Why Phi? – a triple conjunction comparison

Posted: June 14, 2015 by oldbrew in Fibonacci, Phi, solar system dynamics
Tags: , ,

[image credit: imagineeringezine.com]

[image credit: imagineeringezine.com]

Only two questions are needed here:

(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.
Golden ratio: relationship to Fibonacci sequence


Other ‘Why Phi’ posts

Reference data

Comments
  1. oldbrew says:
    June 14, 2015 at 8:15 pm

    The beat frequency (BF) or conjunction period is calculated as follows:

    BF = (Po x Pi) / (Po – Pi)

    where Po is the orbital period of the outer (slower) body, and Pi is that of the inner (and faster) one.

    As explained here under ‘Jupiter and the sunspots’.

  2. Ian Wilson says:
    June 17, 2015 at 3:00 pm

    oldbrew,

    Very interesting post as usual.

    Just being the Devil’s advocate here but what if you took four planets moving in four separate circular orbits whose sizes are randomly assigned (say from 0.3 to 10.0 A.U.). The resulting orbits would have to obey the following relationships:

    V(orbit)^2 * R (orbit) = G * M(sun) and T(orbit) = 2 * PI * R(orbit)

    where V(orbit) = the velocity of the planet in its orbit
    _____R(orbit) = the radius of the planets orbit
    _____M(Sun) = the mass of the Sun
    and __T(orbit) = the period of the planetary orbit.

    This means that:

    T(orbit) = 2 * PI * G * M(sun) / V(orbit)^2 OR

    T(orbit) = c / V(orbit)^2 AND

    R(orbit) = c’ / V(orbit)^2

    where c and c’ are just constants.

    So what determines the orbital velocity of a planet? The last equation suggests that it is just a constant multiple of either the inverse square root of the distance from the Sun or the inverse square root of the orbital period about the Sun.

    Under such circumstances, would you always get a phi ratio between the triple synodic periods and double synodic periods – as you have explained in your post.

  3. Kelvin Vaughan says:
    June 21, 2015 at 6:03 pm

  4. oldbrew says:
    June 25, 2015 at 10:16 am

    Ian Wilson: some time ago I found this to be true…

    Venus:Earth orbital speed ratio is 1.176:1 which is the same as the fourth root of the solar irradiance ratio (1.911:1).
    http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

    So as you suggest all these factors are interlocked as it were, as Kepler’s laws implied.

    ‘would you always get a phi ratio between the triple synodic periods and double synodic periods – as you have explained in your post.’

    I don’t know about ‘always’ but there are other examples e.g. 89 J-S = 39 S-U = 128 J-U, therefore that’s a ‘JSU’ triple conjunction and 89 is a Fibonacci number, so JSE:JSU ratio is 89:21 = phi³:1.

    ‘The last equation suggests that it is just a constant multiple of [either] the inverse square root of the distance from the Sun’

    I also came to that conclusion [‘it’ being orbit speed].