FURTHER CORRESPONDENCE WITH DR. JEAN MEEUS REGARDING TRANSITS & NODE MOVEMENT
July 23, 2004
Dear Monsieur Meeus,
Please forgive me for my avid interest in the orbital mechanics of the Venus and Mercury transit phenomenon. I do not mean to bother you with all my questions, but it seems to me that you are the only person who clearly comprehends this matter.
It is my understanding that you have recently written an article on possible simultaneous transits of Venus and Mercury, explaining how the longitudes of the orbital nodes of both planets are slowly increasing. This may very well answer most of my concerns. If it does not cause too much trouble, would it be possible for you to send me a copy of your paper?
According to the latest issue of Sky & Telescope magazine, "Mercury’s nodes will catch up to and pass those of Venus in about 11,000 years."
If that statement reflects a correct interpretation of your calculations, does this mean that in about 11,000 years the descending node or the ascending node of Mercury will pass between the Sun and the descending node of Venus?
Based on observations, what is the approximate yearly rate of Mercury's orbital plane movement realtive to inertial space and relative to the equinox?
Thank you so much for your time and understanding.
With kind regards,
Uwe
cc: Walter Cruttenden, Binary Research Institute
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July 25, 2004
Dear Mr Homann,
< ... but it seems to me that you are the only person who clearly comprehends this matter.
This certainly is not true. There are hundreds "mathematical" astronomers over trhe world who understand the matter. I send you as an attached message a pdf file containing the article about simultaneous transits which I wrote in collaboration with Dr. Vitagliano.
< According to the latest issue of Sky & Telescope magazine, "Mercury’s nodes will catch up to < and pass those of Venus in about 11,000 years." < If that statement reflects a correct interpretation of your calculations, does this mean that < in about 11,000 years the descending node or the ascending node of Mercury will pass between < the Sun and the descending node of Venus? Based on observations, what is the approximate < yearly rate of Mercury's orbital plane movement realtive to inertial space and relative to < the equinox?
As mentioned in the article, in A.D. 12720, the longitude of Mercury's ascending node will be equal to that of Venus' ascending node. At that time, yes, the asc.node of Mercury will be "between" the Sun and Venus' ascending node. By about the year 67730, the line of nodes of Mercury will have turned over an additional 180 degrees with respect to that of Venus, and at that time the longitudes of the two ascending nodes will differ by 180 degrees, and so on.
In your "Open Letter" of July 1, you asked "What causes the presumed retrograde motion of the orbital plane of Venus relative to inertial space?" First, it is not a "presumed" retrograde motion, but a real motion. Secondly, the cause of this motion is simply the gravitational action (the attraction) by the other planets. This is known since more than two centuries, and experts in celestial mechanics can calculate exactly the speed of the motion of that line of nodes, just by using Newton's law of attraction.
So, it was not necessary to write an Open Letter to ask that question, as if you had discovered something new, something that cannot be explained by the astronomical science!
There are other incorrect or incomplete statements in your Open Letter. For instance, you write that "three transit pairs occur in a period of 251 years". The actual periodicity is 243 years, not 251. For instance, the transit of June 1769 will be "repeated" in June of the years 2012, 2255, 2498, 2741, etc. There will be no transit in the year 1769 + 251 = 2020.
You wrote that "When a transit occurs, a second one follows five mean synodic periods later." Actually, this is the situation between the years 1500 and 3000. Sometimes the Venus transits are "isolated". For instance, there was a Venus transit in the year 910, but none in 902 and none in 918.You can find the complete list of the transits of Venus between the years -2000 and +4000, and all Mercury transits between +1600 and +2300, in my booklet "Transits" which was published by Willmann-Bell in 1989 and which is still available.
You also wrote "The orbital planes of the other planets do not show a similar motion...". This is not correct. The lines of nodes, eccentricities and inclinations of *all* planets are subject to so-called secular variations. For instance, relative to inertial space, while the line of nodes of Venus rotates by -0.278 degree per century (the negative sign indicating retrograde motion), that of Mercury's orbit moves by -0.125 degree per century, that of Mars by -0.295 degree, that of Jupiter by +0.177 degree, and so on.
Sincerely Yours.
Jean Meeus
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July 26, 2004
Dear Mr. Meeus,
Thank you for sending me your paper on "Simultaneous Transits" and for your comments regarding my Open Letter of July 1, 2004.
If the line of nodes of Venus increases by about +0.90112 degree per century and Mercury's by about +1.18619 degree per century relative to the mean equinox of date, are you sure about the figures that the line of nodes of Venus rotates by -0.278° degree per century and Mercury's by -0.125 degree per century relative to inertial space?
Do these motions imply that the fundamental position of the mean equinox, which remains fixed in the dynamic reference frame of a 360-degree equinoctial year (Earth's orbit around the Sun), also rotates (by approx. 5026" per tropical century) relative to inertial space?
Sincerely,
Uwe
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July 27, 2004
Dear Mr Hohmann,
< If the line of nodes of Venus increases by about +0.90112 degree per century < and Mercury's by about +1.18619 degree per century relative to the mean < equinox of date, are you sure about the figures that the line of nodes of < Venus rotates by -0.278° degree per century and Mercury's by -0.125 degree < per century relative to inertial space?
Yes, certainly. You can find the various expressions for the mean elements of the planetary orbits, referred to both the mean equinox of the date and to the fixed equinox of 2000.0, on pages 212-215 of my book "Astronomical Algorithms" (2nd ed., Willmann-Bell, ed.; 1995). Those expressions are not mine, but were taken from still more "complete" expressions given by Simon, Bretagnon, Chapront, Francou and Laskar in "Astronomy & Astrophysics", vol.282, pages 663-683 (1994).
For instance, T being expressed in Julian centuries (of 36525 days) from 2000 January 1.5 Dynamical Time, and T2 indicating T*T and T3 the cube of T, and the coefficients being given in degrees and decimals, we have for the longitude of the ascending node of Venus :
Referred to the mean equinox of the date : 76.679920 + 0.9011206 * T + 0.00040618 * T2 - 0.000000093 * T3
Referred to the fixed equinox of 2000.0 (that is, to inertial space) : 76.679920 - 0.2780134 * T - 0.00014257 * T2 - 0.000000164 * T3
There are similar expressions for the orbital elements for all planets, from Mercury to Neptune.
< Do these motions imply that the fundamental position of the mean equinox, < which remains fixed in the dynamic reference frame of a 360-degree < equinoctial year (Earth's orbit around the Sun), also rotates (by approx. < 5026" per tropical century) relative to inertial space?
I don't understand what you mean. The mean equinox (of the date) does NOT remain fixed in the dynamic reference frame. The dynamic reference frame (of 2000.0) is inertial, of course, and doesn't rotate.
Regards.
Jean Meeus
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July 27, 2004
Dear Mr Hohmann,
Sorry, I made an error at the end of my previous message.
The dynamical equinox is NOT fixed in space. It is defined for instance by Bretagnon's VSOP planetary theory. The equinox is the intersection of the celestial equator and the plane of the ecliptic. Both planes are changing, the first due to the precession, the second because the orbital planes of all planets are changing in the course of the millennia.
The dynamical equinox is the "true" equinox, defined by the rotation of the Earth (the equator) and the dynamics of the planetary motions. The dynamical equinox differs very slightly from, for instance, the standard FK5 system used for the stars (which is slightmy in error), but this is irrelevant to your questions.
Planetary positions or orbital elements can be referred to the mean equinox of the date (that is, neglecting the periodic terms of the nutation), or to a standard (fixed) equinox. The latter is simply the (mean) equator and the ecliptic at a certain date, for instance 2000.0. The latter is, of course, fixed.
Regards.
Jean Meeus
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August 2, 2004
Dear Mr. Meeus,
I had been away for a week and your last two emails caught me just before leaving, so I apologize for responding so late.
Thank you for clarifying the difference between the reference frame of the dynamic equinox and the fixed (historic) equinox (2000.0), which represents inertial space (at least for a certain period of time). My previous question was meant to get a general understanding of which reference frame astronomer use in order to determine the real motion of the line of nodes and to calculate and predict transits. I think you will agree with me that if precession is caused by a wobble of the Earth's axis of rotation, the historically observed forward or prograde motion of the lines of nodes (e.g. Mercury, Venus, Mars) relative to the Sun would merely be an "apparent" motion.
In the case of Venus, for instance, very accurate formula exists that describes the "apparent" forward motion of the line of nodes (+0.90112°/century or approx. -32.44"/year) relative to the dynamic equinox. It seems, however, that the "real" retrograde motion of about -17.82"/year relative to inertial space (50.26"/year real motion of the equinox minus the 32.44"/year apparent motion of the nodes) is not well defined or expressed by a formula. If the line of nodes have retrograded by -0.278°/century or approx. -10.01"/year, as measured with respect to a "fixed" equinox, then over the last few centuries from which inertial point of reference has the line of nodes been observed to retrograde by about 17.82"/year or about -0.495°/century?
Best regards,
Uwe
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