Chapter 12: Relative motions of Sun & Earth
Each year, planet Earth (travelling at the tranquil speed of 1.6 km/h) covers a distance of ≈14036 km along its PVP orbit. This distance equals amounts to 0.0039457% of the PVP orbit’s circumference of 355 724 597 km. From one year to the next, the Earth and the Sun will thus meet up at a slightly 'earlier' point in space, by an amount corresponding to a 0.0039457% “slice” of the solar orbit’s circumference.
Earth moves each year by 14036 km (which represents 0.0039457% of its PVP orbit's circumference of 355 724 597km).
If we multiply 14036km by 2.64233 (so as to obtain the equivalent circle-sectional distance projected onto the Sun’s larger orbit) we obtain:
14036 X 2.64233 ≈ 37088 km (which is, in fact, 0.0039457% of the Sun's orbital circumference of 939 943 910km).
Here is how this can be illustrated:
In our epoch (2000AD), the firmament is observed to drift 'eastwards' by about 50.3" arcseconds annually. However, as I shall expound and illustrate further on in this chapter, the true annual eastward drift of the firmament is 51.136" arcseconds (i.e. about 1.68% more than the observable drift).
Please take note of this 51.136" value which I will henceforth refer to as the 'ACP'(Annual Constant of Precession)
Note that 51.136(periodic) arcseconds equals 0.0039457% of 1 296 000 arcseconds (i.e. 0.0039457% of 360°)
And in fact, 51.136" X 25344 (the number of years in a TYCHOS Great Year) = 1 296 000" - or 360°
Hence, the so-called "precession of the equinoxes" is caused by the Earth’s clockwise motion around its PVP orbit.
We shall now see how the TYCHOS model accounts for the so-called "sidereal and solar days".
In the TYCHOS, the sidereal day occurs as follows: the Earth employs 23h56min to complete one 360° revolution around its axis - and to realign with a given star"X" (this is what is known as the "sidereal day"). During that time, the Sun will move “eastwards” in relation to the stars by about 1°. Hence, an earthly observer will have to wait another 4 minutes or so to realign with the Sun: a 'solar day' (of 24 hours) will then be completed. Here's how this can be illustrated:
You may agree that the TYCHOS model accounts for the sidereal & solar days in the simplest imaginable manner. As we shall see shortly, the Copernican theory’s explanation for the solar and sidereal days is not only complicated; it is inherently unphysical.
'Unphysical': "Not supported by, or contrary to, the laws of physics." "unphysical" - Wiktionary's definition (opens in a new tab)
The Sun moves every day by about 1° (or 4 minutes of RA) in relation to the background stars
This fact alone suggests that it is indeed the Sun (and not Earth) that moves each day by 2 573 424km, for this value equals roughly 2X the Sun’s diameter of 1 392 000 km. Since the Sun’s apparent size in the sky subtends about 0.5°, it makes perfect optical sense that its observed daily displacement (of about 1°) amounts to about twice its visible diameter.
Let us now verify if the notion that the Sun orbits around Earth (and not vice versa) can be further confirmed.
As we divide the Sun’s orbital circumference (939 943 910km) by 1 296 000 arcseconds we obtain:
939 943 910km / 1 296 000" ≈ 725.265 km
Here is what we can read on Wikipedia’s entry for "Angular diameter" - Wikipedia (opens in a new tab):
“An object of diameter 725.27 km at a distance of 1 AU (average Earth > Sun distance) will have an angular diameter of one arcsecond”.
(Incidentally, this is why the observed solar parallax value is 8.794″, because 8.794 X 725.265 km = 6378 km, i.e. the Earth's radius. Therefore, as viewed from the Sun, the Earth's angular diameter will subtend 8.794" X 2 = 17.588". "The currently accepted value of solar parallax is 8".794 143"- source: Wikipedia).
We may now verify how many kilometers of the Sun’s orbital circumference will be subtended by our ACP of 51.136 arcseconds:
51.136(periodic) X 725.27 km ≈ 37087.6km
This is excellent confirmation of our above estimate of 37088km for the annual drift of the Sun’s position against the starry background (caused by Earth’s yearly 14036-km motion), representing 0.0039457% of the Sun’s orbital circumference.
One could also put it this way: since 51.136” equals 0.05681 minutes of time...
...we find that: 0.05681 min. X 25344 = 1440 min. (360°)
(Remember, we measure our celestial sphere by means of a spherical ruler divided in 1440 minutes - or 24 hours.) Once more, it would seem that our 25344-year reckoning of the Great Year holds up quite nicely. Next, let us unpack how the solar & sidereal days are (poorly) explained by the Copernican heliocentric theory.
Solar versus Sidereal Day - as of Heliocentric theory
Here follows a classic diagram illustrating the proposed Copernican explanation for the different lengths of the solar and sidereal day. Keep in mind that Earth is supposedly travelling for over 2.5 Million km every day - yet no parallax whatsoever is observed between the Sun and the stars at the completion of one sidereal day (i.e. 23h56min). Once again, the Copernican explanation for this incontestable fact is that “the stars are almost unimaginably distant”. However, as we take a closer look, this proposed explication doesn’t make physical / optical sense.
Here’s how the occurrence of the sidereal day and the solar day is conventionally illustrated:
In the above graphic, Earth is supposed to have moved some 2.5 Mkm between the positions of “Day 1” and “Day 2”. Let us examine what this classic Copernican diagram is trying to say, quite carefully.
To think that Earth would be moving by over 2.5 Million km each day without the background stars drifting by any noticeable amount besides these last 4 minutes of earthly rotation has to be among the most surreal aspects of the Copernican model! To put this problem into due perspective, let´s see how the sidereal and solar day unfold in the below 3-frame sequence.
The Sidereal vs. Solar Day (23h56min versus 24h00 min) as depicted by the NEAVE planetarium
The following description is what is observed, in reality, from one day to the next:
In 23h56m, an earthly observer will line up again with the same given star. At such a point and time, the Sun will already have moved eastwards by approximately 4 minutes of RA. Four minutes later, we see the stars drifting by 4 min of RA westwards.
Ergo, the entire amount of our daily, Westward stellar drift will appear (to an earthly observer) to occur in the last 4 minutes of earthly rotation! In other words, Earth might just as well be stationary while only rotating around its axis. Many astronomers in ancient times believed this to be true - and understandably so. This wasn’t because they were stupid, but because this is what matches careful and patient observation of the Sun's behaviour.
Of course, the TYCHOS model submits that Earth moves by a mere 38.4 km per day, which is hardly a noticeable amount of lateral displacement to the naked eye. Those 4 min of RA are the consequence of Earth having rotated by 360° in 23h56min, thus needing another 4 minutes to line up again with the Sun. Meanwhile, the Sun has been moving Eastwards by about 4 minutes of RA.
Instead, the Copernican theory would have you believe that Earth is moving each day by 2.5 Million kilometers with no amount of the observed, daily 4-minute stellar drift that can be optically attributed to this distance. It is as if the Earth’s rotation is the only thing that changes the star positions, but Earth’s (supposed) 2.5-Mkm daily displacement has no effect!
We shall now see how Earth’s 1-mph-motion accounts for the 20.41-min. difference between the solar (a.k.a. tropical) year and the sidereal year. You may have asked yourself (rightfully) this good & fair question: “Why-oh-why is the sidereal DAY shorter (by 4 min.) than the solar DAY, whereas the sidereal YEAR is 20.41 min. longer than the solar YEAR?”
The Copernican model offers yet another incredibly convoluted explanation for this conundrum. If you are not familiar with it, you can go to sources like the Wikipedia or browse the example data compiled by Michael J. White, an Arizona State University professor of philosophy. "Sidereal, tropical, and anomalistic years" - by Michael J. White (opens in a new tab)
In any case, the riddle is nicely summarised in this discerning question raised by the Binary Research Institute:
“Sidereal vs. Solar Time: Why is the delta (time difference) between a sidereal and solar day attributed to the curvature of the Earth’s orbit (around the Sun), but the delta between a sidereal ‘year’ and solar year is attributed to precession? (...) The burden of proof lies with those who support the current lunisolar precession theory which requires a different explanation for the two deltas.”
Let us first take a look at these 20.41 min. which, in fact, represent the time difference between a solar and a sidereal year.
Average duration of a solar (or “tropical”) year: 365.24219 days - or 525948.753 minutes
Average duration of a sidereal year: 365.256363 days - or 525969.163 minutes
Ergo, a discrepancy of 20.41 minutes – or a difference of 0.00388%
We see that 20.41 min. is 0.00388% of 525960 min. (i.e.; 365.25 days). And in fact, the currently-observed amount of annual “equinoctial precession” (50.29 arc seconds) amounts to 0.00388% of 1 296 000 arcseconds (remember: 1 296 000″ = a full 360° circle). Hence, those 20.41 minutes are, manifestly, a direct consequence of the so-called equinoctial precession which, in the TYCHOS, is simply caused by Earth's orbital motion.
Earlier on, we determined that our ACP (Annual Constant of Precession) amounts to:
51.136 arc seconds-per-year
i.e.; a value 1.68% larger than the currently-observable annual precession rate of 50.29″.
Note that the official estimate of the duration of one full 360° equinoctial precession (i.e.; one Great Year) is ca. 25771 years. This is, in fact, about 1.68% longer than the TYCHOS reckoning of 25344 solar years.
My next graphic should help visualizing why a small portion (about 1.68% in our epoch) of the equinoctial precession will always remain unobservable from Earth. The reason for this “hidden” angle of precessional drift is something that, yet again, can be demonstrably attributed to Earth’s orbital motion. As we saw earlier, the Sun’s orbit’s “sectional equivalence” of Earth’s yearly 14036-km displacement is 37088 km.
In 20.41 min. (or 0.3401667 hours), the Sun - travelling at 107226 km/h - will cover about 36475 km, i.e about 1.68% less than 37088 km.
It thus becomes plainly evident what causes this 20.41-min difference between the solar and sidereal year: it simply represents the extra time needed for the Sun to realign with a given star (as viewed from Earth). These 20.41 minutes will effectively reset the Earth-Sun-star alignment which, in actuality, has been offset by Earth’s motion around its PVP orbit.
Let me describe in due detail the above graphic by using the solar positions marked A, B and C:
As Joe, our earthly observer, moves from A to B (i.e.; from June 21, 2001 to June 21, 2002) he will have experienced a “Solar year”. Since Earth (travelling at 1 mph) has, in that time period, moved along by 14036 km, Joe will “meet up” with the Sun at an earlier point (compared to the previous year) of the solar orbit. As the Sun’s orbit is 2.64233 X larger than Earth’s PVP orbit, Joe’s lateral displacement will be proportionally equivalent to a 37088km-section of the solar orbit (14036 km X 2.64233 ≈ 37088 km). This is the distance between A and B, as of the above graphic.
This small angular offset (with respect to the Sun-stars alignment) that Earth’s motion causes will then quickly be “regained” by the Sun’s speedy motion. In only 20.41 minutes, the Sun will once again line up (at point “C” ) with the same star that it faced one year earlier. Travelling at 107226 km/h, in 20.41 minutes the Sun covers 36474 km (the distance between B and C) which is, in fact, 1.68% less than 37088 km.
This explains why our earthbound Joe will not realize the full extent of the true annual stellar precession; a small (ca. 1.68%) portion of it will remain unobservable to him. This, because Joe is unaware of Earth’s 1-mph-motion and he therefore (wrongly) believes that Earth has returned at the same physical place as the previous year. Joe will thus conclude that the annual stellar precession rate amounts to 50.29” instead of the true constant rate of 51.136” per year (our ACP).
Thus, the TYCHOS model explains why the sidereal year is longer than the solar year.
Note that, as viewed under Earth’s rotational frame of reference, in 20.41 min. Earth will ROTATE by 18369” arcseconds. This, because 1 min. of our daily 1440-min rotation corresponds to 900 arcseconds (1 296 000”/ 1440 = 900”). Hence, 20.41 X 900 = 18369”. If we now divide this value by 365.25, we obtain:
18369”/ 365.25 = 50.29” (the currently observed annual “equinoctial precession”). This further corroborates what we saw in Chapter 10, namely that the observed precession has nothing to do with any sort of wobble of the Earth's polar axis.
ABOUT THE "ANOMALISTIC" YEAR
Astronomy describes the so-called “anomalistic year” as follows: .
“The anomalistic year is usually defined as the time between perihelion passages. Its average duration is 365.259636 days (or 365 d 6 h 13 min. 52.6 s – at the epoch J2011.0).” "Anomalistic Year" - Wikipedia (opens in a new tab)
The oddly-named “anomalistic year” (the period in which the Sun returns to its closest or furthest point from Earth) lasts on average for 365.259636 days. Incidentally, this is approx. 4.7min more than the sidereal year of 365.256363 days. It is defined as “the time interval between perihelion passages”.
In the TYCHOS, a more aptly-worded description would be “the time interval between the Sun’s perigee transits”. In our current epoch, the Sun’s perigee transit occurs around January 3.
In short, the “anomalistic year” is defined from Sun’s perigee procession, and lasts for about 4.7 minutes longer than a sidereal year. We see that in the course of 4.7 minutes a given point on Earth’s equator will rotate (within the terrestrial rotational reference frame) by 4230”. This is because 1 minute of our celestial sphere of 1440 min. (24 hours) corresponds to 900 arcseconds.
Thence, 4.7 min. X 900” = 4230” (arcseconds)
Now, let us imagine two hypothetical signposts (“S” and “A”) being moved around Earth’s equator year by year with the following parameters.
Signpost “S” (Sidereal) is kept pointing towards a given star.
Signpost “A” (Anomalistic) is kept pointing towards the celestial spot of each year’s passage of the Anomalistic year.
Since signpost “S” is conceptually always being kept oriented towards a given fixed star (in this thought exercise we disregard Earth’s daily rotations around its axis), signpost "S" will complete 1 revolution around Earth’s equator in 25344 years.
On the other hand, signpost “A” will have to be moved each year by an extra 4230” (arcseconds) in relation to signpost “S”.
By the end of one TYCHOS Great Year (of 25344 solar years) signpost “A” will have revolved around Earth by 4230” X 25344 = 107 205 120”
107 205 120” / 1 296 000” (i.e.; 360°) = 82.72
Hence, in 25344 years signpost “A” will complete 82.72X as many revolutions as signpost "S" (for a spin ratio between “A” and “S” of 82.72 : 1)
Since we know that signpost “A” moved by 4230” annually, we can now find out by how much signpost “S” was moving annually:
4230” / 82.72 = 51.136" arcseconds-per-year (i.e. our previously-determined 'ACP'!)
Hence, the so-called “anomalistic” year (which is observed to be 4.7min longer than the sidereal year) further corroborates our Annual Constant of Precession of 51.136 arc seconds (representing Earth’s 1-mph-motion as posited by the TYCHOS model).
We may further confirm our ACP by using the value of 11.75” arcseconds which is described in astronomy literature as 'the progressive motion of the apogee in a year':
“On the anomalistic year: the year called the anomalistic year is sometimes used by astronomers, and is the time from the sun’s leaving its apogee till it returns to it. Now, the progressive motion of the apogee in a year is 11.75", and hence the anomalistic must be longer than the sidereal year, by the time the sun takes in moving over 11.75" of longitude at its apogee.”
In 25344 solar years (or one "Great Year" - as of the TYCHOS), the Sun’s apogee will thus precess by:
25344 X 11.75” = 297 792 arcseconds
The Sun’s apogee will therefore employ slightly more than 4.35 Great Years to complete a full 360° precession:
1 296 000” / 297 792” = 4.35203094777
And once again, we can find our ACP by multiplying this factor of 4.35203094777 by 11.75” :
4.35203094777 X 11.75” ≈ 51.136"
Lastly, we see that the duration of a full, 360° precession of the Sun's apogee will amount to:
4.35203094777 X 25344 ≈ 110297.87 years
This value is close (only 0.6% smaller) to what heliocentrists refer to as "the 111000-year precession of the perihelion of Earth's orbit."
“The perihelion of the earth’s orbit, and of all the planets, is moving around the sun, and completes its revolution in 111,000 years.”
— "Foot Steps of the Ancient Great Glacier of North America: A Long Lost Document of a Revolution in 19th Century Geological Theory" - by Harold W. Borns Jr. and Kirk Allen Maasch (2015) (opens in a new tab)
Note the underlying absurdity of the above sentence. Why would the perihelion of Earth and of all the planets" share such a period, and return to their respective perihelions in unison? It is obviously the orbital motion of the Sun (i.e. of its apsides) that has this periodicity - and not the other way around.
In the following two chapters, we shall be "howling at the Moon" and its many mysteries that have befuddled astronomers for millennia. As it is, the complexity of the Moon's motions is, still today, a subject of intense study and unceasing re-appraisal. So let's gather courage and see how the TYCHOS paradigm may help answering the innumerable questions surrounding the Moon's complex and perplexing motions.