Why Phi? – the Kepler-47 circumbinary system

Kepler Space Telescope [credit: NASA]

A headline at Phys.org today reads:
‘Astronomers discover third planet in the Kepler-47 circumbinary system’

The report starts:
‘Astronomers have discovered a third planet in the Kepler-47 system, securing the system’s title as the most interesting of the binary-star worlds. Using data from NASA’s Kepler space telescope, a team of researchers, led by astronomers at San Diego State University, detected the new Neptune-to-Saturn-size planet orbiting between two previously known planets.

With its three planets orbiting two suns, Kepler-47 is the only known multi-planet circumbinary system. Circumbinary planets are those that orbit two stars.’

In this system the two stars orbit each other about every 7.45 days.

What can the latest information tell us about these planets, including newly discovered planet ‘d’?

Why Phi? – Resonances of exoplanetary system Kepler-107

Orbital (top line) and synodic relationships of Kepler-107, plus cross-checks

The system has four planets: b,c,d, and e.

The chart to the right is a model of the close orbital relationships of these four recently announced short-period (from 3.18 to 14.75 days) exoplanets.

It can be broken down like this:
b:c = 20:13
c:d = 13:8
d:e = 24:13 (= 8:13 ratio, *3)
b:d = 5:2
c:e = 3:1
(1,2,3,5,8, and 13 are Fibonacci numbers)

Why Phi? – a luni-solar link

This was a surprise, but whatever the interpretation, the numbers speak for themselves.

‘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.

What happens if we relate this period to the lunar draconic year?

Why Phi? – a lunar evection model, part 2

Image credit: interactivestars.com

It turns out that the previous post was only one half of the lunar evection story, so this post is the other half.

There are two variations to lunar evection, namely evection in longitude (the subject of the previous post) and evection in latitude, which ‘generates a perturbation in the lunar ecliptic latitude’ (source).

It’s found that the first is tied to the full moon cycle and the second to the draconic year.

Why Phi? – a lunar evection model

Apogee = position furthest away from Earth. Earth. Perihelion = position closest to the sun. Moon. Perigee = position closest to Earth. Sun. Aphelion = position furthest away from the sun. (Eccentricities greatly exaggerated!)

Lunar evection has been described as the solar perturbation of the lunar orbit.

One lunar evection is the beat period of the synodic month and the full moon cycle. The result is that it should average about 31.811938 days (45809.19 minutes).

Comparing synodic months (SM), anomalistic months (AM), and lunar evections (LE) with the full moon cycle (FMC) we find:
1 FMC = 13.944335 SM
1 FMC = 13.944335 + 1 = 14.944335 AM
1 FMC = 13.944335 – 1 = 12.944335 LE

Since 0.944335 * 18 = 16.9983 = 99.99% of 17, and 18 – 17 = 1, we can say for our model:
18 FMC = 233 LE (18*13, -1) = 251 SM (18*14, -1) = 269 AM (18*15, -1)
See: 3 – Matching synodic and anomalistic months.

Why Phi: is the Moon a phi balloon? – part 2

Credit: universetoday.com

Picking up from where we left off here…

Three well-known aspects of lunar motion are:
Lunar declination – minimum and maximum degrees
Orbital parameters – perigee and apogee distances (from Earth)
Anomalistic month – minimum and maximum days

Standstill limits due to the lunar nodal cycle

‘The major standstill limit of the moon can be reached if the lunar node is near the vernal (or autumnal) point, and with the moon at its max. distance from the equator, equal to a declination at present days of 23.44° + 5.1454°= 28.59°.

The minor standstill limit of the moon can be reached if the lunar node is near the vernal (or autumnal) point, and with the moon at its min. distance from the equator, equal to a declination at present days of 23.44°- 5.1454° = 18.29°.’
http://iol.ie/~geniet/eng/moonperb.htm#nodes

28.59 / 18.29 = 1.5631492
4th root of 1.5631492 = 1.11815
This number leads to the key to the puzzle.

Why Phi? – Saturn’s inner moons and exoplanets of Kepler-223

Three of Saturn’s moons — Tethys, Enceladus and Mimas — as seen from NASA’s Cassini spacecraft [image credit: NASA/JPL]

This is a comparison of the orbital patterns of Saturn’s four inner moons with the four exoplanets of the Kepler-223 system. Similarities pose interesting questions for planetary theorists.

The first four of Saturn’s seven major moons – known as the inner large moons – are Mimas, Enceladus, Tethys and Dione (Mi,En,Te and Di).

The star Kepler-223 has four known planets:
b, c, d, and e.

When comparing their orbital periods, there are obvious resonances (% accuracy shown):
Saturn: 2 Mi = 1 Te (> 99.84%) and 2 En = 1 Di (> 99.87%)
K-223: 2 c = 1 e (>99.87%) and 2 b = 1 d (> 99.86%)

Why Phi? – a long-term Jupiter-Saturn-Uranus model

Here we find a match between the orbit numbers of Jupiter, Saturn and Uranus and see what that might tell us about certain patterns in the solar system.

715 U = 60072.044 years
2040 S = 60072.895 years
5064 J = 60072.282 years
Data source: Nasa/JPL – Planets and Pluto: Physical Characteristics

The Jupiter-Saturn part of the chart derives directly from this earlier Talkshop post:
Why Phi? – Jupiter, Saturn and the de Vries cycle

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).