Why Phi? – resonant moons of Uranus

A montage of Uranus’ large moons and one smaller moon: from left to right Puck, Miranda, Ariel, Umbriel, Titania and Oberon. Size proportions are correct. [image credit: Vzb83 @ Wikipedia (from originals taken by NASA’s Voyager 2)]

The five major moons of Uranus in ascending distance from the planet are:
Miranda, Ariel, Umbriel, Titania and Oberon

Of these, the first three exhibit a synodic resonance similar to that of Jupiter’s Galilean moons, as we showed here:
Why Phi? – the resonance of Jupiter’s Galilean moons

Quoting from that post:
The only exact ratio is between the synodic periods which is 3:2:1.
It isn’t necessary to have an exact 4:2:1 orbit ratio in order to get a 3:2:1 synodic ratio.

Why Phi? – Jupiter, Saturn and the ice giants

Giant planets of the solar system [image credit: universetoday.com]

This post on the ice giants Uranus and Neptune follows on from this one:
Why Phi? – Jupiter, Saturn and the inner solar system

The main focus will be on Uranus. A planetary conjunction of three bodies (e.g. two planets and the Sun, in line) is also known as a syzygy.

Here’s the notation for the table shown below:
J-S = Jupiter-Saturn conjunctions
S-U = Saturn-Uranus conjunctions
U-N = Uranus-Neptune conjunctions

Each of the columns: U, S-U, J-S shows a Fibonacci progression.

Accuracy of best match is between 99.965% and 99.991%.

Quoting Wikipedia: ‘The mathematics of the golden ratio and of the Fibonacci sequence are intimately interconnected.’
The Greek letter φ (phi) represents the golden ratio.

Why Phi? – resonant exoplanets of star YZ Ceti

Exoplanet – NASA impression

YZ Ceti is a recently discovered star with three known planets (b,c and d) orbiting very close to it. Although some types of mean motion resonance, or near resonance, are quite common e.g. 2:1 or 3:2 conjunction ratios, this one is a bit different.

The orbit periods in days are:
YZ Ceti b = 1.96876 d
YZ Ceti c = 3.06008 d
YZ Ceti d = 4.65627 d

This gives these conjunction periods:
c-d = 8.9266052 d
b-c = 5.5204368 d
b-d = 3.4109931 d
(Note the first two digits on each line.)

Nearest matching period:
34 c-d = 303.50457 d
55 b-c = 303.62403 d
89 b-d = 303.57838 d

34,55 and 89 are Fibonacci numbers.
Therefore the conjunction ratios are linked to the golden ratio (Phi).

Phi = 1.618034
(c-d) / (b-c) = 1.6170106
(b-c) / (b-d) = 1.618425

Data source: exoplanets.eu

APOD: Fibonacci relations appear in IR at Jupiter’s poles – Why Phi?

So to clarify here (because I messed it up earlier) what was presented this morning at #AAS231 were infrared images of Jupiter's north pole (left, via @lrebull) and south pole (right, via @Stardustspeck) showing respectively eight and five cyclones circling central vortices. 😲🤯 pic.twitter.com/lFkfl25wam

— Jason Major (@JPMajor) January 9, 2018

Why Phi? – Jupiter, Saturn and the inner solar system

From left, Mercury, Venus, Earth and Mars. [Credit: Lunar and Planetary Institute]

The planetary theory aspect appears a bit later, but first a brief review of some relevant details.

In this Talkshop post: Why Phi? – a triple conjunction comparison we said:
(1) What is the period of a Jupiter(J)-Saturn(S)-Earth(E) (JSE) triple conjunction?
JSE = 21 J-S or 382 J-E or 403 S-E conjunctions (21+382 = 403) in 417.166 years (as an average or mean value).

(2) What is the period of a Jupiter(J)-Saturn(S)-Venus(V) (JSV) triple conjunction?
JSV = 13 J-S or 398 J-V or 411 S-V conjunctions (13+398 = 411) in 258.245 years (as an average or mean value).

Since JSV = 13 J-S and JSE = 21 J-S, the ratio of JSV:JSE is 13:21 exactly (in theory).

As these are consecutive Fibonacci numbers, the ratio is almost 1:Phi or the golden ratio.

Why Phi? – solar and lunar rotation

This started as a search for a period when the Sun and the Moon would both complete a whole number of rotations.
The result was:
Solar: 25.38 days * 197 = 4999.860 d
Lunar: 27.321662 * 183 = 4999.864 d
(data sources: see reference notes at end)

Taking these as equivalent, we have 197-183 = 14 ‘beats’.
197 = 14*14, +1
183 = 13*14, +1
4999.864 / 14 = 357.13314 days
357.13314 days * 45/44 = 365.2498 days
45 * 14 (630) beats = 44 * 14 (616) calendar years, difference = 0.022 day

So the beat period of the two rotations is 44/45ths of a year, i.e. the difference in number of rotations is exactly 1 in that length of time.
630 beats = 616 years (630 – 616 = 14)
616/45 = 13.68888 calendar years = 4999.8663 days
184 lunar sidereal months (rotations) = 4999.864 days

Then something else popped up…

The Phi factor:
‘We recover a 22.14-year cycle of the solar dynamo.’ (2016 paper)
See: Why Phi? – modelling the solar cycle

Solar Hale cycle = ~22.14 years (est. mean)
13.68888 * Phi = 22.149~ years
22.14 / 13.68888 = 1.61737 (99.96% of Phi)
(55/34 = 1.617647)

From the same post:
Jupiter-Saturn axial period (J+S) is 8.456146 years.
That’s when the sum of J and S orbital movement in the conjunction period = 1
13.68888 / 8.456146 = 1.618808
Phi = 1.618034

Conclusion:
This cycle of solar and lunar sidereal rotation (SRC) sits at the mid-point of the Phi²:1 ratio between the J+S axial period and the mean solar Hale cycle, i.e. with a Phi ratio to one and inverse Phi to the other.
SRC = (J+S) * Phi
SRC = Hale / Phi
SRC = Hale – (J+S)
(Mean Hale value is assumed)

In a period of 616 years there are 45 SRC.
The period is 44 * 14 years = 45 SRC = 45 * 14 beats.
SRC * (45/44) = 14 years.

Cross-checks:
Carrington rotations per 616 y = 8249
8249 CR / 45 = 4999.865 days

Synodic months per 616 y = 7619
7619 SM / 45 = 4999.856 days
8249 – 7619 = 630 = 45 * 14

45*183 sidereal months = 8235
8235 – 7619 = 616
8249 CR – 8235 Sid.M = 14
Beat period of CR and Sid.M = 616/14 = 44 years = 45 * (13.6888 / 14)
Every 44 years there will be exactly one less lunar rotation (sidereal month) than the number of Carrington rotations.

8249 CR – 7619 synodic months = 630 = 45 * 14
630 – 616 = 14
– – –
The anomalistic year

The beat period of the tropical month and solar sidereal rotation * 45/44 = the anomalistic year.
(27.321582 * 25.38) / (27.321582 – 25.38) = 357.14265 days
45 * 357.14265 = 16071.419 days
44 * 365.259636 = 16071.423 days

The anomalistic year is the time taken for the Earth to complete one revolution with respect to its apsides. The orbit of the Earth is elliptical; the extreme points, called apsides, are the perihelion, where the Earth is closest to the Sun (January 3 in 2011), and the aphelion, where the Earth is farthest from the Sun (July 4 in 2011). The anomalistic year is usually defined as the time between perihelion passages. Its average duration is 365.259636 days (365 d 6 h 13 min 52.6 s) (at the epoch J2011.0).
http://en.wikipedia.org/wiki/Year#Sidereal.2C_tropical.2C_and_anomalistic_years
– – –
Data sources

— Carrington Solar Coordinates:
Richard C. Carrington determined the solar rotation rate by watching low-latitude sunspots in the 1850s. He defined a fixed solar coordinate system that rotates in a sidereal frame exactly once every 25.38 days (Carrington, Observations of the Spots on the Sun, 1863, p 221, 244). The synodic rotation rate varies a little during the year because of the eccentricity of the Earth’s orbit; the mean synodic value is about 27.2753 days.
http://wso.stanford.edu/words/Coordinates.html

— The standard meridian on the sun is defined to be the meridian that passed through the ascending node of the sun’s equator on 1 January 1854 at 1200 UTC and is calculated for the present day by assuming a uniform sidereal period of rotation of 25.38 days (synodic rotation period of 27.2753 days, Carrington rotation).
http://jgiesen.de/sunrot/index.html

The sidereal month is the time between maximum elevations of a fixed star as seen from the Moon. In 1994-1998, it was 27.321662 days.
http://scienceworld.wolfram.com/astronomy/SiderealMonth.html

Lunar precession update

Credit: NASA

I found out there’s an easy way to simplify one of the lunar charts published on the Talkshop in 2015 on this post:
Why Phi? – some Moon-Earth interactions

In the chart, synodic months (SM) and apsidal cycles (LAC) are multiples of 104:
79664 / 104 = 766
728/104 = 7

The other numbers are not multiples of 104, but if 7 is added to each we get this:
86105 + 7 = 86112 = 828 * 104 (TM)
85377 + 7 = 85384 = 821 * 104 (AM)
5713 + 7 = 5720 = 55 * 104 (FMC)
6441 + 7 = 6448 = 62 * 104 (TY)

TM = tropical months
AM = anomalistic months
SM = synodic months
LAC = lunar apsidal cycles
FMC = full moon cycles
TY = tropical years

Here’s an imaginary alternative chart based on these multiples of
104. [Cross-check: 828 – 766 = 62]

In reality, 55 FMC = just over 62 TY and 7 LAC = just short of 62 TY.
For every 7 apsidal cycles (LAC), there are 766 synodic months (both chart versions).

In the real chart:
For every 104 apsidal cycles, all numbers except SM slip by -1 from being multiples of 104. So after 7*104 LAC all the other totals except SM are ‘reduced’ by 7 each.

In the case of tropical years, 6448 – 7 = 6441 = 19 * 339
19 tropical years = 1 Metonic cycle

If the period had been 6448 TY it would not have been a whole number of Metonic cycles.
Also 6441 * 4 TY (25764) is exactly one year more than 25763 synodic years i.e. the precession cycle, by definition.

Fibonacci: 104 is 13*8, and the modified FMC number is 55 (all Fibonacci numbers).

Phi: we’ve explained elsewhere that the number of full moon cycles in one lunar apsidal cycle is very close to 3*Phi².
We can see from the modified chart that the FMC:LAC ratio of 55:7 is 3 times greater than 55:21 (55/21 = ~Phi²)
– – –
Note – for more discussion of the ~62 year period, try this search:
site:tallbloke.wordpress.com 62 year
[see Google site search box in grey zone on left of this web page]

Why Phi? – the rainbow angle

The rainbow angle [credit: Hong Kong Observatory]

The minimum deviation angle for the primary bow [of a rainbow] is 137.5° according to Wikipedia. This is known as the rainbow angle. A circle is 360 degrees, so the ratio of the rainbow angle to the circle is therefore the square of the golden ratio i.e. 137.5:360 = 1:2.61818~.
– – –
Hong Kong Observatory has some useful explanatory text and graphics (rounding 137.5 to 138 degrees) titled:
Why is the region outside the primary rainbow much darker than that inside the primary rainbow?
Written by : SIU Kai-chee (summer intern) and HUNG Fan-yiu

Let’s first look at Figure 1, which shows sun rays entering a water drop and going through refraction and reflection.

The ray (ray no. 1) passing through the centre goes directly backward on reflection, i.e. a change in direction of 180 degrees.

For ray no. 2, this angle becomes smaller, following the rules of refraction and reflection.

For the next (ray no. 3) the angle continues to decrease, so on and so forth. This trend does not continue for long, however.

Why Phi? – a lunar ratios model

Lunar ratios diagram

The idea of this post is to try and show that the lunar apsidal and nodal cycles contain similar frequencies, one with the full moon cycle and the other with the quasi-biennial oscillation.

There are four periods in the diagram, one in each corner of the rectangle. For this model their values will be:

FMC = 411.78443 days
LAC = 3231.5 days
LNC = 6798.38 days
QBO = 866 days (derived from 2 Chandler wobbles @ 433 days each)
The QBO period is an assumption (see Footnote below) but the others can be calculated.

Why Phi? – Pluto’s eccentric orbit

Pluto’s non-standard orbit [credit: Wikipedia]

‘Pluto’s orbital period is 248 Earth years. Its orbital characteristics are substantially different from those of the planets, which follow nearly circular orbits around the Sun close to a flat reference plane called the ecliptic. In contrast, Pluto’s orbit is moderately inclined relative to the ecliptic (over 17°) and moderately eccentric (elliptical). This eccentricity means a small region of Pluto’s orbit lies nearer the Sun than Neptune’s.’ – Wikipedia

Why Phi? – lunar eclipses at Stonehenge

Bluestone Horseshoe at Stonehenge – 19 Stones

Stonehenge Visitors Guide – under ‘Eclipse Cycles’ – says:

‘Now, it’s widely accepted that Stonehenge was used to predict eclipses. The inner “horseshoe” of 19 stones at the very heart of Stonehenge actually acted as a long-term calculator that could predict lunar eclipses. By moving one of Stonehenge’s markers along the 30 markers of the outer circle, it’s discovered that the cycle of the moon can be predicted. Moving this marker one lunar month at a time – as opposed to one lunar day the others were moved – made it possible for them to mark when a lunar eclipse was going to occur in the typical 47-month lunar eclipse cycle. The marker would go around the circle 38 times [2 x 19] and halfway through its next circle, on the 47th full moon, a lunar eclipse would occur.’

Why Phi? – solar rotation notes

Carrington Rotations = CarRots [credit: dreamstime.com]

Tallbloke recently acquired a book by Hartmut Warm called ‘Signature of the Celestial Spheres: Discovering Order in the Solar System’ which offers this gem:
588 solar Carrington rotations (CarRots) = 587 lunar sidereal months
We’ll call this the HW cycle, about 43.91 years.

‘Richard Christopher Carrington determined the solar rotation rate from low latitude sunspots in the 1850s and arrived at 25.38 days for the sidereal rotation period. Sidereal rotation is measured relative to the stars, but because the Earth is orbiting the Sun, we see this period as 27.2753 days.’ – Wikipedia

Picking this ball up and running with it, we find there are 308 CarRots (27.2753 d) per 331 solar sidereal days (25.38 d) in 23 years (331 – 308). This period, or a multiple of it, can be found in certain identified solar-planetary cycles (as discussed below).

Why Phi? – a unified precession model

Combined precession cycle [credit: wikipedia]

‘Because of apsidal precession the Earth’s argument of periapsis slowly increases; it takes about 112000 years for the ellipse to revolve once relative to the fixed stars. The Earth’s polar axis, and hence the solstices and equinoxes, precess with a period of about 26000 years in relation to the fixed stars. These two forms of ‘precession’ combine so that it takes about 21000 years for the ellipse to revolve once relative to the vernal equinox, that is, for the perihelion to return to the same date (given a calendar that tracks the seasons perfectly).’ – Wikipedia

Here we’ll fit the three precession cycles into one model and briefly examine its workings.

Why Phi? Sylvia & Sons Swing Serenely by Shining Stars

Out at the unfashionable end of the Asteroid Belt, lies a seldom seen squashed spud of rock known as Sylvia. NASA has this:

Composite image showing the two moons at several locations along their orbits (shown by red dots). Image Credit: NASA

Discovered in 1866, main belt asteroid 87 Sylvia lies 3.5 AU from the Sun, between the orbits of Mars and Jupiter. Also shown in recent years to be one in a growing list of double asteroids, new observations during August and October 2004 made at the Paranal Observatory convincingly demonstrate that 87 Sylvia in fact has two moonlets – the first known triple asteroid system. At the center of this composite of the image data, potato-shaped 87 Sylvia itself is about 380 kilometers wide. The data show inner moon, Remus, orbiting Sylvia at a distance of about 710 kilometers once every 33 hours, while outer moon Romulus orbits at 1360 kilometers in 87.6 hours. Tiny Remus and Romulus are 7 and 18 kilometers across respectively. Because 87 Sylvia was named after Rhea Silvia, the mythical mother of the founders of Rome, the discoverers proposed Romulus and Remus as fitting names for the two moonlets. The triple system is thought to be the not uncommon result of collisions producing low density, rubble pile asteroids that are loose aggregations of debris.

Why Phi? – the resonance of Jupiter’s Galilean moons

Io, Europa and Ganymede – three of Jupiter’s four Galilean moons

The resonance of three of the four Galilean moons of Jupiter is well-known. Or is it?

We’re usually told there’s a 1:2:4 orbital ratio between Ganymede, Europa and Io, but while this is not far from the truth, a closer look shows something else.

Phi and the Great Pyramid of Khufu

Great Pyramid of Giza from a 19th-century stereopticon card photo [credit: Wikipedia]

Let’s have a look at some numbers for the Great Pyramid.

Source: Building the Great Pyramid (aka Cheops)
Copyright 2006 Franz Löhner and Teresa Zuberbühler

Dimensions as designed (in Egyptian royal cubits):
Length: 440
Height: 280
Slope: 356

Original dimensions as built (a,h and c in the pyramid diagram below):
Length: 230.36m (half = 115.18m)
Height: 146.59m
Slope: 186.42m

Why Phi? – some Moon-Earth interactions

Tallbloke writes: Stuart ‘Oldbrew’ has been getting his calculator warm to discover the congruences in various aspects of the Lunar orbit around Earth, and its relationship to Earth-Moon orbit around the Sun. Emerging from this study are some useful insights into longer periods, such as the ‘precession of the equinoxes‘.

Some matching periods of lunar numbers:
86105 tropical months (TM) @ 27.321582 days = 2352524.8 days
85377 anomalistic months (AM) @ 27.55455 days = 2352524.8 days
79664 synodic months (SM) @ 29.530589 days = 2352524.8 days

These identical values are used in the chart on the right (top row). The second row numbers are the difference between the numbers in the first row (TM – AM and AM – SM).
The derivation of the third row number (6441) is shown on the chart itself [click on the chart to enlarge it].

The period of 6441 tropical years (6440.75 sidereal years) is one quarter of the Earth’s ‘precession of the equinox’.
Multiplying by 4: 25764 tropical years = 25763 sidereal years.
The difference of 1 is due to precession.

Why Phi? – Neptune’s moon Triton and its neighbours

The orbit of Triton (red) is opposite in direction and tilted −23° compared to a typical moon’s orbit (green) in the plane of Neptune’s equator [image credit: Wikipedia]

Triton is the seventh largest moon in the solar system. Not only that, it has over 99% of the mass of all Neptune’s moons combined. Its retrograde orbit makes it unique among the large moons of the solar system, and it is also the coldest known planetary body at -235° C (-391° F).

Turning to the orbit numbers, and looking at Triton’s closest ‘inner’ (nearer to Uranus) neighbour Proteus and the next two ‘outer’ moons, we find these values (in days):
1.122d Proteus
5.877d Triton
360.13d Nereid
1879.08d Halimede

We’ll treat Proteus and Triton as a pair, and the same for Nereid and Halimede.
Nereid is over fifteen times further from Uranus than Triton is, so hardly a neighbour at all.

Looking at the orbit ratios (which are also the rotation ratios, as usual with moons):
T/P = 5.877 / 1.122 = 5.238
H/N = 1879.08 / 360.13 = 5.218

The first thing to say is that the two results are very similar. One is about 99.62% of the other.

Why Phi? – a Jupiter-Neptune model

Neptune (top), Uranus, Saturn, Jupiter (bottom)

Continuing our long-term series researching Fibonacci and/or Phi based ratios in planetary conjunction periods, it’s time for a look at the inner- and outer-most gas giants of our solar system: Jupiter and Neptune.

Initial analysis shows the period of 14 Jupiter orbits is close to that of one Neptune orbit of the Sun, and even closer to the period of 13 (14 less 1) Jupiter-Neptune (J-N) conjunctions.

It also turns out that there’s a multiple of 13 J-N that equates to a whole number of Earth orbits:
Jupiter-Neptune(J-N) average conjunction period = 12.782793 years
221 J-N = ~2825 years (2824.9972y)
(221 = 13 x 17)

But this period is not a whole number of either Jupiter or Neptune orbits.
This is resolved by multiplying by a factor of 7.

Why Phi? – a Jupiter-Uranus-Neptune model

From the top: Neptune, Uranus, Saturn, Jupiter
[image credit: NASA/JPL]

Continuing our quest to understand more about planetary frequencies, we turn to links between the largest planet Jupiter and the two ‘outer’ giant planets, Uranus and Neptune.

This model is based on a match of synodic periods, which is found to be:
22 Uranus-Neptune (U-N) = 273 Jupiter-Uranus (J-U) = 295 Jupiter-Neptune (J-N)

The period is just under 3771 years (~3770.93y).
To find a link to Fibonacci numbers we can look first at Jupiter-Uranus:
273 J-U = 13 x 21 (13 and 21 are Fibonacci numbers)