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Sunday, March 18, 2012

Short Comings of the Planetary Spin-Orbit Coupling Model


SIMPLE TIDAL TORQUING MODEL
(UPDATED 20/03/2012)

Imagine that Venus and Earth are aligned directly above a point A that is on the surface of the Sun. The combined tidal force of Venus and Earth produced two tidal bulges upon the surface of the Sun located at A and B.



Jupiter is located at an angle θ to the line joining the two tidal 
bulges at A and B. Let Rs be the radius of the Sun = OA = OB.

Now by the cosine rule:

JB2 = RJ2 + Rs2 – 2 × RJ × Rs × cos(θ)                         and
JA2 = RJ2 + Rs2 – 2 ×RJ × Rs × cos(π - θ)
      = RJ2 + Rs2 + 2 ×RJ × Rs × cos(θ)                                                             

Let angle JAO = φ 
and angle JBO = ψ                                                                  

then by the sine rule:

RJ / sin (φ) = JA / sin (π – θ)                                           and
RJ / sin(ψ) = JB / sin (θ)

therefore:

sin (φ)  = (RJ / JA) × sin (π – θ)
sin (ψ) = (RJ / JB) × sin (θ)

 By Newtons Law of Universal Gravitation

Ff = G MJ MS / JA2 = const / JA2                 const = G MJ MS
Fn = G MJ MS / JB2 = const / JB2

Now

Ffp  = Ff × sin(φ)

hence:

Ffp  = (const / JA2) × (RJ / JA) x sin (π  − θ) 
       = (const’ / JA3) × sin (θ)       
                                                                   const’= const × RJ
and

Fnp = Fn × cos (ψ – π/2) 
       = Fn × cos (-(π/2 – ψ)) 
       = Fn × cos (π/2 – ψ) 
       = Fn × sin (ψ))

hence:

Fnp = (const / JB2) × (RJ /JB) × sin (θ) 
      = (const’ / JB3) × sin (θ)

Finally:  ΔF = The net tangential force of Jupiter's 
gravitational on the Sun's tidal bulges 

ΔF = Fnp – Ffp 
     = const’ × [(1 / JB3) – (1 / JA3)]  × sin (θ)

ΔF = const’ × {(1 / [RJ2 + Rs2 – 2 × RJ × Rs × cos(θ)]3/2
         − (1 / [RJ2 + Rs2 + 2 ×RJ × Rs × cos(θ)]3/2)} × sin(θ)

The following graph shows the net tangential acceleration of the  
Sun's surface due to Jupiter's gravitational force acting upon the 
tidal bulges that are induced by Venus/Earth upon the Sun's surface 
as a function of Jupiter's angle θ (please refer to diagram above)
Note: It is assumed that Jupiter's force only acts upon one percent 
of the mass of the convective zone of the Sun (=0.0002 % of the 
mass of the Sun). 


Even under these ideal assumptions, the maximum peak acceleration 
only reaches ~ 3.0 micro-metres per second^2. 
Assuming that half this peak acceleration is applied to 0.02 % of the 
Sun's mass for one full day at each of the roughly seven alignments 
of Venus and Earth over the 11.07 years it takes for θ to change from 
0 to 90, the net change to to the Sun's velocity should be 


acceleration x delta time ~1.5x10^(-6) x 7 x 24 x 3600 ~ 0.91 m/sec

Given these highly optimistic assumptions, it could be argued that 
if Jupiter's gravitational force only had to change the rotational
velocity of one % of the mass of the convective zone of the Sun 
(~ 0.02 % of the mass of the Sun) it would produce a significant
change rotational velocity of this small amount of mass.


Of course, the assumptions used in these calculation require 
that the one % of the Sun's convective layer mass that is affected 
by Jupiter's gravity is dynamically decoupled from the remaining 
0.998 % of the Sun's mass. At this stage, there is no region of 
the Sun's convective layer that is known to be effectively 
dynamically decoupled from the rest of the Sun. In addition, 
even if such region did exist within the Sun's convective zone, 
we have no idea of its relative mass. 


Given these large uncertainties, all we can say is that the 
0.91 m/sec change in rotational velocity is most likely a 
loose upper bound to the real value.   


(N.B. All the arguments given above assume that there are 
no other induced asymmetries in the spherical shape of the Sun 
other than those produced by the combined tidal forces of 
Venus and Earth at the time of alignment).

CONCLUSION

The simple planetary spin-orbit coupling model does not
appear to produce a significant change in the velocity of 
rotation in the outer layers of the Sun.

Hence, in order for it to taken seriously, the planetary 
spin-orbit coupling model would require a considerable 
and as yet unknown amplification mechanism .     


One possible amplification mechanism is discussed here:



6 comments:

  1. Hello! Is there anybody out there?

    ReplyDelete
  2. It's not clear what you mean by spin orbit coupling.

    First, you've only calculated 2 points on the surface of the Sun facing Jupiter (ignoring the mutual rotations.)

    Since the force equation is a second order differential equation in time, you still have to integrate 6 equations for roughly a billion years.

    Second, shouldn't the spin direction be in the direction opposite to the direction of the objects angular momentum is rotating since tidal forces are dissipative?

    This is assuming the spin axis is the same axis as the angular momentum.

    Gravity is weak to begin with - tidal dissipation is even weaker.

    ReplyDelete
  3. Agile Aspect,

    These calculations are not meant to thorough, they are just back-of-the-envelope.

    There is a momentary tidal bulge that appears on the surface of the Sun for a brief period when Venus and Earth align once every 1.6 years. I am assuming that Jupiter's gravity tugs on this tidal bulge over the brief time that it exists (i.e. there is tidal torquing which affects rotation rate of the Sun).

    Let's say that enhanced tidal bulge only lasts while Venus moves from being two degrees behind the Earth in Helio-centric latitude, to two degrees ahead.

    Venus moves at ~ 360/224.70 degrees per day.
    i.e. 1.602 degrees per day.

    Earth moves at ~ 360/365.26 degrees per day.
    i.e. 0.986 degrees per day.

    So each day, Venus moves ahead of the Earth by 0.616 degrees per day.

    So it takes Venus roughly 6.5 days for it to move from being 2 degrees behind the Earth to 2 degrees ahead. This is short compared to the 27 rotation period of the Sun. [note: even if you extended the alignment period from 5 degrees behind to 5 degrees ahead the tidal bulge would only last for 16.2 days, which is still ~ 1/2 the rotation period of the Sun).

    I am not sure about your concerns about the dissipative nature of the tides induced by the alignments of Venus & Earth, since I am only using the tidal bulge as an asymmetric handle that Jupiter's gravity can grasp with its gravity.

    ReplyDelete
  4. I should add that while I am claiming that Jupiter, the Earth and Venus affect the rotation rate of the Sun, the orbital motion of the Sun about the Barycentre of the Solar system is produced by the changing orientation of the Jovian planets.

    Hence, I am not claiming a classical spin-orbit coupling mechanism is responsible.

    It's more like a tidal torquing, produced by the gravity of Jupiter acting upon the momentary tidal bulge produced by alignments of Earth and Venus every 1.6 years that changes the rotation rate of the outer layers of the Sun (e.g. near the tacho-cline).

    I am claiming that this means that variations in the Sun's rotation rate are synchronized with the rate of the Sun's motion about the Solar system's Barycentre for the simple reason that the orbits periods of terrestrial planets are synchronized with the orbital periods of the Jovian planets (particularly Jupiter).

    ReplyDelete
  5. As we have noted above:

    If we assume that the enhanced tidal bulge only lasts while Venus moves from being two degrees behind the Earth in Helio-centric latitude, to two degrees ahead.

    Venus moves at ~ 360/224.70 degrees per day.
    i.e. 1.602 degrees per day.

    Earth moves at ~ 360/365.26 degrees per day.
    i.e. 0.986 degrees per day.

    So each day, Venus moves ahead of the Earth by 0.616 degrees per day.

    Hence, it takes Venus roughly 6.5 days for it to move from being 2 degrees behind the Earth to 2 degrees ahead.

    Given that the maximum peak acceleration reaches ~ 3.0 micro-metres per second^-2, if we assume that half this peak acceleration is applied to 0.02 % of the Sun's mass for 6.5 full days at each of the roughly seven alignments of Venus and Earth over the 11.07 years it takes for θ to change from
    0 to 90, the net change to to the Sun's velocity should be

    acceleration x delta time

    = 1.5x10^(-6) x 7 x x 6.5 x 24 x 3600

    = 5.9 m/sec ~ 6 m/sec

    This is remarkable close to the amplitude of the torsional oscillations seen on the surface of the Sun of ∼6 ms^−1 (0.0086 μrad ^s−1) observed by Howe et al. (2000).

    This raises the intriguing possibility that the torsional oscillations in the outer layers of the Sun may be a product of the mechanism that is responsible for the tidal torquing between Jupiter and the Sun.

    [N.B. It is interesting to note that the average OBSERVED change in the equatorial rotation rate between solar cycles of 0.0062 μrad s^−1, corresponds to a change in speed at the Sun’s equator of 4.3 ms^−1. Also, remarkably close to the 6 ms^-1 cited above]

    ReplyDelete
  6. Ninderthana Apr 12, 2012 11:01 AM


    It's more like a tidal torquing, produced by the gravity of Jupiter acting upon the momentary tidal bulge produced by alignments of Earth and Venus every 1.6 years that changes the rotation rate of the outer layers of the Sun (e.g. near the tacho-cline).


    I had not realized the Earth-Venus pulse every 1.6 years. I also remember Howe's inner solar pulse at around the same interval which you have now given an option for. I think you are making some worthwhile headway here.

    I wrote an article in 2009 on Carl's blog re Howe's paper that might interest some.

    landscheidt.wordpress.com/2009/02/25/latest-solar-differential-rotation-information/

    ReplyDelete