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Joe Keller

USA
957 Posts

Posted - 06 Dec 2010 :  17:31:12  Show Profile  Reply with Quote
An empirical relation between Earth's mass, Earth's radius, Earth's semimajor orbital axis, the speed of light, and the Golden Ratio

Sconzo (Astronomical Journal 67:19-21, 1962) noted that if one chooses the unit of length to be Earth's radius, and chooses the unit of mass to be Earth's mass, then the gravitational constant will be exactly one, if the unit of time is 806.832 sec. That is, the (average) acceleration of gravity at Earth's surface, is one Earth radius per 807sec, per 807sec.

The time for light to travel one Astronomical Unit ( = semimajor axis of Earth's orbit) is 1.4959787*10^13 / (2.9979246*10^10) = 499.0048sec. The Golden Ratio is (1+sqr(5))/2 = 1.618034.

806.832/499.0048 = 1.616882

The ratio, 1.000712, of the Golden Ratio, to this number, differs from one, by about the present relative uncertainty of the gravitational constant. So, the numbers are the same, as nearly as they have been determined.
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Jim

1825 Posts

Posted - 06 Dec 2010 :  17:55:40  Show Profile  Reply with Quote
Dr Je, It seems dreadfully complex to set the gravity constant at one-at least to me. What is to be gained by this project?
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Joe Keller

USA
957 Posts

Posted - 07 Dec 2010 :  13:59:23  Show Profile  Reply with Quote
quote:
Originally posted by Jim

...complex to set the gravity constant at one...


Here is an equivalent way of saying what I said in my previous post:

The acceleration of gravity at the surface of the Planet (not counting centrifugal force due to rotation) times T squared (where T is the light time from the Sun to the Planet, at its semimajor axis, i.e. approx. its average distance from the Sun) times the "Golden Ratio" ( 1.618034 = (1+sqrt(5))/2 ) to some whole power, equals the radius of the Planet. Like the NASA "Fact Sheets", I use the "volumetric average" radius, i.e. the geometric mean of the three semi-axes of the Planet's ellipsoidal shape (two of which are equal). Using the current figures on NASA's Fact Sheets, I find that all the rocky planets, except Venus, and spherical minor planets, on which data are available, conform to this law.

I approximated Ceres as an ellipsoid, used the larger of the two diameters (customary for asteroids) given by NASA as the equatorial, then found the volumetric average radius myself. For Charon, only an equatorial radius was given, so I used it as the average, since for Pluto the equatorial and polar radii are given as equal. Other minor planets are not very round and/or lack moons and have unknown compositions, so their surface gravity is unknown or undefined. The gas giant planets have surface accelerations which NASA gives arbitrarily as that at a depth of one bar pressure in their atmosphere.

The formula is:

(surf. accel.)*(Sun light time)^2 * 1.618...^n = radius of Planet

Here is a list of the values of "n" found:

Mercury 5.968
Venus 3.441
Earth 1.994
Mars 0.951
Jupiter -1.812
Saturn -2.932
Uranus -7.209
Neptune -9.604
Pluto -10.89
Charon -11.04
Ceres 2.818

For the outer planets (Jupiter & beyond) "n" is negative.

For the minor planets Pluto & Ceres the last digits above are not significant, and for Charon the next to last digit is hardly significant. For these planets, NASA gives the masses or surface accelerations only to two, or not quite three, decimal places. Agreement is excellent for Earth; good for Mercury, Mars, and Saturn; and at least fair for Jupiter, Uranus, Pluto, Charon and Ceres.
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Jim

1825 Posts

Posted - 07 Dec 2010 :  14:47:16  Show Profile  Reply with Quote
Dr Joe, I'm not able to determine what "n" stands for-can you explain what it is and what units it has? It looks to me it could be a calculated ratio different for every body in the system. Your latest post sidesteps my question regarding setting the gravity constant at unity.
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Joe Keller

USA
957 Posts

Posted - 07 Dec 2010 :  17:31:50  Show Profile  Reply with Quote
Luna also conforms to the relationship given in my previous post. Using the NASA Fact Sheet figure for Earth's volumetric average radius, but the figures in the 2008 Astronomical Almanac, p. E4, for Earth's mass and for Luna's mass and radius,

mLuna / mEarth / (rLuna / rEarth )^3 = 0.60650,

so if Earth is changed to Luna, in the formula, an extra factor of 1/0.60650 = 1.6488 is needed, that is,

n = 1.994 + log(1.6488)/log(1.618034) = 3.033

so the exponent, n, for Luna, is 3, vs. n=2 for Earth and n=1 for Mars.

Furthermore, the Sun itself conforms fairly well:

(rSun /1AU)^2 * mSun/mEarth / (rSun/rEarth)^3 = 1/1.618...^25.1605

so the exponent, n, for the Sun, is 1.994+25.1605 = 27.15.

Edited by - Joe Keller on 30 Dec 2010 01:44:33
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Stoat

United Kingdom
964 Posts

Posted - 10 Dec 2010 :  16:31:09  Show Profile  Reply with Quote
First thoughts on this Joe, how do you think it relates to the fission v collision models of planetary formation?
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shando

Canada
208 Posts

Posted - 28 Dec 2010 :  15:00:03  Show Profile  Visit shando's Homepage  Reply with Quote
There is supposed to be a post by Joe Keller stamped: 28 Dec 2010 12:21:43. Why can't I view it?
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Stoat

United Kingdom
964 Posts

Posted - 29 Dec 2010 :  05:17:07  Show Profile  Reply with Quote
Hi Shando, it's probably that Joe has edited one of his posts, that shows as a last posting.
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Joe Keller

USA
957 Posts

Posted - 30 Dec 2010 :  02:30:57  Show Profile  Reply with Quote
quote:
Originally posted by Stoat

...Joe, how do you think it relates to the fission v collision models of planetary formation?



I agree, that this fact, that usually a body in our solar system conforms to the formula in my first Dec. 7, 2010, post here, must result from the mechanism of formation. The golden ratio, 1.618... , satisfies the quadratic equation

x^2-x-1=0

and the equations satisfied by powers of a root, are related to the original equation (see Fritz John, "Lectures on Advanced Numerical Analysis", sec. 2.9, "Graeffe Root-Squaring Process", pp. 53-56). Somehow, one of this special family of equations must be satisfied.
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Joe Keller

USA
957 Posts

Posted - 30 Dec 2010 :  02:40:14  Show Profile  Reply with Quote
quote:
Originally posted by Jim

..."n"...what it is and what units it has?



"n" is the exponent of 1.618... . It's dimensionless (it has no units, that is, its value is the same no matter what the units of measurement are).

Edited by - Joe Keller on 30 Dec 2010 03:15:31
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Joe Keller

USA
957 Posts

Posted - 30 Dec 2010 :  02:44:36  Show Profile  Reply with Quote
quote:
Originally posted by shando

...a post by Joe Keller stamped: 28 Dec 2010 12:21:43. Why can't I view it?



Stoat's explanation is right: I submitted my long article on Mars' orbital period, to a journal early this month, then on Dec. 28 improved it.
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Joe Keller

USA
957 Posts

Posted - 30 Dec 2010 :  03:08:11  Show Profile  Reply with Quote
The July 15 & 22, 2008, Avebury Crop Circles: 2012 & Barbarossa

For pictures of the crop formations, see www.realufos.net, blog article dated Aug. 1, 2008 (find by typing "Avebury 2008 2012 July" in that site's search window).

The July 15 crop formation, I've discussed in earlier posts to this thread. The consensus of crop circle expert opinion is that this formation shows the positions of the planets (except that Pluto is perturbed) for Dec. 23, 2012. I myself confirmed that this is true to within a few days.

The less well known nearby second formation, of July 22, was, according to commentary on www.realufos.net, unexplainedly censored by usually reliable sources. It shows, among other things, a small ellipse.

If I measure from the centers of the small circles along that small ellipse, I find that its eccentricity is about the same as the eccentricity of Barbarossa's orbit. Using the apparent (perspective) axis ratio of Neptune's orbit on the same photo (in the July 15 circle), to correct for perspective by multiplying the minor axis of the small ellipse by that factor, I get eccentricity 0.68 based on the small ellipse's axis ratio. Based on the position of the presumed "Sun" mark, i.e. focus, on the long axis of the small ellipse, I get eccentricity 0.53. Thus the best estimate is e=0.605+/-0.053 (SEM). Barbarossa's eccentricity calculated from the sky survey positions, is e=0.6106.

The large circle (of the second formation, July 22) has a small circle attached to it, on the side near the original July 15 formation; the small circle might signify Neptune's orbit and the large circle Barbarossa's. The ratio of the circles' radii (measuring all the way to the edge of the thick white line demarcating the small circle) is about 11.8::1. The ratio of Barbarossa's to Neptune's major axes, is 343.84/30.07 = 11.4::1.

Edited by - Joe Keller on 31 Dec 2010 01:45:37
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Stoat

United Kingdom
964 Posts

Posted - 31 Dec 2010 :  03:46:51  Show Profile  Reply with Quote
Hi Joe, and the "pairing" of planets? The idea that the proto Venus fissions off Mercury, the proto Earth fissions off Mars. This is for distances out to about earth orbit. Fission of a proto planet at about Jupiter distance would throw the fission particle out of the solar system. In this model the fission process would create a filament of droplets, which could explain our moon. Venus could have had a moon but the sun's tidal effects could make it fall into Venus. The smaller fission body would not have moons but could have orbiting collision fragment, caused by impacts of "droplets."

(edited) just for fun I added pairs of n values and checked the ratio, close to three.

Edited by - Stoat on 31 Dec 2010 03:54:48
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Jim

1825 Posts

Posted - 31 Dec 2010 :  14:30:03  Show Profile  Reply with Quote
Dr Joe, On Dec7,2010 you posted a list of values for n indicating every body in the solar system has a different value for n. Then on Dec30 you said n is a constant. Am I ever confused now-can you edit these two posts so they are consistent? Your second value for n is very near the constant called Phi so I wonder if it is in fact that number?
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Stoat

United Kingdom
964 Posts

Posted - 13 Jan 2011 :  06:03:50  Show Profile  Reply with Quote
Hi Joe, a while ago now someone put up a link to a gravity simulator program, i downloaded and intended to learn it but never got round to it. I decided a couple of days ago to take a look at it again. Looking through the simulations that you can download I found this one which has to be of interest to you. http://www.orbitsimulator.com/gravity/articles/sedna.html

I did try and add a "Barby" to the solor system model, and bearing in mind that I'm still learning this app, it looks as though Barby has the most effect on Mercury's orbit. Though that's really down to Jupiter.
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Stoat

United Kingdom
964 Posts

Posted - 19 Jan 2011 :  10:48:25  Show Profile  Reply with Quote
Playing with the Sedna simulation, it struck me that this program could be used to simulate the EPH idea. A quick try out, I made Mars have a Neptune mass and then stuck a bunch of stuff in orbit round it. One of them got nicked by Jupiter. Then I dropped the Neptune mass Mars, down to a mass of zero. Its planets went into a bunch of orbits with varying eccentricities. Anyway, I think it looks quite promising, is there anyone here who knows the program well?
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Stoat

United Kingdom
964 Posts

Posted - 26 Jan 2011 :  06:41:36  Show Profile  Reply with Quote
A little update on that gravity simulator. If you go to the website and click on the messageboard, there's a topic for a newer beta version. Scroll down to the bottom of the thread and download and save the three files there. Create a new folder in the grav sim folder and copy the new files into that. There's a lot more functionality to the beta version, including being able to save out stills of the orbits, create a bunch of orbiting objects with a Gaussian distribution etc.

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Larry Burford

USA
2115 Posts

Posted - 27 Jan 2011 :  19:23:55  Show Profile  Reply with Quote
Stoat - that's an interesting kludge to get this simulator to do stuff it was not designed to do.

Kudos

If you are interested in making a more realistic explosion, try these suggestions:

{
According to Tom's most recent thinking, Mars and at least one other object (called Body C, and about the same size as Mars) were moons of a larger planet called Body V that exploded way back when. Mars and Bocy C then orbited their common center of mass until about 3.5 MYA, when Body C exploded.

Body V was a helium class planet of about 5 Earth masses (m_e). When it exploded almost all (99.99%) of it would have vaporizd.
}

To simulate this one -

  • Add a 5 [m_e] mass at about the orbit of Mars

  • Adjust it's radius to about 1 [km]

  • Adjust Mars to orbit at about 5 E5 [km]

  • Add and Adjust Body C to orbit at about 3 E5 [km]

  • Add a large number of very small masses (as many as you can stand or as the simulator will allow) (Their masses should be tiny, say a thousand tonnes?)(These will simulate the vaporized interior of the planet. In reality these pieces will be a lot smaller [perhaps one gram each, on average?] but for a simulator like this we can take some poetic license.)

  • Set them to orbit between 1 E2 [km] and 2 E3 [km] (This causes their orbital velocities to be very large, so that when they are released they fly off like explosion fragements. Some experimenting will probably be needed to get this to look right.)



Now (if you can get this conglomeration to actually stay in orbit) when you turn down the mass of Body V all those tiny masses will fly out and create something like a blast wave. It should sweep the solar system and escape, with very little left behind.

Mars and body C should end up about where they are, orbiting each other.

{
Body C is small enough that its explosion will be less energetic and should leave a little of its crustal material (1 or 2 or 3% ?) unvaporized.
}

To simulate this one -

  • Start with Mars and Body C orbiting each other

  • Adjust Body C's radius to about 1 [km]

  • Add a large number of very small masses (as many as you can stand or as the simulator will allow)

  • Set them to orbit Body C between 60 [km] and 2 E3 [km]

  • Divide the mass of Mars by 100

  • Divide this mass by 1000

  • Add 1000 or 2000 or 3000 masses of this value orbiting Body C at about 6 E3 [km] (to simulate the unvaporized pieces of the crust)



Now (if you can get this conglomeration to actually work) when you turn down the mass of Body C all those tiny masses will fly out and create something like a blast wave with chunks.

If you experiment you might get some of the chunks to orbit the sun sort of like asteroids ... and some others like comets

(Can this simulator save anything that can be posted here?)


Regards,
LB
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Jim

1825 Posts

Posted - 28 Jan 2011 :  13:09:57  Show Profile  Reply with Quote
LB, Do you know the background data that TVF was using to develop the idea about these bodies? I never have found data TVF said indicated an explosion let alone several.
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Larry Burford

USA
2115 Posts

Posted - 28 Jan 2011 :  18:01:54  Show Profile  Reply with Quote
Jim,

I sympathize. I do in fact know parts of it. But not all. It is called a PhD in astrophysics with specialty in celestial mechanics. Mixed in with about 25 years of mainstream work experience. (That's right, Tom used to be one of THEM. So did I.)

The numbers I use in my posts come partly from the various books and papers Tom wrote over the years (talking about his exploded planet theory in this case). And partly from various standard reference books. And they come partly from my own mind. I've spent years reading and re-reading his stuff. And then re-reading my old physics textbooks. And thinking about what he said and did not say.

I rarely dig up raw observational data and process it myself. It takes decades. If someone is not paying you (and several coworkers) to do it you will probalby never finish.

Wish I could show you the secret of the universe, but honestly I have not found it. (Yet. But if you track down all the stuff I've written here over the years you can get a feel for how my search is going and where it is going.)

I doubt that it will ever be possible to condense everything down to one theory. One equation. That's the holy grail of modern physics - the quest for a TOE (theory of everything) or GUT (grand unified theory). The universe is infinite in five dimensions. One size fits all? Not likely.

I found an interesting quote recently. It seems to fit here.

"Always trust those searching for The Truth, never those who have found it."
~ Jordan Maxwell

Anyone know who he is? (Probably religious. It comes from a hippy website.)

LB
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