A Worked Out Example of Forces and Torques (Part 2--Instrumental)

Now we will demonstrate that the Sun exerts a torque on the Earth-Moon subsystem. First, we review the coordinates and forces:

• Sun: ( 0 , R )
• Earth: ( +(M * r) / (E + M) , 0 )
• Moon: ( -(E * r) / (E + M) , 0)

Where:

• E, M, and S are the mass of the Earth, Moon, and Sun, respectively;
• r is the distance from the Earth to the Moon; and,
• R is the distance from the Sun to the Earth-Moon barycenter.

The force vectors are:

• Gravitational force of Sun acting on the Earth: ( -(G*S*E) * (M / (E+M)) * (r / d_E^3) , +(G*S*E) * (R / d_E^3) )
• Gravitational force of Sun acting on the Moon: ( +(G*S*M) * (E / (E+M)) * (r / d_M^3) , +(G*S*M) * (R / d_M^3) )

The signs are reversed from the previous post because now we are considering the attraction of the Sun on the Earth and Moon.

We will now consider the torque on the Earth-Moon subsystem about the Earth-Moon barycenter. The moment arm from the center to the Earth is (M / (E+M)) * r and the component of force perpendicular to it is +(G*S*E) * (R / d_E^3). By the right-hand rule it contributes +(G*S*E*M/(E+M)) * (R * r / d_E^3) to the torque. By a similar computation we see that the force on the Moon contributes -(G*S*E*M/(E+M)) * (R * r / d_M^3) to the torque.

Once again, to lowest order d_E^3 = d_M^3 = R^3 and the torques cancel to this order. However, to the next order we find:

(G*S*E*M/(E+M)) * (R * r / R^3) * ( (-3/2) * (M / (E+M))^2 * (r/R)^2 - (-3/2) * (E / (E+M))^2 * (r/R)^2 )

= +(3/2) * (G*S*E*M/(E+M)) * ((E-M)/(E+M)) * (r^3 / R^4).

There is a torque. What is the meaning of this torque? Because of the malicious choice of coordinates, we see that the configuration chosen was for the moon near first quarter. If we reverse the signs of the x-coordinates of the Earth and Moon we see that the torque will have the opposite sign at third quarter. If we had chosen the Sun, Earth and Moon to be collinear there would be no torque, so there is no torque at new or full moon. The values of the torque that we have computed here imply the Moon is speeding up from new to full, reaching maximum velocity at first quarter, and then decelerates from full to new, reaching a minimum velocity at third quarter.

We have shown the origin of one of the many perturbations of the Moon, a term called the parallactic inequality, which has a period of one synodic month. It is not the largest of the perturbations, that honor goes to one called the evection. However, the evection is a bit too complicated to demonstrate.

This is only a rough calculation in which a lot of the perturbations have been ignored. The best reference to the lunar theory is An Introductory Treatise on the Lunar Theory by Ernest W. Brown, republished by Dover in 1960.

__________________
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.

Whew! And I thought I knew math!

Lovely example of calculation of a torque vector! I gnaw on my own ankle in abject jealousy and admiration!

And an enthusiastic kudos for the "mild chiding" having to do with the calculation of ephemerides tables!

Silas

roidspop
Quote:
=======================================
I wonder why they keep pushing that example at you?

Something to do with conservation of angular momentum?

Maybe that has to do with this problem?

And...replace "arms" with "moon" and you're getting someplace.
========================================

Do you think!

Could Be!

I'm sure of it!

Yep, that's it!

OK, now what?
As I said, shoot the skater to eliminate the external force. The computer program the runs in our head, is an external source. We are not born with it and when we die we leave it behind.
Listen, I know all that stuff it's wrong.
Or at least my way is just as good.
During the last almost 4 months, no one has listened long enough to find a problem with my idea.
Whenever I say something that is against the book, they jump up and down, and wave the book at me.
All I ask is a fair shake. If I say something that is different from what's in the astronomy book, it needs to be verified with physics. If the laws of physics say I'm right then the astronomy book might be wrong. There are times when both can be correct but different. If I make a mistake I'd like a chance to correct it nobody's perfect especially me. Silas and others don't seem to like the way I use some of the terms. That's fine I don't mind being corrected, but beating around the bush or lecturing isn't going to help. I don't believe I make very many mechanical errors they are all here on this keyboard.

This past week has been really weird and I'm still not sure where we are with any of this stuff.
I'm quite certain my interpretation is close to being correct, but I sure don't get any positive feed back.

roidspop
Quote:
=======================================
You mentioned something about bad teaching...this
is a hard medium to teach in! Besides, the ONLY
teacher a person ever has is himself.
Follow?
=======================================
I think I follow!
I really don't want to teach; I just think I have a lot to share.
I'm no genius I just think the astronomy books are wrong.
I'd like to share that if you'd let me.

What if we were to try it my way instead? I mean we've wasted 4 months already what's a couple of weeks?


Gary

Celestial Mechanic,

I just went on line and found what the you had done for me. That was very thought full. I certainly hope you didn't do that on company time, and I think it would have been far more productive if you had looked in the Ephemeris and found out whether the Moon goes around at the Equator or some other latitude. As you may remember I predicted an 8 minute difference in latitude. I still stick by that prediction.
I think you may have misunderstood when I said the Astronomy books; I didn't mean the collected data was wrong. Nor the predictive calculations.
I'm specifically in reference to the "tidal forces" hypothesis. You more or less summed up my conclusions with the following:

Celestial Mechanic
Quote:
-------------------------------------------------------------------
There is a torque. What is the meaning of this torque? Because of the
malicious choice of coordinates, we see that the configuration chosen
was for the moon near first quarter. If we reverse the signs of the
x-coordinates of the Earth and Moon we see that the torque will have the opposite sign at third quarter. If we had chosen the Sun, Earth and Moon
to be collinear there would be no torque, so there is no torque at new
or full moon. The values of the torque that we have computed here imply
the Moon is speeding up from new to full, reaching maximum velocity at
first quarter, and then decelerates from full to new, reaching a minimum velocity at third quarter.
-------------------------------------------------------------------
I would say in laymans terms that the net effect is zero. I found the tides to be likewise. The tide comes in the tide goes out, the ocean heats up the ocean cools down. Again the net effect is zero.


Gary

I predict that the average latitude and the average declination are both zero. Your homework assignment (should you choose to accept it!) is to input the declinations (I think you are more likely to find declinations than latitudes) from an ephemeris for several months into a spreadsheet. Start at the first maximum or minimum for the year, group it into months running from maximum to maximum (or minimum to minimum). Take the average of each month. They should be nearly zero, some positive, some negative. Average over the entire year. This should be about the same, maybe even smaller in magnitude than the monthly averages.
But the predictive calculations are based on the very things that you say are "wrong".You are correct that the net effect of the parallactic inequality is zero. It is a periodic perturbation and the time average of any periodic perturbation is zero. The intent of my example was to show that the Sun causes a torque on the Earth-Moon system and that this torque is a cause of many of the perturbations of the Moon. The Sun is the source of the largest perturbations.

The homework assignment that I have given myself is to find a calculation similar to the one I did above in which I take the inclination of the Moon's orbit into account. I have to show that there is a component of torque that is always perpendicular to the angular momentum vector of the Moon and that it has a constant part. The result of this torque is to cause the angular momentum vector to precess just like a top. This assignment will require that numbers be plugged in so that the period can be calculated. I expect to get very nearly 18.6 years.

__________________
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.

Celestial Mechanic

Or maybe 8.85 years.

Gary

No, I should get something like 18 years. The perigee precesses once every 8.85 years. This motion is notoriously difficult to calculate. (It gave Newton headaches, so imagine what it will do to us!)

It turns out that, ignoring eccentricities, inclinations and the ratio of the Earth-Moon distance to the Solar distance the ratio of the lengths of the various months are series in the ratio of the mean motions, m = n' / n. These ratios are:

• Anomalistic/Sidereal = 1 + 3/4 * m^2 + 225/32 * m^3 + ...
• Nodical/Sidereal = 1 - 3/4 * m^2 + 9/32 * m^3 + ...

The interesting thing is that, to the lowest order in m, the precession rates of the perigee and node are the same. Higher order terms and terms dependent on eccentricities, inclination and ratio of distances changes this. In particular, the value for the node changes very little from the value found in the first approximation, but higher approximations have to be taken before you begin to get a good value for the perigee motion.

The values above are from Newton's Clock--Chaos in the Solar System by Ivars Peterson (published by Freeman), p. 137. I highly recommend this book.

_________________
Microsoft is over if you want it.

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-06 00:58 ]</font>

Celestial Mechanic

Oh! Ok, your useing cheats. I thought you were going to just do a gravity, mass, distance, precession thing.
I was sure that wouldn't work for the nodes because they don't precess.

Gary

I hope everyone had a great holiday.
I certainly did.
I did my homework that everyone seemed to think I should have done.
Celestial Mechanic, thank you for all the suggestions. Never having used the Ephemeris before it took me a little while. Perhaps 20 minutes.
The results exceeded my expectations, so I would rather not brag to much.

I would rather one of you check my work. To make sure I didn't contaminate the results.

I think you will be as unpleasantly surprised as I was pleased.

OK. The Earth is going faster than the Moon around the Galaxy. They are on the same freeway. When did the Earth pass the Moon and how did it happen?


Gary

Elementary Calculation of the Nodal Precession (Part 1--Vocal)

As promised, here is an explanation of the nodal precession in terms of the Solar torque on the Earth-Moon subsystem. Enjoy!

As in the torque example in an earlier post, we ignore general relativistic corrections, other planets, and the distribution of mass in the component bodies, reducing the problem to that of three point-masses. We also ignore the eccentricities, but NOT the inclination.

We define the following longitudes and arguments:

• L, mean longitude of the Moon. Since we are ignoring the eccentricities, this is also the true longitude.
• L', mean longitude of the Sun, also its true longitude for our purposes here.
• Omega, the longitude of the node.
• D, mean elongation, equal to L-L'. This is 0° near new moon, 90° at first quarter, etc.
• F, mean argument of latitude, equal to L-Omega . It is equal to 0° when crossing the ascending node, etc.

The Earth-Moon barycenter is at the origin and the x-y plane is the ecliptic. The coordinates of the Sun are therefore ( R * cos(L') , R * sin(L'), 0 ) = R * R_u, where R_u is a unit vector directed towards the Sun. The coordinates of the Earth and Moon are as follows:

• Earth: -(M / (E+M)) * r * r_u
• Moon: +(E / (E+M)) * r * r_u

The unit vector for the Earth-Moon separation is a little tricky, but we can start with the unit vector in the x-direction (1, 0, 0) and apply three rotations to bring it into position.

1. Start: ( 1 , 0 , 0 )
2. Rotate about z-axis through angle F: ( cos(F) , sin(F) , 0)
3. Rotate about x-axis through angle I: ( cos(F) , cos(I) * sin(F) , sin(I) * sin(F) ). Here I is the inclination of the Earth-Moon orbital plane to the ecliptic. For convenience define c = cos(I) and s = sin(I) so that the above unit vector becomes ( cos(F) , c * sin(F), s * sin(F) ).
4. Finally, rotate about z-axis through angle Omega: r_u = ( ((1+c)/2) * cos(L) + ((1-c)/2) * cos(L-2*F) , ((1+c)/2) * sin(L) + ((1-c)/2) * sin(L-2*F) , s * sin(F) ).

Now the angular momentum vector and the line of nodes vector:

• Orbital angular momentum of Earth-Moon subsystem: h = (E*M/(E+M)) * n * r^2 * h_u. Here n is the mean motion of the Moon.
• Unit vector h_u = ( s * sin(Omega) , -s * cos(Omega) , c)
• Unit vector in direction of line of nodes: N_u = ( cos(Omega) , sin(Omega) , 0 )
• Unit vector perpendicular to angular momentum and line of nodes: ( - c * sin(Omega) , c * cos(Omega) , s ).

The three unit vectors in the above list are mutually perpendicular ("verification left as an exercise for the student!") and form a basis. For reasons that will become apparent later, we are mainly interested in the line of nodes vector.

Now the distances:

• Sun-Earth: D_E^2 = (R * R_u + (M / (E+M)) * r * r_u)^2.
• Sun-Moon: D_M^2 = (R * R_u - (E / (E+M)) * r * r_u)^2.

Once again we will only be concerned with the first order correction in (r/R), so by a process similar to my previous example:

• D_E^n = R^n * (1 + n * (M / (E+M)) * R_u · r_u (r/R) ).
• D_M^n = R^n * (1 - n * (E / (E+M)) * R_u · r_u (r/R) ).

Notice the n this time instead of (n/2) and the presence of a dot product. This dot product is:

R_u · r_u = ((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F) ("verification left as an exercise for the student!")

We can now write:

• D_E^n = R^n * (1 + n * (M / (E+M)) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) (r/R) ).
• D_M^n = R^n * (1 - n * (E / (E+M)) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) (r/R) ).

The forces are then:

• Sun on Earth: (G*S*E) * ( R * R_u + (M / (E+M)) * r * r_u ) / D_E^3
• Sun on Moon: (G*S*M) * ( R * R_u - (E / (E+M)) * r * r_u ) / D_M^3

Next: The torques.

_________________
Microsoft is over if you want it.

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-08 00:51 ]</font>

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-08 00:53 ]</font>

Elementary Calculation of the Nodal Precession (Part 2--Instrumental)

The forces that we just computed are the sums of multiples of two unit vectors, R_u and r_u. Since the cross-product of a vector with itself is zero and our moment arm vector only contains r_u, the torques are simply:

• Earth: -(G*S*E*M/(E+M)) * ( r * R / D_E^3) * r_u &times; R_u.
• Moon: +(G*S*E*M/(E+M)) * ( r * R / D_M^3) * r_u &times; R_u.

So the total torque is:

(G*S*E*M/(E+M)) * ( r * R ) * (D_M^-3 - D_E^-3) * r_u &times; R_u.

The cross-product (also left as an exercise!) is:

( - s * sin(F) * sin(L') , +s * sin(F) * cos(L') , -((1+c)/2) * sin(D) - ((1-c)/2) * sin(D-2*F) ).

Now, if we were developing the full lunar theory we would be considering much more than these few terms, but here we are only interested in a constant part, not the periodic perturbations. We note the following properties of sine and cosine series:

• The product of a sine series with a sine series is a cosine series.
• The product of a cosine series with a cosine series is a cosine series.
• The product of a sine series with a cosine series is a sine series.
• A constant part can only appear in a cosine series.

The cross-product calculated above has three components that are cosine, sine, and sine series respectively. The series for D_E^n and D_M^n are cosine series. Of the three unit basis vectors considered, only the line of nodes unit vector is also of type cosine, sine, sine (0 may be considered as either), so only that component of the torque that projects onto the line of nodes can have a constant part.

This dot product (also left as an exercise!) is:

-s * sin(F) * sin(L'-Omega) = -(s/2) * cos (D) + (s/2) * cos(D-2*F).

The projection of the total torque onto the line of nodes is then:

(G*S*E*M/(E+M)) * ( r * R ) * (D_M^-3 - D_E^-3) * (-(s/2) * cos (D) + (s/2) * cos(D-2*F)) * N_u.

As in the previous example, D_M^-3 = D_E^-3 = R^-3 at the lowest order so that it cancels at that order. To the next order, however we find:

3 * (G*S*E*M/(E+M)) * ( r * R ) * (((1+c)/2) * cos(D) + ((1-c)/2) * cos(D-2*F)) * (r / R^4) * (-(s/2) * cos (D) + (s/2) * cos(D-2*F)) * N_u.

We are only interested in the constant parts, which arise from cos(D) * cos(D) = 1/2 + 1/2 * cos(2*D) and similarly for cos(D-2*F). Therefore, throwing away the periodic terms we are left with:

3 * (G*S*E*M/(E+M)) * ( r^2 / R^3 ) * (-s*(1+c)/8 + s*(1-c)/8) * N_u

= -3/4 * (G*S*E*M/(E+M)) * s * c * ( r^2 / R^3 ) * N_u.

As claimed, there is a constant part of the torque directed along the line of nodes. Because it is perpendicular to the angular momentum it does not change the magnitude of the angular momentum, only the direction.

Now let's calculate the torque another way. The angular momentum of the Earth-Moon subsystem is (E*M/(E+M)) * n * r^2 * h_u, where n is the mean motion of the Moon. In our model we can consider n and r to be constant, so remembering that the torque is also defined as the time derivative of the angular momentum, we have:

(E*M/(E+M)) * n * r^2 * (d/dt)h_u

as the torque. But (d/dt)h_u = (d/dt) ( +s * sin(Omega) , -s * cos(Omega) , c )

= dOmega/dt * ( s * cos(Omega) , +s * sin(Omega) , 0 ) = dOmega/dt * s * N_u. Equating the two expressions for the torque projected along the line of nodes gives:

(E*M/(E+M)) * s * n * r^2 * dOmega/dt = -3/4 * (G*S*E*M/(E+M)) * s * c * ( r^2 / R^3 ) and some cancellation yields:

dOmega/dt = -3/4 * (G*S) * c / n / R^3.

We can massage this a bit more: because the motion of the Sun about the Earth-Moon barycenter is a Keplerian ellipse, we can write Kepler's Third Law for it as N^2 * R^3 = G*(S+E+M), where N is the mean motion of the Earth-Moon about the Sun, and finally obtain:

dOmega/dt = -3/4 * (S/(S+E+M)) * c * N^2 / n = -3/4 * N^2 / n (approximately),

where I'm ignoring (S/(S+E+M)) and c since they are both nearly one.

Next: Number crunching time!

__________________
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.

Elementary Calculation of the Nodal Precession (Part 3--Dance Mix)

In our previous two posts we calculated a first approximation to the nodal precession as:

dOmega/dt = -3/4 * N^2 / n,

where N and n are the mean motions of the Sun and Moon. According to Chapront-Touz&eacute; and Chapront [1] the mean motions of arguments L and D for one year (Julian year of 365.25 days) are:

Mean motion of L (n): 4812.6788&deg;/yr,
Mean motion of D (n-N): 4452.6711&deg;/yr,

so the mean motion of L' (N) is 360.0077&deg;/yr.

Thus dOmega/dt = -0.75 * (360.0077)^2 / 4812.6788 &deg;/yr = -20.1975 &deg;/yr.

The node takes 360/20.1975 = 17.8 years to complete a revolution.

Of course this is only the first approximation, but still this is about 95 percent of the observed value. Not bad for a first approximation!

If we include the next term, +9/32 * N^3/n^2 [2], we find dOmega/dt = -20.1975 + 0.5666 = -19.6309 &deg;/yr, which gives 360/19.6309 = 18.34 years. This is about 98 percent of the result.

Also from [1], the mean motion of F (n-dOmega/dt) is 4832.0202&deg;/yr, so that dOmega = 4812.6788 - 4832.0202 = - 19.3414&deg/yr, which gives 360/19.3414 = 18.6129 years.

As claimed in an earlier post the Sun (not the Earth's equatorial bulge!) is the main source of perturbations in the Moon's motion, accounting for all but the tiniest bit of the precession of the nodes.

The astronomy books do not "lie".

References:
[1] Chapront-Touz&eacute; , Michelle and Chapront, Jean, 1991. Lunar Tables and Programs from 4000 BC to AD 8000, p. 12, Willmann-Bell, Inc.
[2] Peterson, Ivars, 1993. Newton's Clock, Chaos in the Solar System, p. 137, W.H. Freeman and Company, New York.

__________________
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.

Celestial Mechanic,
Quote:
================================================
The astronomy books do not "lie".
================================================
I hope you mean "intentionally". The astronomy books seem to indicate that the Earth is at a focus of the ellipse formed by the Moon's orbit.
I was instructed by yourself and others to do the homework that would prove that. My proof shows that to be wrong. I have not published that proof I think you need to check my work and see for yourself.

As I said it took me about 20 minutes. I assume from what you have demonstrated so far that you could do the job in 5 minutes or less.
As I've said the collected data is correct, it's just that no one has ever looked at it in this manner before.

The truth is this:
The Earth-Moon barycenter lies at the apex of the two cones of which make up the ellipses, through which the Earth and Moon travel, not at the focus of either ellipse. I think you need to check my work and see for yourself.

The distance between the ellipses is (much) greater than what I had predicted.

Gary

<html>Jeff,

Welcome back.



Quote:

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

"Weight" is resisted acceleration. To use the terms in the standard

physics way, you shouldn't use "force" and "weight" interchangeably. It

just leads to confusion. If the Moon were sitting on the Earth, it would

be reasonable to talk about its weight. But in orbit, not.

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

OK, Now what? In my uneducated brain I have certain thoughts that are correct, but I don't seem to be able to express them to you the way you have been taught. For example, in the following I use the word (weights), I mean it as a resistance to the gravitational acceleration. For some reason, inertia doesn't seem appropriate. And centrifugal force is not a force.

How would you say it?



Your explanation of "R cross F", was very good. I am somewhat confused on one or two points though.

When you say: "For every point on the Earth, you measure its distance from the Earth's center of gravity.", do you mean the center of gravity that the Earth gives itself due to its distribution of mass. Or do you mean the center of gravity given the Earth by the Moon, or the one given by the sun. . .



It sounds to me as if you meant the gravitational center caused by the distribution of mass, which is somewhat different from the center of mass/momentum around which the Earth rotates. If that was the center meant, then the remainder of you description is clear in that it establishes the center of gravity given to the Earth by the Moon.

In other words the point around, which, the (weights) of the masses, of the particles, of the Earth, as given by the Moon, are equally distributed.



Whew, now that was a mouth full.

Simply stated, the force exerted by the Moon on the Earth is at a point other than the Earth's center of mass.



If that is so, then "R cross F" establishes the location of that point, which is my "red X" high and to the right just as I had shown in my drawing.

The red X .gif




Now then you have added a new center to the confusion, the Earth's own center of gravity. Somehow I don't see how that can change the outcome.



Quote:

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

As near as I can figure out the "Tidal Forces"

hypothesis was conceived to explain the "Leap Second"

and the "Regression of Nodes".

Isn't that correct?

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

I don't think so. I'm looking right now for info to back me up. Newton

himself was looking at the tidal system, and I don't think that leap

seconds were on his mind.

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

Perhaps not. I do know that as soon as the astronomical pendulum clock was perfected, that it started to show the Earth as having an apparent slowing effect. The exact amount wasn't established until the atomic clock. I do remember when I was a kid the scholastic magazine made a big deal about the first declared "Leap Second". But it had been known for a long time that the Earth appeared to slow down.

George Darwin was really into the "Tidal Force" thing, around 1850's ish.



Quote:

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

But the tides are always in the same place in relation to the Moon. As

the Earth spins through the tidal bulge, it is constantly feeling the

tug of torque. It's a constant braking maneuver.

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

OOPS, that is wrong. Yes, the books lead you to believe that, but it's wrong.



For this
Honolulu tide
chart, I first established the time the Moon crossed the Honolulu meridian or was at zenith. Then using the tide tables I plugged in the time of the associated high tide. The difference in hundredths of hours gives the chart.

The peaks and troughs show the monthly and yearly accelerations.





Gary




<font size=-1>[ This Message was edited by: Gary Redmond on 2002-07-09 14:18 ]</font>

And I must concede my own error: I had thought it was the earth's bulge that was the primary source, simply because it is so much closer than the sun is. I thought it was essentially a "tidal" effect, i.e., varying in strength as the inverse cube of distance, not the inverse square.

(Which is why lunar tides are more significant than solar tides...)

So, as nothing more than a good moral example, I abjure, abhor, and reject my error, and promise to try to do better next time!

Silas

<HTML>Silas

Quote:

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

So, as nothing more than a good moral example, I abjure, abhor,

and reject my error, and promise to try to do better next time!

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

Silas,



I believe I've spoken to you about this before, but I think I need to do it again. You have "no" reason to apologize and back down from these guys.

I've not heard you say anything that isn't taught or written in a book somewhere.

If you remember I said there were several "tidal force" hypotheses.
You learnt one of them. Those guys learnt a different one. Both of those hypotheses contradict each other. That is one of the reasons I chose to develop my own. I figured that if the worlds most learned minds were at odds as to what was going on perhaps they all were wrong.




Please listen to yourself, I'm not sure how long ago you studied this stuff, but you seem to do fine.



Celestial Mechanic,

I'm about to say some unkind things.

Please don't get angry and leave, as you have a very important talent that we (I) need here.



I once worked with a guy who lived by the code, "If you can't dazzle them with brilliance, baffle them with B _ _ _ S _ _ _ ."

It looks like the Celestial Mechanic has some of that philosophy.
You must look past or put on the solar filter when reading his posts.
When you do, you can see that he is human and makes as many errors as the rest of us.



Three months ago we argued and argued about the difference between regression and recession. I lost, and I have corrected my error. It appears the Celestial Mechanic needs to go back and learn that lesson. The Moon's nodes do not precess, or recess. They regress.

He then needs to go study the physics books. No where in the physics book does it teach anything about regression of anything. Therefore the "regression of nodes" must be a unique thing, with its own explanation.
If that is so then the copying of mathematical terms written down by someone else who derived those terms by observation not calculation can hardly be considered brilliant.



Celestial Mechanic,

If you didn't get angry and leave I would like to assign you homework.
As you saw I wasn't impressed with what you gave yourself. My math is only slightly above the high school level so I need help with this.

To prove the latitude of the Moon's ellipse, I used longitude "0", latitude "0", and altitude "0".

That however does not give the true lunar ellipse latitude, it only shows it to be south of the equator.

With some manipulation, it appears to me that the Moon's elliptical plane aligns approximately with the Earth's 7.5 degree southern latitude. That latitude would correspond with the direction correction by DoctorDon.



What I feel would be a true test of the Celestial Mechanic talent, would be a proof of the Earth-Moon ellipse misalignment and the amount. I don't know as that will earn you a Nobel, but it seems as if it would be rewarding to prove the books wrong on that one point.



Here is the
Ephemeris Generator complements of jpl.nasa.gov





Gary

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Mr. Redmond:
I assume that you were dazzled by the brilliance of my calculation of the principal part of the nodal precession. If not, I am going to have to take umbrage and demand an apology. I have done more than merely wave my hands--I have produced a calculation at a fairly elementary level (I don't even use any calculus until I differentiate the angular momentum vector). It is up to you