Why Phi? – resonances of exoplanets LHS 1140 b and c

Wikipedia says:
LHS 1140 is a red dwarf in the constellation of Cetus…The star is over 5 billion years old and has 15% of the mass of the Sun. LHS 1140’s rotational period is 130 days…LHS 1140 is known to have two confirmed rocky planets orbiting it, and a third candidate planet not yet confirmed.

Planet b was in the media spotlight in 2017:
LHS 1140b: Potentially Habitable Super-Earth Found Orbiting Nearby Red Dwarf – Sci-News.

“This is the most exciting exoplanet I’ve seen in the past decade,” said Dr. Jason Dittmann, an astronomer at the Harvard-Smithsonian Center for Astrophysics and lead author of the Nature paper.
. . .
“The LHS 1140 system might prove to be an even more important target for the future characterization of planets in the habitable zone than Proxima b or TRAPPIST-1,” concluded co-authors Dr. Xavier Delfosse and Dr. Xavier Bonfils, both at the CNRS and IPAG in Grenoble, France.

The Why Phi Pi Slice goes camping.

The leaves of some plants grow on shoots that form 222.5 degrees round the main stem from their predecsessor

Back in 2013 I wrote a post about the relationship between our favourite number, phi (1.618…) and the famous circularity constant Pi (3.141…).

If we divide the circle of 360 degrees by phi, we get 222.5 degrees, leaving 137.5 degrees as the remainder. In that post I noted that:

The area ‘A’ of a sector of a circle is given by the simple formula: A=angle/360*Pi*R²
For a Radius ‘R’ of 1 and angle 137.507764 this is simply 137.5/360*Pi = 1.19998
The area of the whole circle is simply Pi, since R² = 1
The ratio of Pi to 1.19998 is phi²
The ratio of the smaller sector to the larger is Pi-1.19998:1.19998 which is simply phi itself.

Friday Fibonacci Fun

Someone make this with cheese and tortilla chips so I can have a fibonacho pic.twitter.com/FyAir9GUgJ

— jayynecobb (@JayyneCobb) April 10, 2020

Such a cleverly made 12 second clip. Fullscreen it and enjoy

Neptune, Uranus, Pluto and the Grand Synod

Image credit: beforeitsnews.com

The aim here is to show how the synodic periods and orbits of these three planets align with the so-called Grand Synod, a period of about 4628 years which has 27 Uranus-Neptune conjunctions and almost 233 Jupiter-Saturn conjunctions. Its half-period is sometimes referred to as the Hallstatt cycle (2314 years +/- a variable margin).

1. U-N ‘long period’
1420 Uranus-Neptune conjunctions = 1477 Neptune orbits
(for calculations, see Footnote)
1477 – 1420 = 57
Uranus-Neptune 360 degrees return is 1420/57 U-N = 24.91228 U-N long period = 4270.119 years

2. GS : U-N ratio
Grand Synod = 27 U-N = 4627.967 years (= ~233 Jupiter-Saturn conjunctions)
27 / 24.91228 = 1.0838028
1.0838028 * 12 = 13.005633
Therefore the ratio of 4627.967:4270.119 is almost exactly 13:12 (> 99.956% true)

3. Orbital data
Turning to the orbit periods nearest to the Grand Synod:
28 Neptune = 4614.157y
55 Uranus = 4620.927y
(Data: https://ssd.jpl.nasa.gov/?planet_phys_par )

4. Factor of 12
These periods fall slightly short of the 27 U-N Grand Synod (~4628 years).
However, multiplying by 12 and adding one orbit to each, gives:
28*12,+1 (337) Neptune = 55534.67y
55*12,+1 (661) Uranus = 55535.14y
27*12 (661 – 337) U-N = 55535.61y

Now the numbers match to within a year +/- 55535 years.
Also, the period is 12 Grand Synods (12*4628 = 55536y), or 13 U-N ‘long’ periods.

5. Pluto data
Pluto’s orbit period is 247.92065 years.
55535 / 247.92065y = 224.003
So 224 Pluto orbits also equate to 12 Grand Synods.

Therefore, a U-N-P synodic chart can be created for that period of time.

6. Neptune:Pluto orbits
Neptune has one more orbit in the period than an exact 3:2 ratio with Pluto – a planetary resonance.
224 P = 112*2
337 N = 112*3, +1
113 N-P = 112, +1

7. Phi factor
Uranus and Neptune both have one more orbit than this ratio:
660:336 = (55*12):(21*16)
55/21 = Phi²
12/16 = 3/4
Therefore the U:N ratio is almost (3/4 of Phi²):1

The U-N-P chart should repeat every 12 Grand synods i.e. every 55,535 years or so.
– – –
Footnote
360 / Neptune orbit (164.79132) = 2.184581
2.184581 * U-N conjunction (171.40619) = 374.4507
374.4507 – 360 = 14.4507

Obtain nearest multiple of 360 degrees:
1420 * 14.4507 = 20519.9994
20520 / 360 = 57
1420 + 57 = 1477
1420 U-N = 1477 Neptune orbits
1420 + 1477 = 2897 Uranus orbits

Update (Dec. 2020): the number of occurrences of the Uranus-Neptune conjunction precession in the period is 337 – 324 = 13. Therefore the ratio of that period to the Grand Synod (GS) is 13:12, because the GS is 27 U-N and 324 = 27*12.

Why Phi? – Jupiter-Venus harmonics and the Fibonacci/Lucas series

Jupiter – the dominant planet in the solar system

The aim here is to show a Lucas number based pattern in five rows of synodic data, then add in a note on Mercury as well.

There’s also a strong Fibonacci number element to this, as shown below.

The results can be linked back to earlier posts on planetary harmonics involving the Lucas and Fibonacci series (use ‘search this site’ box on our home page).

Why Phi? – the three planets of the Kepler-47 circumbinary planet system

Kepler-47 system [Image Credit: NASA/JPL Caltech/T. Pyle]

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, says NASA’s Exoplanet Exploration team.

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.

Continued here.
– – –
Now at the Talkshop let’s take a quick look at the data.

Why-Phi? – some Uranus-Neptune long-term resonances

Continuing our recent series of posts, with Uranus-Neptune conjunction data an obvious starting point for the table is where the difference between the number of Neptune orbits and U-N synods is 1.

647 U-N takes a long time (~110,900 years) but the accuracy of the whole number matches is very high.

Lucas no. (7 here) is fixed, and Fibonacci nos. follow the correct sequence (given their start no.).
Full Fib. series starts: 0,1,1,2,3,5,8,13,21…etc.
Multiplier: 0,1,1,2,3
Addition: 1,1,2,3,5

The Neptune orbits are multiples of 26 with the same Fibonacci adjustment:
Add 0,1,1,2,3 to the Neptune column numbers to get an exact multiple of 26 (which will be the pattern number in the last column).

Why Phi? – the Saturn-Uranus connection

Distances not to scale.

This is an easy data table to interpret.

The Uranus orbits are all Fibonacci numbers, and the synodic conjunctions are all a 3* multiple of Fibonacci numbers.
[Fibonacci series starts: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …etc.]

In addition, the difference between the two is always a Lucas number. And that’s it for Saturn-Uranus, which would make for a very short blog post.

But it’s possible to go further.

Why Phi? – Jupiter-Earth harmonics and the Lucas series

Jupiter – the dominant planet in the solar system

The aim here is to show a Lucas number based pattern in seven rows of synodic data.
There’s also a Fibonacci number element to this, as shown below.
The results can be linked back to an earlier post on planetary harmonics (see below).

The nearest Lucas number equation leading to the Jupiter orbit period in years is:
76/7 + 1 = 11.857142 (1, 7 and 76 are Lucas numbers).
The actual orbit period is 11.862615 years (> 99.95% match).
[Planetary data source]

It turns out that 7 Jupiter orbits take slightly over 83 years, while 76 Jupiter-Earth (J-E) synodic conjunctions take almost exactly 83 years. One J-E synod occurs every 1.09206 years. (83/76 = 1.0921052).

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