# Thread: Why do these planetary 'Spin Ratios' work out to 'unity'?

1. ## Why do these planetary 'Spin Ratios' work out to 'unity'?

QUESTION: Why do planet ratios employing Orbital Energy, Black Body Kelvin, orbital periodicity, and square root of AU, gives an answer that appears to be something of a constant?

(This had been discussed on another forum, posted (Jan. 29, 2010): http://www.humancafe.com/discus/mess....html#POST5413 about Spin Ratios where the ratio of about "one" keeps coming up, if divided (multiplied in the denominator) by a spin-ratio (Earth's) of SR =~2.32, when figuring the interrelations of Planet's Kelvin black-body and orbital Energy.)

In fact, the long original equation can be reduced and better written as:

Spin = [PK^2 x Ev^2 x 365 x (AU^0.5)] x 1/ [(231.7K^2) x (PE * 17.33E+16J) x orbit days x ~SR] = ~1

In words, it means:

Planet Kelvin squared, times Venus orbital Energy squared, times 365 Earth days, times square root of AU distance from the Sun; all divided by Venus Kelvin squared, times planet Energy, times Venus orbit Energy, times orbit days, times (inverse) Spin Ratio 2.32, equals aprox =~1.

Note: Planet orbital Energy is a value derived from Solar radiance received by planet (W/m^2), times 1/2 mean distance from the Sun, times mean orbital velocity squared, to give us Energy E' for that planet.

When calculated for Earth's orbital Energy, taking data from "Nasa Planetary Fact Sheet" [ http://nssdc.gsfc.nasa.gov/planetary/planetfact.html ] per equation:

E' = solar irradiance x 1/2 Rv^2

Earth's E' is:

Mean distance from Sun: 149.6E+9 meters
Mean orbital velocity: v = 29.78 km/s

(1367.6) (149.6E+9) = 204592.96E+9 = 2.046E+14 W/m = solar radiance energy

KE = (1/2) (1) (29.78)^2 = (1/2)(1)(886.85) = 443.4 m^2.kg.s^-2 (Joules)

KE * W/m = ( 443.4 J) (2.046E+14 W/m) = 9.07e16 Joules (times Newton force N) = Earth's total orbital Energy. (Please note m = 1 is a net function of planet mass already in orbital motion, and the planet KE is 'template' function only.)

Earth's E = ~9.07E+16 J (which is close to E = mc^2 = 90 petajoules). This Energy level (for Earth) then sets the basic template for like Energy E' levels for the other planets, using the same methodology:

MERCURY: 60.55E+16 J
VENUS: 17.33E+16 J
EARTH: 9.0E+16 J
MARS: 3.66E+16 J
JUPITER: 0.335E+16 J
SATURN: 0.1004E+16J
URANUS: 0.0247E+16J
NEPTUNE: 0.01E+16 J

This was then found true for all the planets, from Mercury to Neptune (Pluto was left out as an odd-ball, not in same orbital plane with the Sun as the others). When these ratios are worked out for Earth, the spin-ratio is SR=2.32, a dimensionless number. When all the other planets have their ratios worked out by same equation, if divided (1/multiplied) by Earths 2.32 ratio, they then approximate SR=~1. Why?

Maybe there is something that nets out I can't see? (BTW, I had mentioned this 5-6 years ago briefly on this forum when I first calculated it, but nothing was resolved then.)

That is the question posed here for readers of Against the Mainstream. Nothing like this exists in current astronomical/astrophysical literature, to my knowledge, so it is obviously NOT Mainstream. However, it is not even Against Mainstream, since it is not a theory nor hypothesis, merely a play on numbers using known planetary data that just happen to stack up this weird way, where all the SR spin-ratios for all the planets approximate unity (except Venus). Why would it be?

Here is an example of how those numbers work:

Taking the stats for the planets (from NASA page), with above equation for Spin:

AU; PLANET; P-KELVIN; P-ENERGY; ORBIT DAYS; SPIN; (AU)^1/2;
0.39 ; Mercury; 442K ; 60.55E16 J; 88 days; 58.8 ; 0.624 ;
0.72 ; Venus ; 231.7K ; 17.33E16 J; 244 ; -245 (retro?); 0.850 ;
1.0 ; Earth ; 254.3K ; 9E16 J ; 365 days; 1 ; (2.32 base SR)
1.5 ; Mars ; 210.1K ; 3.86E16 J; 687 ; 1.03 ; 1.225 ;
5.2 ; Jupiter ; 110K ; 0.335E16 J; 4329 ; 0.415 ; 2.28 ;
9.5 ; Saturn ; 81.1 K; 0.1004E16J; 10753 ; 0.455 ; 3.08 ;
19.2; Uranus ; 58.1K ; 0.024E16J; 30660 ; 0.718 ; 4.38 ;
30 ; Neptune ; 46.6K ; 0.01E16J ; 60225 ; 0.673 ; 5.48

By the numbers, where planet's orbital Energy is in bold:

Spin = [PK^2 x Ev^2 x 365 x (AU^0.5)] x 1/ [(231.7K^2) x (PE * 17.33E+16J) x orbit x ~SR]

Mercury: [442K^2*17.33E16J^2*365*0.624] x 1/[231.7K^2*60.55E16J*17.33E16J*88*2.32 =?
[195364K*300.3E32J*227.8] x 1/[53684.9K*1049.33E32J*204.2] = ?
[13354530 / 11503230 = 1.16 (slightly high)

Venus: [231.7K^2*17.33E16J^2*365*0.85] x 1/[231.7K^2*17.33E16J*17.33E16J*244*2.32=?
canceling likes = [310.25] / [566.1] = 0.548 (off unity by ~ half)*

Earth: [(254.3K^2*17.33E16^2*365*1)] x 1/[231.7K^2*(9E16 J*17.33E+16J)*365*2.32] = Spin
[64668.5*17.33E16J^2*365] x 1/[53684.9*9E16J*17.33E+16J*365*2.32] = Spin
[64668.5*300E32J] x 1/[53684.5*9E16J*17.33E+16J*2.32] = Spin
[20005500E32J/ 19425898E32J = 1.03 (vs. Earth = 1 spin)

Mars: [210.1K^2*17.33E16J^2*365*1.225] x 1/[231.7K^2*(3.86E16J*17.33E16J)*(687?)*2.32 = ?
[44142K*300E16J*447.1] x 1/[53684.9K*3.86E16J*17.33E16j*687*2.32] = 5920766.5 / 5723777.43 = 1.034

Jupiter: [110K^2*17.33E16J^2*365*2.28] x 1/[231.7K^2*(0.335E16J*17.33E+16J)*4329*2.32 =?
[12100*300.3E32J*832] x 1/[53684.9K*5.8E32J*10043.3] = 3023180160 / 3127106626 = 0.967

Saturn: [81.1K^2*17.33E16J^2*365*3.08] x 1/[231.7K^2* (0.1004E16J*17.33E16J)*10753*2.32 =?
[6577.2K*300.3E32J*1124] x 1/[53684.9K*1.74E32J*24947] = 2220049672 / 2330342329 = 0.95267

Uranus: [58.1K^2*17.33E16J^2*365*4.38] x 1/[231.7K^2*(0.024E16J*17.33E+16J)*30660*2.32 =?
[3375.6K*300.3E32L*1598.7] x 1/[53684.9K*0.416E16J*71131] =?
[1620590488 / 1588562819 = 1.02

Neptune: [46.6K^2*17.33E16J^2*365*5.48] x 1/[231.7K^2*(0.01E16J*17.33E16J)*60225*2.32] =?
[2171.6*300.3E32J*2000] x 1/[53684.9K*0.1733E16J*139722] = 1302600000 / 12999916645 = 1.00

Why would this happen?

I think the most puzzling aspect of this exercise is that of all variables, I would not expect the square root of planet's distance from the Sun, expressed in AUs, to be part of this 'unity' result. Again, why this variable (AU^1/2) need be included? Wouldn't a simple ratio of planet orbital energy and planet's back body Kelvin give similar results? But they don't. That appears a real mystery to me! Hope it is to you too.

Ps: I will be leaving in a few days for a two month trip in Italy and Europe, so may not always be able to answer queries, it depends on Wi-Fi availability on my iPhone. I surely will not be back within the 30 day window this thread might run, but hope to look in and try to respond as I am able.
Last edited by nutant gene 71; 2012-Jul-22 at 05:21 AM. Reason: fixed BB code, deleted failed smiley, added parenthetical

2. How can Planet Kelvin be 231 K for Venus?
Secondly you may want to define your variables a little better

Why is E' equal to solar irradiance (Watts per square meter) times 0.5 R v^2? (radius times velocity squared, or is that Rv, it is unclear from what you calculate under that line where it seems only to be a velocity) what does that represent 0.5 R v^2?
KE seems to be kinetic energy per Earth unit or what? Or do you really take a 1 kg mass?
Then what is KE * W/m supposed to be? I guess you mean KE * E'?

Please rewrite this in an understandable way, that we can follow your math and reasoning for calculating these values etc.

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Where to start...

This can be broken down into a series of (arbitrary) ratios and constants. You use Venus's orbital energy term twice as well, it cancels out. Your equation becomes:

r_temperature_venus^2 . sqrt(k_radius) / (2.32 . r_energy_venus . r_period_earth)

(using Stephan Boltzmann I = KT^4 so t = (I/k)^(1/4), taking the artios cancels out the constants)
(orbital velocity is proportional to r^(1/2), you multiply by r making your term proportional to r^3/2)
(using standard Keplerian circular orbits T = 2.pi (r^3 / GM)^(1/2) reducing as a ratio to r^(3/2))

Replacing the ratios with these we get:

Remembering that one AU is defined as radius_p / radius_earth we can write this as:

So what are we left with? You are effectively working out something very close to the ratio between the ratio of the Earth's orbital distance and Venus's over and over again. Effectively you are working out 1 / (2.32 * Venus's distance in AU) which works out to be 1.3. Deviations from the perfect and simple formulae I have used probably account for your variability.

So since everything cancels it is more of a surprise that you didn't get the same number every time. And also I hope I have explained why you need your square root of distance in AU term - it is to cancel out your ratio of orbital periods better.

Mystery solved.

4. Originally Posted by tusenfem
How can Planet Kelvin be 231 K for Venus?
Secondly you may want to define your variables a little better

Why is E' equal to solar irradiance (Watts per square meter) times 0.5 R v^2? (radius times velocity squared, or is that Rv, it is unclear from what you calculate under that line where it seems only to be a velocity) what does that represent 0.5 R v^2?
KE seems to be kinetic energy per Earth unit or what? Or do you really take a 1 kg mass?
Then what is KE * W/m supposed to be? I guess you mean KE * E'?

Please rewrite this in an understandable way, that we can follow your math and reasoning for calculating these values etc.

Thanks tusenfem, for quick response on my puzzling equation. The origin of this was trying to fit some sort of pattern explaining why planets spin (other than conserved momentum) relating internal planetary/black-body heat with the Sun's energy density where planet's orbit resides. Well, that didn't go anywhere, but this curious formula came out of it, which I called "spin ratios", SR, which always seemed to balance out at unity, or approximately. I didn't think I had duplicated Venus in both the numerator and denominator, perhaps I unknowingly had (as per Shaula, above), so this would be an obvious error, if Venus cancels out. The reason I picked Venus (in my dim logic 6 years ago) was because it had virtually no spin (other than slight retro) in relation to its orbital period, so used her as a base planet. As such, this equation was formed out of misconceived 'failed' concept regarding planet spins, but I looked at it once in a while, out of curiosity. So shared it here as a puzzle, to see what's with it.

About Venus black-body 231K, that's what it was those years ago. Now I see it had been changed to 184K per NASA fact sheet. Others have it at 251K, so don't know if we really have a fix on her Kelvin. Venus with its thick cloud cover and high albedo is something of mysterious mystique, so don't know why it was 231K when I looked it up then. That was the value I found... Without crunching the numbers, I think no matter what value K we use for Venus, the end result would be the same, except the interim value would change (Earth's SR would be other than 2.32, but end brings them back to ~1).

The E' I designed then was to incorporate both solar irradiant energy E with planet's kinetic energy KE, so two forms of energy were working together, under principle that Energy is Energy no matter from what source. The equation was to take radiance as W/m^2 time the planet's kinetic energy as Joules, as a product of both. KE is simply 1/2 m*v^2, if I remember my high school physics. But here I reasoned that since the planet is already in a balanced orbit, it's mass cancels to m=1, as it is no longer a factor. (I.e., no matter the size mass of the planet its stable orbit is already factored in, so it effectively cancels out, same as an astronaut's lower mass is canceled out while on space walk near ISS of much greater mass, they are both m=1, so move in unison.) So revising KE=1/2 (1) v2 is how how that was arrived at. Then I re-introduced the distance R from Sun to planet as a 'density' factor, meaning that the Sun's radiant energy is concentrated along that whole distance, so W/m^2 now becomes times distance R in meters, and reduces to W/m, in effect. So total E' is a product of these two forms of energy, E and KE, for each planet.

Yes, E*KE(R)= E' is how I saw it.
Last edited by nutant gene 71; 2012-Jul-22 at 02:10 PM.

5. Originally Posted by Shaula
Where to start...

This can be broken down into a series of (arbitrary) ratios and constants. You use Venus's orbital energy term twice as well, it cancels out. Your equation becomes:

r_temperature_venus^2 . sqrt(k_radius) / (2.32 . r_energy_venus . r_period_earth)

(using Stephan Boltzmann I = KT^4 so t = (I/k)^(1/4), taking the artios cancels out the constants)
(orbital velocity is proportional to r^(1/2), you multiply by r making your term proportional to r^3/2)
(using standard Keplerian circular orbits T = 2.pi (r^3 / GM)^(1/2) reducing as a ratio to r^(3/2))

Replacing the ratios with these we get:

Remembering that one AU is defined as radius_p / radius_earth we can write this as:

So what are we left with? You are effectively working out something very close to the ratio between the ratio of the Earth's orbital distance and Venus's over and over again. Effectively you are working out 1 / (2.32 * Venus's distance in AU) which works out to be 1.3. Deviations from the perfect and simple formulae I have used probably account for your variability.

So since everything cancels it is more of a surprise that you didn't get the same number every time. And also I hope I have explained why you need your square root of distance in AU term - it is to cancel out your ratio of orbital periods better.

Mystery solved.

Nice response Shaula, thanks for looking at it. I recall ~6 years ago when this was first mentioned, one poster (forget who) did a math analysis and found no duplication, so I didn't bother to look further. But he did find that it all reduces to "unity", something I had not realized until then. So this was really not my idea! So if Venus is duplicated, and cancels, the whole idea is moot, of course. However, planet Kelvin squared times Venus Energy Ev squared in numerator, is not the same as planet energy PE times Venus Energy (17.33E_16J) in denominator, though I see the redundant Venus Energy is above and below. So they do cancel out as you show. Then, if we cancel them out, one less constant above and below, there is still the rest of it. That simplifies the equation further, a good thing, but still leaves consistent results, if I am not mistaken. So it should still all reduce to approx. unity once again. It would take some numbers crunching to prove that, of course, but I suspect nothing is changed.

I had not figured on Stephan Boltzmann's constant, as a black-body constant in thermodynamics, so new to me. But in your other, "you are working out 1 / (2.32 * Venus's distance in AU) which works out to be 1.3", I don't get the same 1.3 when I tried it; got ~0.6 instead. Don't think this represents a constant ratio of Earth to Venus distances, or at least I can't see it. That said, it can be that temperature and energy factors do related to distance from Sun factors, as expected. But why the square root of AU? That is still the main question, one I don't have an answer to, so still a Mystery.

[BTW, an aside, the square root of AU, taken as inverse, 1/AU^1/2, works out to be 'almost' the Pioneer Anomaly, where -a = -8.505E-7.5 cm/ s^2 in centimeters, vs. 8.74E-8 cm/ s^2 observed. A coincidence, perhaps?]
Last edited by nutant gene 71; 2012-Jul-22 at 02:23 PM. Reason: added [BTW..]

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There is no mystery. I have explained it - the quantities you have calculated are all ONLY dependent on the orbital radius. When you do the maths all the radius parts cancel out leaving you with a constant value. You are basically saying it is amazing that 2x / 3x is a constant. Because you take ratios the constant of proportionality all cancel leaving you simple radius ratios. Which then cancel thanks to the way you are multiplying them.

I may need to go back to work it all through again to prove it for certain but I think the onus is on you to do that. It is your theory. So:

Can you please substitute in all the basic astrophysical equations for orbital speed, period, blackbody temperature and so on and show that the radius terms do not cancel? Show all working.

7. Originally Posted by Shaula
There is no mystery. I have explained it - the quantities you have calculated are all ONLY dependent on the orbital radius. When you do the maths all the radius parts cancel out leaving you with a constant value. You are basically saying it is amazing that 2x / 3x is a constant. Because you take ratios the constant of proportionality all cancel leaving you simple radius ratios. Which then cancel thanks to the way you are multiplying them.

I may need to go back to work it all through again to prove it for certain but I think the onus is on you to do that. It is your theory. So:

Can you please substitute in all the basic astrophysical equations for orbital speed, period, blackbody temperature and so on and show that the radius terms do not cancel? Show all working.
So you say Kelvin black-body temps and square root of AU are irrelevant? Sorry, can't see it. The way it was written originally, except for canceling out constant Venus orbital E above and below, once, is the only modification I can see for now. But results are basically same anyway. I will grant you that these results are distance to orbit related, as expected. The Sun's total energy, both as radiant energy and its gravity, are indeed distance related. The planets are in orbit around it, so stay in equilibrium vis-a-vis this solar energy. No mystery there. But the AU square-root factor is still Mystery, as far as I understand it... or don't.

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What mystery? Really. Do the maths. Please. You will see that if you had put in a not rooted term there then the ratio would have grown with increasing radius and not be anywhere near unity. There is no mystery. You had to put it in to approximately balance the other terms. Simple. Really no mystery at all.

As I have explained, ALL your terms are basically functions of R. Because you have written them as ratios then all the constants cancel. All you are left with are powers of R which cancel. See for yourself. I found the solution to your mystery with a piece of paper and about twenty minutes doodling. While trying to beat a headache from hell. Don't try to see it, work it out. This is an example of bundling random numbers together until you get a 'surprising' result without trying to see where the result comes from.

Sorry Shaula about 'headache from hell' ... I empathize. The above is where you lost me on first line. How does "temperature ratio reduces to radius of planet/ radius of Venus," together taken to fourth square? This was a leap I couldn't take to understand it. My bad, no doubt... I'm afraid this lost me.

(PS) That said, I can see how if you net everything out as you suggest, it would lead to progressively larger values based on distance. But that would be ignoring an important part of the equation in the denominator, which in effect divides the results by the "square-root of AU" for the orbital distance examined. Without that one important function, the whole thing then is only a "function of distance" as you suggest. So this is an important component. What remains a "mystery" to me is why is this so? What is it about (AU)^1/2 that makes it work like this?.... So I am no nearer an answer than five years ago!
Last edited by nutant gene 71; 2012-Jul-22 at 11:05 PM. Reason: added (PS)

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I made some basic errors due to said headache. Here is another try:

Blackbody temperature equation:
Flux = sigma T^4
This means that for a body at radium R from the Sun its blackbody temperature is given by
I / 4pi.R^2 = sigma T^4
or to put it another way
T = k_t . R^(-2/4)
T = k_t . R^(-1/2)

Orbital period works out as:
P = 2.pi (R^3 / GM)^(1/2)
Or
P = k_p . R^(3/2)

Your orbital energy term is more complicated:
You work it out as
E = I . R . 0.5 . V^2
But V = (GM / R)^(1/2)
So E = k_e
The reason E is fairly constant across the planets is that it is really just a residual factor coming from your awful mixing of units and dimensions. It varies by a factor of a thousand despite the huge range of velocities and ranges. So it looks like it is a junk parameter serving only to add noise and highlight where the basic assumptions of the simple orbital models are breaking down.

So what are we left with?

k_t . r_planet^(-1) . k_e . k_p . r_earth ^(3/2) . r_planet^(1/2)
--------------------------------------------------------------
k_t . r_earth^(-1) . k_e . k_p . r_planet^(3/2) . r_earth^(1/2)

Simplifying:
r_planet^(-1) . r_earth ^(3/2) . r_planet^(1/2)
----------------------------------------------
r_earth^(-1) . r_planet^(3/2) . r_earth^(1/2)

Simplifying more
r_planet^(-1/2) . r_earth ^(3/2)
-------------------------------
r_earth^(-1/2) . r_planet^(3/2)

Or:
r_earth^2 / r_planet^2

Now that just upset me. Didn't cancel. So I dug. Plot your Energy parameter against orbital radius. When you plot a power fit to it you get a VERY good match for E = k_e . r^(-2). I have no idea why this is the case, something I have missed in your analysis. Anyway we can reinsert the energy parameter and redo our simplification:

k_t . r_planet^(-1) . k_e . r_venus_(-2) . k_p . r_earth ^(3/2) . r_planet^(1/2)
----------------------------------------------------------------------------
k_t . r_earth^(-1) . k_e . r_planet^(-2) . k_p . r_planet^(3/2) . r_earth^(1/2)

Simplifying:
r_planet^(-1) . r_venus_(-2) . r_earth ^(3/2) . r_planet^(1/2)
-------------------------------------------------------------
r_earth^(-1) . r_planet^(-2) . r_planet^(3/2) . r_earth^(1/2)

Simplifying more
r_planet^(-1/2) . r_earth ^(3/2) . r_venus^(-2)
-----------------------------------------------
r_earth^(-1/2) . r_planet^(-1/2)

Or:
r_earth^2 / r_venus^2

Whoo hoo! All the Rs cancel. We are left with a factor 1 / 2.32 . r_venus_au or about 0.6. Throw in the residuals from all the horrible unit mixing and it is hardly surprising that you do not get 0.6 but instead get some other number. The only mystery is what the scale factors from the mixed units are and without looking more into the tangle that is your energy parameter I cannot really say much about that.

The basic answer is the same: The ratio is constant because ALL the variable terms cancel out. The square root AU is just what is required to make this exact. Without this factor you would be left with a r_planet ^ (1/2) in there and the ratio would not be constant.

11. ## I think we got it!

Originally Posted by Shaula
I made some basic errors due to said headache. Here is another try:

Blackbody temperature equation:
Flux = sigma T^4
This means that for a body at radium R from the Sun its blackbody temperature is given by
I / 4pi.R^2 = sigma T^4
or to put it another way
T = k_t . R^(-2/4)
T = k_t . R^(-1/2)

Orbital period works out as:
P = 2.pi (R^3 / GM)^(1/2)
Or
P = k_p . R^(3/2)

Your orbital energy term is more complicated:
You work it out as
E = I . R . 0.5 . V^2
But V = (GM / R)^(1/2)
So E = k_e
The reason E is fairly constant across the planets is that it is really just a residual factor coming from your awful mixing of units and dimensions. It varies by a factor of a thousand despite the huge range of velocities and ranges. So it looks like it is a junk parameter serving only to add noise and highlight where the basic assumptions of the simple orbital models are breaking down.

So what are we left with?

k_t . r_planet^(-1) . k_e . k_p . r_earth ^(3/2) . r_planet^(1/2)
--------------------------------------------------------------
k_t . r_earth^(-1) . k_e . k_p . r_planet^(3/2) . r_earth^(1/2)

Simplifying:
r_planet^(-1) . r_earth ^(3/2) . r_planet^(1/2)
----------------------------------------------
r_earth^(-1) . r_planet^(3/2) . r_earth^(1/2)

Simplifying more
r_planet^(-1/2) . r_earth ^(3/2)
-------------------------------
r_earth^(-1/2) . r_planet^(3/2)

Or:
r_earth^2 / r_planet^2

...[SNIP]...

Or:
r_earth^2 / r_venus^2

Whoo hoo! All the Rs cancel. We are left with a factor 1 / 2.32 . r_venus_au or about 0.6. Throw in the residuals from all the horrible unit mixing and it is hardly surprising that you do not get 0.6 but instead get some other number. The only mystery is what the scale factors from the mixed units are and without looking more into the tangle that is your energy parameter I cannot really say much about that.

The basic answer is the same: The ratio is constant because ALL the variable terms cancel out. The square root AU is just what is required to make this exact. Without this factor you would be left with a r_planet ^ (1/2) in there and the ratio would not be constant.
Thanks Shaula, that was beautifully done! And very clear. But in your last you say "The square root AU is just what is required to make this exact." And yes, that is how both the top and bottom equal out to unity. But that is the WHOLE question! Why does that need to be there to make it balance out to =1? Why does AU^1/2 match for ALL the planets to make it work? I don't know if you really appreciate what this is saying. And everybody else reading has remained silent. So I truly appreciate your examining this (mine much messier than yours, but that was how I reasoned it out originally, so simplifying as you did is the better equation) because you too confirm that including AU^1/2 is needed balances out to unity, same as I found. My crunched numbers sometimes were off from =1, especially Venus, but they were ball park. BTW, by my strange equation to make it come to unity for Venus, I figure out Venus's black-body came out to 312K (about human body temperature, that's warm goddess Venus!) which was a surprise too. Venus, at 0.72 AU, is a mystifying planet indeed. in any case, I still wonder why this R^1/2 value is needed, what purpose it serves, does it correlate with anything else we know? ..... The only answer I found is that, if taken in centimeters, it sort of approximates Pioneer Anomaly's acceleration towards the Sun, and possibly MOND parameters. But beyond that, I have no idea.

Regardless, thanks for giving this a go, I learned couple of things from it: 1) redundant Venus orbital Energy above and below cancel, merely canceling two equal constants; 2) it all reduces to planet distance relationships, though to balance this to unity, we need AU^1/2 in denominator.

No small feat, IMHO. Will be traveling in couple of days, so may not be able to respond to further queries after that, unless find free Wi-Fi. Cheers.
Last edited by nutant gene 71; 2012-Jul-23 at 07:48 PM. Reason: smilies don't work? :D

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in any case, I still wonder why this R^1/2 value is needed, what purpose it serves, does it correlate with anything else we know? ..... The only answer I found is that, if taken in centimeters, it sort of approximates Pioneer Anomaly's acceleration towards the Sun, and possibly MOND parameters. But beyond that, I have no idea.
I thought it was obvious why it was needed from what was worked out above. Your quantities are arbitrary, there is no reason to multiply them in the way you have. The sqrt(r) is just there because your other terms happen to lead to an imbalance of powers of R. There is no reason for these terms to balance, so there is no reason for there not to be some arbitrary power of R needed to balance things. It really all hinges on the energy term. Why it happens to have an r^(-2) dependency. If it was something different (and I see no reason for it to have the form it does) then you might need an R, R^2, R^7/2, whatever was needed to balance the books.

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Originally Posted by nutant gene 71
The only answer I found is that, if taken in centimeters, it sort of approximates Pioneer Anomaly's acceleration towards the Sun, and possibly MOND parameters.
It is my understanding that the Pioneer orbital motions have been understood since May 2011.

http://arxiv.org/abs/1104.3985

Are you arguing that some further aspect needs explaining?

14. Originally Posted by utesfan100
It is my understanding that the Pioneer orbital motions have been understood since May 2011.

http://arxiv.org/abs/1104.3985

Are you arguing that some further aspect needs explaining?
No, not at all. Simply noted a 'coincidence', that Pioneers Anomaly mimics the (AU)^1/2 factor needed, as Shaula pointed out, to balance both top and bottom of equation, to give us unity. That was the surprising thing, this coincidence, not hypothesis or theory. It seems conclusive Rievers and Lammerzahl (including Turyshev) found the anomaly in heat radiation pressures, per your reference. Chalk it up to coincidence.

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I am still not sure what your equations mean, or the terms used in them, or what they represent.

I do notice that 1AU is the radius of our orbit, and for some kind of spin process the circumference might make more sense. Using the circumference in your equation gives a AU, or a factor of 2.51.

This appears to agree with the observation at least as well as your 2.32 based on Earth's parameters. It might be interesting, then, to arrange the planets by eccentricity and see if there is a correlation with your ratio.

It also appears to me that your equation is equivalent to arguing that the following expression is a constant for all planets:

where:
T is the temperature of the planet
P is the orbital period
E is the orbital energy (apparently the radiant energy of the sun at the average orbital distance, not the kinetic energy of the mass)
r is the radial distance from the Sun

Kepler's third law relates P and r, with , reducing the relation to .

If you want to argue that this is against the mainstream, you will need to determine the expression for the equilibrium temperature of a planet as a function of distance from a point source with a radiant energy E, and show that this differs from , for some constant k.

I don't know the relation, so I can't say if your equation is mainstream or not.

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It also appears to me that your equation is equivalent to arguing that the following expression is a constant for all planets:
That is the other way to do it. Removing all constant terms you get:
T^2 . r^1/2
---------------
E . P

T is proportional to r^(-1/2)
E is proportional to r^(-2)
P is proportional to r^3/2

Subbing:
K . r^(-1) . r^(1/2)
-------------------
r^(3/2) . r^(-2)

gives
K . r^(-1/2)
------------
r^(-1/2)

or K...

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Originally Posted by Shaula
That is the other way to do it. Removing all constant terms you get:
T^2 . r^1/2
---------------
E . P

T is proportional to r^(-1/2)
E is proportional to r^(-2)
P is proportional to r^3/2

Subbing:
K . r^(-1) . r^(1/2)
-------------------
r^(3/2) . r^(-2)

gives
K . r^(-1/2)
------------
r^(-1/2)

or K...
Are you saying that the temperature of bodies at thermal equilibrium with a central star is known to be proportional to r^(-1/2)? If so, than this justifies the OP's observation as being understood by mainstream science.

As the OP also observed, this relation would not be exact because the planets have excess heating from residual energy from the collisions that formed them, radioactive isotopes in their composition and tidal frictions that contribute non negligible components to the temperature.

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OP is using the planets black body temperature. It is explicitly stated. That is why the Earth's temperature is given as 254K rather than the more usual 287 average.

The beef I have is that the 'energy' term makes no sense at all. And that the relationship is meaningless. It is just a bunch of things thrown together and then said to be significant. Well, they are not.

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