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

</HTML>

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 to re-do my math until you understand it. If you cannot do it yourself, find someone to explain it to you in person and fill in the gaps in your mathematical knowledge until you do understand it.
You claim that you have seen no explanation of the precession of the node in any of the physics books. That is because of the groundwork necessary to develop a suitable vocabulary for stating and solving the problem. Physics books have other fish to fry and cannot be faulted for wanting to move on to topics more general than the Moon's motion. You will find the explanation in astronomy books, where the time is spent on developing the important prerequisites that make a calculation such as mine possible. You will need to read something at a technical level higher than Astronomy for Dummies to get a complete explanation. Here are three references from books that I own. A visit to a university library would undoubtedly add more.

1. Brown, Ernest W., An Introductory Treatise on the Lunar Theory (1960, Dover, originally printed 1896) pp. 106-109. Still the classic introductory work on the subject. The derivation cited goes as far as -3/4*m^2 + 9/32*m^3 - 273/128*m^4. There are other derivations in the book, including the Hill-Brown method using an infinite determinant.
2. Danby, J.M.A., Fundamentals of Celestial Mechanics, 2nd ed. (1988, Willmann-Bell) pp. 374-5. Excellent introduction to celestial mechanics, period.
3. Moulton, Forest Ray, An Introduction to Celestial Mechanics (1923, MacMillan, New York). Explains source but does not give a number. Classic text, a bit dated.

My derivation is entirely original but I suspect someone must have also done it in nearly the same way. In putting it together I learned something, namely the importance of showing that the constant part lies along the line of nodes.

As for the use of the word precession vs. regression: the Moulton book above uses the word regression so this use goes back at least that far. A more recent book, Solar System Dynamics by C.D. Murray and S.F. Dermott on p. 256 says the following: "Prograde motion of the pericentre (or node) is called precession and retrograde motion is called regression." It is their book and they can write what they like, but I think this is a bad usage. We do not distinguish the precession of a spinning top according to the sense of the precession, we just call it "precession". We also have the usage of "precession of the equinoxes" or "luni-solar precession" for the precession of the Earth's rotation axis, which is in a retrograde direction.

I call the principal secular part of the motion of one axis about another the precession and give the sense if I think it is important, and the periodic deviations from this motion I call "nutation".

See almost any advanced mechanics book about the precession of tops, etc.
My math level is way beyond high school. I tried very hard to keep my demonstration as elementary as possible.

If your math is only slightly above the high school level then you are not qualified to judge my presentation. You will need to learn at least the following in order to be qualified to make statements about celestial mechanics:

• Vector addition, subtraction, multiplication by scalars, and the dot and cross products of two vectors.
• Calculus, both differential and integral. You will need to be able to perform volume integrals.
• Mechanics at the university level, preferably graduate level. Topics include Newton's laws, rotational motion of rigid objects, Lagrangian and Hamiltonian dynamics.
• Some understanding of special and general relativity in order to calculate corrections to the Newtonian mechanics.

Once you have learned these things then maybe you will be able to make as worthy a contribution to science as you wish.

_________________
Microsoft is over if you want it.

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-12 00:38 ]</font>

Oops! Remembered a few more useful things that a practicing celestial mechanic must know:

• Elementary differential equations, in particular about initial and boundary conditions.
• Some of what are called the advanced or engineering functions, such as Bessel functions, hypergeometric functions, elliptic functions.
• A few of the orthogonal polynomials, particularly Legendre polynomials.

_________________
Microsoft is over if you want it.

<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-12 13:26 ]</font>

Celestial Mechanic
Quote:
-------------------------------------------------------------------
I assume that you were dazzled by the brilliance of my calculation
of the principal part of the nodal precession.
-------------------------------------------------------------------
I ***-u-me very little, and it takes a great deal to dazzle me.

Could you please explain to us why you refuse to do the 5 minutes of homework I asked of you?

Do you think that to prove the astronomy books wrong, is such a trivial thing, that it is unworthy of someone such as yourself?


Celestial Mechanic
Quote:
-------------------------------------------------------------------
See almost any advanced mechanics book about the precession
of tops, etc.
-------------------------------------------------------------------
Yes, my point exactly.
In those books it explains what is called a right hand rule. Jeff just told us about "R cross F", and how that calculation used the right hand rule. You yourself, have mentioned the right hand rule several times. Even DoctorDon has told us about the right hand rule.

If we apply the right hand rule to the Earth-Moon system, we find the Moon's perigee precesses (as it should).

The nodes do not precess. OOPS!

If we go back to those books, and work our way from cover to cover, no where do they give a left hand rule to cover regression. OOPS!

The regression of nodes is an "unnatural" motion as far as the physics books are concerned. OOPS!

If you go to most any machine shop and talk to the machinist there, they can give you hundreds of examples of regression.
Those examples involve speed differences.
The Earth goes around the Galaxy at one speed the Moon goes at another (slower) speed. The result is regression.

Hint: If we are going down the freeway one at 55 mph and one at 65 mp, one of us will regress.

That speed difference also causes the Moon's ellipse to trail the Earth's ellipse. The difference is more than 500 miles.

Would you please do your homework. Arguing with a crazy old man is not very becoming of such a brilliant mind.


Gary Redmond

I did a quick Google search this morning. The results:

• "precession of nodes" -- 9
• "regression of nodes" -- 20
• "precession of equinoxes" -- 447
• "regression of equinoxes" -- 0
• "lunisolar precession" -- 133
• "lunisolar regression" -- 0
• precession +"spinning top" -- 1,100
• regression +"spinning top" -- 82

The word regression is misleading and should be replaced in all cases by precession.

__________________
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 word regression is misleading and should be replaced in all cases by precession.
-------------------------------------------------------------------
I sure hope our celestial never breaks down because I sure wouldn't want you working on it.

What you just said made as much sense as if I said the "E" and "F" on the fuel gage meant the same thing, or "CW" and "CCW" motors, are interchangeable, or a lug nut marked with "L" can be replaced with one marked with "R".
I would love to give an automotive example, but it appears that it's no longer relevant whether I put a chalk mark on the vibration dampener or the flywheel, when I use the timing light to precess or regress the ignition timing.

Regression and precession are the exact opposites.
To precess, means to go in front of or before.
To regress, means to go behind or in back of.

Nodic 27.212221 days

Sidereal 27.321662 days

Anomalistic 27.554550 days

Are you telling me there is no difference between the Nodic month and the Anomalistic month?

Are you telling me that if we spin a top or a gyroscope that the right hand rule no longer applies?
Are you saying that if I set that spinning top, down on the floor it will randomly go in which ever direction it feels like?

I think you need to do a lot more homework than what I gave you.
I am completely baffled.


Gary Redmond

Precession is the secular part of the motion of one axis about another, regardless of its sense of of that motion. Nutation is the periodic deviation from that motion. I feel that that is a far more reasonable definition than the inconsistently applied practice of calling one sense precession and the other regression.
You are fond of quoting these. Can you calculate the ratios of anomalistic and nodical to the sidereal with your incoherent theory? I think not. I have demonstrated how the computation of the principal part of the nodical precession can be done. Instead of studying and understanding you have chosen to insult me, and yet you claim to need the talents that I offer. You cannot have it both ways.
If we spin a top one way, it will precess in a certain sense. If we then spin the top the other way, it will precess in the other sense. The physics books that you keep referring me to do not make this artificial distinction that the astronomers seem to do, and inconsistently at that.

I have in front of me at the moment my graduate level mechanics book, Classical Mechanics, 2nd edition by Herbert Goldstein, (1980, Addison-Wesley), and guess what? The index has not a single reference to "regression" of anything. Under "precession", however, we find the following sub-entries:

• of angular velocity vector in Poinsot motion
• charges in magnetic field
• of elliptic orbits
• of the equinoxes
• of heavy symmetrical top (see also Top)
• Larmor (see Larmor frequency)
• lunisolar
• of perihelion of Mercury
• of symmetric body
• Thomas

There are also entries for Precession frequency and Precession vector.

Here is where the astronomers should follow the physicists in order to make the nomenclature clearer, and I follow it.

About my background, I will say that my undergraduate degree is in Mathematics and my master's degree is in Physics. What I know of celestial mechanics I know forward and backward, and what I don't remember I can either derive it myself (such as the principal part of the nodical precession), or I know where to look it up.

You have admitted that your mathematical/physical knowledge is inadequate to the task at hand. You do not know what you are talking about when you talk about celestial mechanics. I do.

__________________
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.

<html>Celestial Mechanic,



Please forgive me, and my insults. It appears I got your attention, and you didn't get angry and leave. This is good. Can we talk without all the baggage? Yes, I can see that you are extremely talented. That is why I need you.



As I have said the astronomy books (not the Ephemeris) are wrong, and I need help from people like yourself to prove it. I can't do it by myself. And you can't do it if you keep looking in those books for the answers.



Quote:

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

I have in front of me at the moment my graduate level mechanics book,

Classical Mechanics, 2nd edition by Herbert Goldstein, (1980,

Addison-Wesley), and guess what? The index has not a single reference

to "regression" of anything. Under "precession", however, we find the

following sub-entries:

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

Very good initiative. That's not the homework I asked for, but it's good.

As you may remember in my automotive example, "I use(d) the timing light to precess or regress the ignition timing." The words regression and precession are not used, and those motions only take place during the time the distributor is being turned. They call it advancing or retarding the spark. It has to do with the differences in speed, as the distributor is turned at a speed independent of the engine.



Note: It's been a few years since cars have had distributors, do you even know what I'm talking about? By rotating the distributor back and forth you impart a different speed (frequency) to the strobe light.



To prove the books wrong (if you choose) we need to look at things a little differently. You guys may be right the books may not be wrong. But, I think you should try it my way.

You know when you first start learning about relativity you have to change the way you think. Then space and time take on a whole different aspect.



The homework I asked of you would as I said only take about 5 minutes. That little exercise would prove to you that the Earth ellipse and Moon ellipse, lie in different planes. Could you please do it just for kicks.



Because the Earth is traveling in a northerly direction (thanks DoctorDon) at a faster speed, its ellipse is preceding (precesses). The Moon is being pulled along because it has a slower speed. Thus it regresses or falls back. Precess and regress are the opposite ends of the same motion.

If the Earth and Moon ellipses had the same focus then there would be no precession and no regression.



Conic orbit02.gif





Quote:

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

You are fond of quoting these. Can you calculate the ratios of

anomalistic and nodical to the sidereal with your incoherent theory?

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

Yes, I believe so. I'm sure that the following is not what you mean, but back on page 6 of this thread I showed how to derive one from the other. Here is the ratio I came up with 6793.460443813 / 365.25 .



Given:



Nodic 27.212221 days.


Sidereal 27.321662 days.


To prove: 18.6 yr.


27.321662 - 27.212221 = 0.109441


27.212221 / 0.109441 = 248.6474081925


248.6040562763 * 27.321662 = 6793.460443813


6793.460443813 / 365.25 = 18.59948102344



Close enough?



To do what I think your asking, we need the galactic and orbital speeds, and your talents of course. That way you could calculate the difference between the two helixes.



Humm? That sounds confusing.



This may not be any better, but I'll try.

The Earth's center of mass travels an elliptical path as it moves around the galaxy; that motion is helical.

The Moon does likewise.

The Earth and Moon don't share an elliptical focus, but the distance between them forms opposing conical planes.

Their speed difference, is the cause of the angle of the cones.



??????????????




Gary

</HTML>

Question marks indeed... What are "opposing conical planes?" Do you at least realize that "conical plane" is oxymoronic? Do you realize that none of us has the slightest idea of how to make your observation... because you haven't defined one.

The very few times you have made actual concrete predictions, they have been wrong. (e.g., you said the moon had a proper motion of 18 mph with respect to the earth, something that would be painfully obvious to everyone on earth in less than a year.)

If you want us to set up a global observation network to "shoot the moon" once a night for the next 30 days and plot the results on a globe...well...we aren't going to. That's just too much work.

However, if one of our antipodeal members is willing to make observations roughly in simultaneity with me, on, say, the 1st, 8th, 15th, and 22nd of some given month, we might be willing.

You and the Celestial Mechanic (who has more math than I do!) can predict what the measurements will be, without telling us in advance (the B.A. might be willing to hold them in escrow) and then we'll see who was closest.

Anybody "Downunda" willing to waste a little time with a surveyor's transit?

Silas

Why the Galaxy Does Not Matter (to Solar System Dynamics)

Some contributors to this thread believe that the motions of the Earth and Moon relative to the galactic center are significant, that somehow the Moon is travelling "farther" and "slower" and is being "left behind". We will examine the actual numbers in a quick'n'dirty little "back of the envelope" type calculation. We will use very round but easy to remember values for most quantities and we will round numbers with a vengeance. Numerical precision is not the goal here, enlightenment is.

The general consensus is that the Solar System orbits the galactic center once every 220 million years at a distance of 30,000 light-years. As an example of ruthless rounding, we will take the number of seconds in a year to be about 10^7 * pi, so that the period is (2.2*pi)*10^15seconds, hence the mean motion (in radians per second) is (2*pi)/(2.2*pi)*10^15s = 10^-15s^-1 (approximately). Likewise, a light-year in round numbers is 10^13km, so 30,000 light-years is 3*10^20m.

By Kepler's Third Law, N^2 * R^3 = G*M_eff, where N is the angular mean motion (10^-15s^-1), R is the radius vector (3*10^20m), G is the gravitational constant (6.67*10^-11m^3/kg/s^2), and M_eff is the "effective mass". The effective mass is our way of setting aside the distribution of mass in the galaxy, which we do not have much data on and the details of which are irrelevant to this discussion. The effective mass is:

M_eff = (10^-15s^-1)^2 * (3*10^20m)^3 / 6.67*10^-11m3/kg/s^2 = (27*10^30 / 6.67*10^-11) kg = 4*10^41 kg, approximately.

Since the mass of the Sun is 2*10^30 kg, this means that M_eff is about 2*10^11 solar masses, which sounds about right.

The velocity of the Solar System in its orbit is N * R = 3*10^5m/s = 300km/s, which is plausible.

The centripetal acceleration is N^2 * R = 3*10^-10m/s^2, which is extremely small. By comparison, the acceleration of the Moon toward the Sun is G*S/R^2 = (6.67*10^-11m^3/kg/s^2)*(2*10^30kg)/(1.5*10^11m)^2 = 6*10^-3m/s^2.

But this is the acceleration experienced by the entire Solar System and thus does not affect the positions of the planets relative to one another. In order to have perturbations, we must have tidal forces, which are weaker by a factor of 2*d/R, where d is the distance between the two bodies experiencing the tidal perturbation. For the Earth and Moon, this means a factor of 2*(4*10^8m)/(3*10^20m) = 3*10^-12 weaker for an acceleration of 9*10^-22m/s^2, a ridiculously small quantity.

Therefore, as far as Solar System dynamics is concerned, the rest of the galaxy might as well not exist. It creates no measurable perturbations in the orbits of the planets.

__________________
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Why the Center of Mass Motion of the Solar System Does Not Matter (to Solar System Dynamics)

Some contributors to this thread believe that the motions of the Earth and Moon relative to the galactic center are significant, that somehow the Moon is travelling "farther" and "slower" and is being "left behind". In this essay we will point to an important fallacy in this theory.

The laws of gravitation (in both the physics and the astronomy books) make no mention of distances traveled. There are distances between pairs of bodies and accelerations of single bodies. (We ignore comets and other bodies that lose mass.) We can add a vector X = At + B to all the coordinates, where A and B are arbitrary vectors and t the time. (Well, A can't be too big, then we have to start using relativistic corrections!) The potential energy only contains differences of coordinates, so X cancels out. X contains the time only as far as the first power, so its second derivative is zero and contributes nothing to the accelerations, therefore it does not affect the equations of motion.

You can draw helixes all you like, but any one of them is just as good as any other. The quantities that differ from helix to helix, such as the arc distance traveled, cannot matter. Only those quantities that are invariant and do not depend on the helix chosen are relevant.

Since all of the helixes are equivalent, we choose the one that is easiest to work with, namely the one in which the center of mass of the Solar System is at rest.

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

On the Proper Use of Conic Sections

Some contributors to this thread believe that the motions of the Earth and Moon relative to the galactic center are significant, that somehow the Moon is travelling "farther" and "slower" and is being "left behind". As evidence of this claim, a further claim is made that the Earth and Moon orbit on the surface of a cone, with the Earth on one half and the Moon on the other, and the Earth-Moon barycenter at the vertex of the cone. The claim may also be worded as the barycenter of the Earth-Moon subsystem does not lie in the plane of the Earth's orbit, but at a distance of 540 miles from it. This is utterly and completely wrong.

As mentioned in a previous post, we can always imagine that at some time we can turn off all the other forces except for the mutual attraction of the Earth and Moon. If we could do this, the Earth and Moon would follow a Keplerian orbit determined solely by the positions and velocities at the moment that the perturbations were cut off. This orbit is called the osculating orbit, and it represents the Keplerian orbit that best approximates the motion, just as a tangent to a curve is the line that best approximates a curve in the vicinity of the point of tangency.

Does the conic surface model represent a viable two-body solution? No, and here are two reasons why.

1. The relative position, r, and relative velocity, dr/dt, both define a plane. The second derivative, d^2r/dt^2 is equal to -G*E*M/r^3 * r, that is, it is in the same direction as r. This is just another way of stating the universal law of gravitation. All the higher derivatives turn out to be linear combinations of r and dr/dt, therefore not only are r and dr/dt in the same plane, but also all future (and past) values of r are in the same plane. GR's conic surface model has r pointing along different generators of the cone, and they do not lie in the same plane.
2. In two-body motion, the only direction along which forces act is the r direction. The angular momentum per unit mass is r dr/dt, and the first derivative of this is:
   
   d/dt(r dr/dt) = dr/dt dr/dt + r d^2r/dt^2, both terms identically zero.
   
   In GR's conic surface model, r points along a generator of the cone and dr/dt is tangent to the surface of the cone. The cross-product of these two vectors points away from the vertical axis in a different direction for each different position r. The result is that not only is GR's angular momentum vector not constant, it (Oh! The Horror!) precesses in a prograde direction with a period of one sidereal month. We all know that that is wrong. The angular momentum precesses in the retrograde direction with a period of 18.61 years.

How can we fix this? The only way to fix number 1 is to flatten the cone until all generators become coplanar, which flattens the surface of the cone into a plane. This also fixes 2 because now the angular momentum vector points perpendicular to this plane and along what had been the axis of the cone, and is now constant.

In short, we can look at the two-body solution in two different ways:

• We can regard the two bodies as each orbiting the center of mass in coplanar elliptical orbits with the center of mass at one focus. These orbits are similar, that is they are the same shape (same eccentricity) just different sizes; or,
• We can regard one body (doesn't matter which) as orbiting the other in a planar elliptical orbit with the first body at one focus. In this frame the center of mass moves along with the orbiting body, always on the radius vector from one body to the other.

Both of these views are useful and I use either one as the situation requires.

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<font size=-1>[ This Message was edited by: Celestial Mechanic on 2002-07-14 23:48 ]</font>

As I have noted previously, the Ephemerides are based on these very concepts that you reject. If "the astronomy books" are wrong, the ephemerides based on them must be wrong too.
Not even in the ballpark, but maybe I was not clear. Given that the sidereal month is 27.321662 days, and the sidereal year is 365.256363 days, and the semimajor axis of the Earth-Moon barycenter's orbit about the Sun is 149.6 million km and the semimajor axis of the Moon's orbit about the Earth is 384,400 km, and only these numbers, prove that the nodal month is something near (you won't get all of it but you should get most of it) 27.212221 days.

Actually, you don't really need the mean distances. If you look at my derivation only the Sun's mean distance was in the answer, and I used Kepler's Third Law to remove it.

In other words, calculate the length of the nodical month from first principles, don't quote it. I did so in that three-part post.

__________________
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.

<html>Silas,

Quote:

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

Question marks indeed... What are "opposing conical planes?" Do you at

least realize that "conical plane" is oxymoronic? Do you realize that

none of us has the slightest idea of how to make your observation...

because you haven't defined one.

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

How do I get myself in these messes? BTW I built a round house with a coned roof. That 2600 square feet was pretty flat I'd call it a plane.



OK, did you look at the drawing?

Conic orbit02.gif



The Earth's mass is gravitationally bound to the Moon's mass. As they rotate, the imaginary line between their gravitational centers passes through the intervening space. As it does, the figures formed are two cones with mutual apexes, which meet at the Earth-Moon barycenter. This may be a stretch of the imagination, but if you roll a "plane" (sheet of paper) it can form a cone. Therefore I would call the surface of a cone a plane. Because those cones meet apex to apex I would call them opposing.

Thus "opposing conical planes".



Quote:

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

The very few times you have made actual concrete predictions, they have

been wrong. (e.g., you said the moon had a proper motion of 18 mph with

respect to the earth, something that would be painfully obvious to

everyone on earth in less than a year.)

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

It wasn't 18; I believe I said 13 mph. I am not completely wrong; you just haven't gotten up to speed so to speak.



I'll quickly run through the math for you however it most likely won't make much sense.



The difference between the Nodic month of 27.212221 days, and the Sidereal month of 27.321662 days, is equal to 0.109441 days.

The circumference of the Moon's orbit is approximately 1,507,000 miles.

If you divide 1,507,000 miles, by the sidereal month of 27.321662 days,
you get 55157.6986788 mile per day or 2298.23744495 miles per hour.

Then 0.109441 days times 24 equals 2.626584 hours.

So we have a total of 6036.513701107 this is the number of miles that the Moon falls short each Nodic month of 27.212221 days.

Then 27.212221 days times 24 hours equals 653.093304 hours per nodic month.

Therefore 6036.513701107 miles divided by 653.093304 hours equals approximately 9.242957574569 miles per hour.

You're correct I've made an error somewhere.

Can we split the difference at 10 mph until we can get a Celestial Mechanics opinion. Now then do we call that a -10 or just 10.


It is the direction of that speed that makes it so inconspicuous.



However:

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

Quote:

The orbit circles westward at a rate of 0.053 degrees a day. So, at this rate,
it takes 6,793.5 days (18.6 years) to travel 360 degrees or a complete

cycle. The tiny change is too insignificant for us to notice with casual

glances at the Moon. However, Hipparchus in the 2nd century B.C. observed

it and so did the Anasazi more than a thousand years ago in the American

Southwest.

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




Quote:

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

Anybody "Downunda" willing to waste a little time with a surveyor's

transit?

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

I don't think that will be necessary. The Ephemeris is more than adequate. If thousands of ships and airplanes, worth millions of dollars each, rely on it for navigation it'll tell us what we need. Using the Ephemeris "With some manipulation, it appears to me that the Moon's elliptical plane aligns approximately with the Earth's 7.5 degree southern latitude."




Gary

</HTML>

Um... I don't get it. One you roll the plane, it is no longer planar. You're once again writing your own dictionary.

My apologies: it doesn't matter, of course: any number much greater than 0.00 is going to result in predicted observations differing from actual observations.

The moon's orbit is *almost* circular, but at the level of numerical precision you're using, you can't treat it as a circle any longer.

Math question: do you know how to calculate the area and periphery of an ellipse? If I give you a lovely egg-shaped ellipse, where the semi-major axis is 10 and the ellipticity is 0.3, can you work out the area and periphery? It isn't a trivial problem. (In fact, I'd probably cheat and use numerical integration.)

Silas