mass distribution of the satellite
• General relativistic corrections

This list is not in order by strength.

The Earth-Moon system is unique in the Solar System in the sense that the direct pull of the Sun on the Moon is greater than the direct pull of the Earth on the Moon. This is true of no other planet's satellite system. As a result, the Sun makes the predominant contribution to the Moon's motion, in particular to the much-argued-about-here precession of nodes and pericenters. The Earth-Moon system shares with the Pluto-Charon system in the lack of other satellites.

For the rest of the satellite systems, the perturbations of other satellites are the main source of perturbations. For the lower-lying satellites the mass distribution of the planet predominates.

_________________
Microsoft is over if you want it.

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-06-27 08:57 ]</font>

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-06-28 01:03 ]</font>

Wow. I had to run the numbers myself before I'd believe it. I get the Sun's direct pull on the Moon to be a little over twice the Earth's direct pull. 4.4e26 N to 2.0e26 N.

The things you learn while hanging about here.

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

Asimov told us this, yea many, many years ago, and it's true. The moon and earth are a double planet. So maybe this whole thread can be disposed of...the moon can do whatever it blinkin' well pleases, 'cause we don't own it.

Don't say that! If we quit our claim, then Martian Jim could annex it!

Silas

GrapesOfWrath
Quote:
------------------------------------------------------------------------
If you can tell us why you're sure, you will have answered all my
questions.
------------------------------------------------------------------------

The astronomy books are wrong.

I've said this over, and over, and over, The astronomy books are wrong.
It took Newton 10-12 years to publish his theories the reason is he had to make sure in his mind that he was correct. Even then it was difficult to break the mold. I have (like Newton) been working on this for 12+ years. I am sure.

It's like DoctorDon and the meter stick. I can't make him put the stick on the floor. I can't make him draw a line I can't make him kick or hit it. He has to do it himself.
This is the strangest thing to sit here day after day and say virtually the same thing and not have a single person listen.

GrapesOfWrath, would you please try to follow this.

DoctorDon
Quote:
================================================== =====================
I still have seen no evidence that you understand torque in a two-body, free-falling orbital situation.
================================================== =====================
I find no place in the physics book that mentions anything like that. I only find it in the astronomy books. It seems contrived to explain away the regression of nodes, and the Earth's apparent slowing.
The universal law of gravitation says that gravity obeys the law of superposition. The law of superposition says that the gravitational attraction between two particles is as if they were the only two particles in the universe. The Earth and Moon are masses of particles. Therefore if the Moon was to try torque or turn any particle within the Earth that torque would be counteracted by an equal internal torque the result would be zero.
Then when you look under centers of gravity you find that all the forces pass through a single point, and all the internal torques equal zero. Free-fall would indicate there is no point of suspension or support. Without a secondary force there can be no torque.

I can explain to anyone on the street how gravity works. But, I can not explain it to astronomers.

Go to the playground and look at the seesaw. If the fulcrum is in the middle it does not turn.
The solution is it needs a fulcrum, and to turn, the fulcrum must be offset. To find or determine the center of gravity you must have a fulcrum or point of suspension. For the Earth or Moon to be torqued they would need a fulcrum or point of suspension. The Earth and Moon fall around each other's, center's of gravity. They are therefore suspended from the very center of their weight. They can not be torqued by their weight if they are suspended from the center of it.

To me that is the most simple and basic of ideas.

Yes, there could be some confusion with the meter stick. I believe I covered all that. My physics book is correct kick the stick falls under impulse and momentum. It has nothing to do with gravity.

So why do we have a "Leap Second" why does the Earth appear to slow down?
Very simple the Earth like the Moon is in a regressive trajectory around the Sun.
From our reference frame with the instruments we have we can not tell the difference. Astronomers have in the past and still do "assume" the Earth in an elliptical orbit with the Sun at a focus of that ellipse. That doesn't have to be and is not so. The Moon regresses about 6040 miles each month. The Earth regresses about 12.5 miles each year. The Earth regression shows itself to us as an axial rotation of .68 seconds.

As the Celestial Mechanic just said, and Geo3gh just confirmed The Sun is the controlling force here. If the Earth were to be slowed by gravity it would be the Sun not the Moon doing the slowing.


Gary

I understand what you are saying. It's wrong.

You must be reading the wrong physics books. I have one in front of me--Physics of the Earth, by Stacey.

That assumes that there is no length or width to the particles. Gravity acts on a mass as if it were a single point only if it is a sphere. If it is deformed, then that is no longer true. The rotation of the Earth can make plumb bobs swing away to point at a point ten kilometers from the center of the Earth--but the oblateness of the Earth makes the plumb bob point away another ten kilometers. Plumb bobs on the other side of the Earth also move away from the center, in the other direction--only half of the effect is due to centrifugal force.

Do you understand now why that is false?

You won't get very far with physicists either. Go ahead, try sci.physics

GrapesOfWrath



I hope you haven't left.


I have always been under the impression that if I went out in the junk pile and grabbed an old tire and a fence post, I could do this.


Bring them to the garage and screw them together like this
first junk .gif drawing. Then hang them from some point as in the second drawing, hang your "Red" plumb line.

Then by doing that a second and or third time, the intersection would be the center of gravity.


Why should the Earth be different? Yes, the Earth is not homogenous it has lumps and holes. My tire has a hole the post is a lump. It still only has one center of gravity as far as the Earth can tell.
I say the Moon presents only one center of gravity to the Earth, and the Earth presents only one center of gravity to the Moon. As the Earth rotates under the Moon as the Moon falls around the Earth the intensity of the field strength may very and their centers of mass/momentum may bounce up and down and jiggle around.

Unlike my tire and post The Earth and Moon gravitational centers are self locating. The reason is, that is the point from which they hang.


The following is a thought experiment.

If during an imaginary instant at the time of a Full Moon, we could stop the Earth, Moon, and Sun, Hang the Earth from the Sun and allow the Moon to pull on the Earth. Rotate the Earth and repeat several times we could thus find the Earth's center of gravity. (In relation to the Moon)


The red X on the blue egg.gif



If we reverse the above to the New Moon phase hang the Moon, etc.
We find that with the Moon's offset core, its center of gravity is proportionately offset.


The red X on the dark and light gray .gif



The Earth and Moon hang by gravity (red line) from each other's centers of gravity (red "X's"). Those red "X's" are free to move at the speed of light from atom to atom.


The Earth and Moon rotate around their center of mass/momentum (barycenter) dark blue dot on red line.

That barycenter is also free to move.

The barycenter moves, because it is the center of the momentum. The ellipse expands and contracts, speeds and slows, the red gravitational line lengthens and shortens, thus proportionately moving the barycenter.



Where have I gone wrong?



Gary

The force of gravity falls off as the inverse square of the distance.

For a sphere, if you do the calculus, gravity acts as if it's acting on a point at the center.

But for a spheroid, that isn't true. You can't simplify the earth into a sphere and explain the moon's orbit.

For the sixth time: you can simplify the earth as a sphere with a "spare tire" around the equator. That "spare tire" has a gravitational pull that *cannot* be considered as a point source at the earth's center.

This is important. But because you've never shown comprehension of the issue of torque, it is difficult for us to explain WHY it is important.

Silas

First, a story...

In 1971, on Hollywood Squares (yes! they had Hollywood Squares back then!) Nanette Fabray was asked, "What is the difference between American ice cream and French ice cream?" Her answer was, "It all depends on how you eat it." Well, this broke up everybody for a good three minutes or so before the program could resume. The answer, by the way, is that French ice cream has eggs in it and American ice cream does not.

What does this have to do with astronomy? Very simple.

• Question: What is the average latitude of the Moon?
• Funny Answer: All depends on how you measure it.
• Real Answer: Zero. In particular, there is no northward drift as Gary Redmond claims in his regression of nodes drawings.

In his drawing, Gary Redmond shows the Moon starting at the ascending node, going around once and crossing at the ascending node one nodical month later and continuing until it has completed a sidereal revolution, which leaves the Moon north of the reference plane. This is the whole basis for his claim that the Moon is drifting north by 540 miles a month.

What if, instead, we had started at the descending node? We would find after one sidereal month that we would have gone around once and then ended up south of our original position.

A simple way to see that there is no drift in the latitude is this--average over the appropriate month, namely the nodical month. If we start at the ascending node, for about 13.6 days the Moon is north of the plane of reference (equator or ecliptic, it doesn't matter which), and then for the remaining 13.6 days in the nodical month after crossing the descending node it is south of the reference plane. The average is evidently zero.

What about perturbations? They may seem to present a problem at first, but consider this: perturbations in latitude can be shown to be periodic. The various periods of all the thousands of terms will not be in synch with the nodical month, so the average over a particular month just might not be exactly zero. It won't be equal to the average the month before or the month after either. Well then, just average over many months. If you do this, you will see that once again the average tends towards zero.

__________________
Microsoft is over if you want it.

The bar has been lowered for the promotion of ATM ideas; the bar for the acceptance of ATM ideas must remain high.

I now officially condemn CM's skits as smartaleck, ignorant, sophomoric, inflammatory and and a poor reflection on the level of discussion in BAUT. -- Bob Angstrom

Silas

Quote:

------------------------------------------------------------------------

The force of gravity falls off as the inverse square of the distance.
For a sphere, if you do the calculus, gravity acts as if it's acting on
a point at the center. But for a spheroid, that isn't true.


------------------------------------------------------------------------

Yes, I certainly know that.

Didn't you look at my drawing?

I placed the Earth's center of gravity caused by the Moon way high and to the right that certainly is not in the middle. The black + is the center of the egg.



Please look at the drawing.

The red X on the blue egg.gif



We must be on the same page before I can make my argument.

The center of mass the black "+" is one thing, and the center of gravity the red "x" is another.



Do you see and understand that?




Gary

Yes, I did.

That has been our goal for quite some time!

Unfortunately, it's wrong. An ellipsoid does not have a "center of [the pull of] gravity." Yes, certainly, it has a center of mass. (The black '+') But it does not have a fixed point that acts as if the body's mass were concentrated there.

Spheres do. A spherical planet can, for purposes of gravitational attraction, be considered a "point mass." It works just fine.

Spheroids do not. The equatorial bulge has its own properties, and the attraction from one side does not "cancel out" the attraction from the other side.

You actually have to do the full numerical integration of the attraction toward every little bit of the spheroid. If you do, you'll find out that it is *not* reducible to an equivalent point.

And this is an experiment you can do for yourself! Whip up a program that takes a spheroid, and do ordinary numerical integration on the inverse square distance between all of its parts to some point (the point must be neither equatorial nor polar.) Say the point is at 45 degrees north. Now do it again where the point is at 20 degrees north. If you sum the attractions up and reduce them to a point...it isn't the same point!

The oblate spheroid will pull "down" (or southwards) on a satellite at 45 Degrees north than it will on a satellite at 20 degrees north. The "center of gravity" -- your red 'x' -- moves around with respect to satellites in different orbital inclinations.

The mathematician and astronomer Colin MacLaurin, a follower of Newton, worked this out in detail. (He also showed that a prolate spheroid, rather a football shape, so to speak, can rotate stably, but is not "stably stable," and will eventually slump towards sphericity. The same is not true of an oblate spheroid.)

Silas

Celestial Mechanic
Posted: 2002-06-30 00:37   
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First, a story...

All depends on how you measure it.

What if, instead, we had started at the descending node? We would find
after one sidereal month that we would have gone around once and then
ended up south of our original position.
A simple way to see that there is no drift in the latitude is
this--average over the appropriate month, namely the nodical month. If
we start at the ascending node, for about 13.6 days the Moon is north of
the plane of reference (equator or ecliptic, it doesn't matter which),
and then for the remaining 13.6 days in the nodical month after crossing
the descending node it is south of the reference plane. The average is
evidently zero.
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Good story. BTW my sister's name is Nannette, with two n's in the middle.

OK, it "All depends on how you measure it".

How true.

I have admitted to one error when DoctorDon set me straight on the direction of the solar system's travel.

You're correct there is no (or insignificant) drift in the latitude.

I had determined that. I'm just not sure what the latitude is.

So.
Let me ask you this:
It appears the Earth-Moon system makes one complete revolution in one sidereal month of 27.321662 days.
That means it has a speed of a certain number of miles per hour or meters per second or what ever at that speed.
The ecliptic plane used to determine the Nodic month passes through the barycenter.

the Moon itself takes one nodic month of 27.212221 days to make a pass once "partially" around the equator. Now we know from the first and second statements that the partial trip around the equator is a shorter distance than a full revolution. If the two months were equal in length it would place the barycenter on the same plane as the Earth mass center and the equatorial plane (sort of). It appears to me that the Moon would have to be at some latitude other than the equator. It looks like a speed distance problem, and it looks to me as if my conic solution is still viable.

What do you think?

Was I headed in the correct direction as to how to prove whether the latitude is other than zero?
Can the Ephemeris be used to show a deviation of latitude from the equator? It certainly seems to me that for navigational purposes it should. Perhaps with the correction DoctorDon has made the Moon's latitude falls in the southern hemisphere. But I don't see how it could possibly be on the equator.


Gary