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The eccentricity constant of solar objects



 
 
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  #1  
Old January 3rd 18, 03:21 AM posted to sci.astro
Peter Riedt
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Posts: 83
Default The eccentricity constant of solar objects

The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:

semi major axis a semi minor axis b
MER …..57,909,175,000 .…56,671,636,475
VEN …108,208,930,000 ...108,206,447,840
EAR …149,597,890,000 ...149,577,002,324
MAR …227,936,640,000 ...226,939,989,085
JUP …778,412,010,000 ...777,500,013,843
SAT 1,426,725,400,000 1,424,632,079,805
URA 2,870,972,200,000 2,867,776,762,478
NEP 4,498,252,900,000 4,498,087,097,722
PLU 5,906,380,000,000 5,720,641,563,568

……….a-b …………..a+b X=
MER …1237538525 .....114,580,811,475 1.0
VEN ……..2482160 .....216,415,377,840 1.0
EAR ……20887676 .....299,174,892,324 1.0
MAR …..996650915 .....454,876,629,085 1.0
JUP …..911996157 ..1,555,912,023,843 1.0
SAT …2093320195 ..2,851,357,479,805 1.0
URA …3195437522 ..5,738,748,962,478 1.0
NEP …..165802278 ..8,996,339,997,722 1.0
PLU 185738436432 11,627,021,563,568 1.0

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  #2  
Old January 3rd 18, 08:12 AM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

..5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #3  
Old January 3rd 18, 08:23 AM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 03/01/2018 v 09:12 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #4  
Old January 3rd 18, 03:18 PM posted to sci.astro
Peter Riedt
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Posts: 83
Default The eccentricity constant of solar objects

On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.


I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X.
  #5  
Old January 3rd 18, 08:00 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 03/01/2018 v 16:18 Peter Riedt napsal(a):
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.


I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X.

Where is the improvement wrt the eccentricity e = sqrt(1-(b/a)^2 ) ?

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #6  
Old January 3rd 18, 08:19 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 03/01/2018 v 21:00 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 16:18 Peter Riedt napsal(a):
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.


I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X.

Where is the improvement wrt the eccentricity e = sqrt(1-(b/a)^2 ) ?

P.S.: As that has the very particular geometrical meaning
of relative position of an ellipse focus on the major semi-axis.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #7  
Old January 3rd 18, 11:35 PM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 03/01/2018 v 21:00 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 16:18 Peter Riedt napsal(a):
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' Stř*ž napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the formula
.5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and b = the semi minor axis.
The eccentricity constant X of nine planets is equal to 1.0 as follows:


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.


I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X.

Where is the improvement wrt the eccentricity e = sqrt(1-(b/a)^2 ) ?

P.S.: As that has the very particular geometrical meaning
of relative position of an ellipse focus on the major semi-axis.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.


I have simplified the formula by removing the sqrt part.
  #8  
Old January 4th 18, 07:38 AM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 04/01/2018 v 00:35 Peter Riedt napsal(a):
On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' Stř*ž wrote:

I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1. Orbits are subject to the Law of X.

Where is the improvement wrt the eccentricity e = sqrt(1-(b/a)^2 ) ?

P.S.: As that has the very particular geometrical meaning
of relative position of an ellipse focus on the major semi-axis.



I have simplified the formula by removing the sqrt part.


It is oversimplified in a not useful way.

Similarly as if you would like to get rid of sqrt
in the formula of Pythagoras.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #9  
Old January 4th 18, 09:36 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Thursday, January 4, 2018 at 3:39:00 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 04/01/2018 v 00:35 Peter Riedt napsal(a):
On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' Stř*ž wrote:

I have improved the formula to read 1-3(a-b)^2/(a+b)^2). All objects in
closed orbits around the sun have an X constant between .8 and 1. Halley's comet has an X of about .85. A circular orbit has an X constant of 1.. Orbits are subject to the Law of X.

Where is the improvement wrt the eccentricity e = sqrt(1-(b/a)^2 ) ?

P.S.: As that has the very particular geometrical meaning
of relative position of an ellipse focus on the major semi-axis.



I have simplified the formula by removing the sqrt part.


It is oversimplified in a not useful way.

Similarly as if you would like to get rid of sqrt
in the formula of Pythagoras.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.


sqrt is not necessary for my formula but irreplaceable in Pythagoras
  #10  
Old January 4th 18, 07:20 PM posted to sci.astro
Anders Eklf
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Posts: 98
Default The eccentricity constant of solar objects

Peter Riedt wrote:

On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' St??

wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?? napsal(a):
Dne 03/01/2018 v 04:21 Peter Riedt napsal(a):
The eccentricity constant of solar objects

The eccentricity constant X of solar objects can be calculated by the
formula .5*sqrt(4-3(a-b)^2/(a+b)^2) where a = the semi major axis and
b = the semi minor axis. The eccentricity constant X of nine planets
is equal to 1.0 as follows:


I'm curious:

Where does that formula come from?
What *is* the eccentricity constant in the first place)
How does it relate to e = sqrt(1-(b/a)^2) ?

You listed the semi minor axes (b) of the planets with 11 to 13
significant digits. Impressive!. Where did you get those numbers?
(I have a hunch...)

Even 8 digits for the semi major axes is quite a feat.

Why do you round off to one decimal? Just to prove a point?
Isn't it more interesting to explore the differences? 3 decimals?


So you say a circle has the eccentricity constant 1.0.
Interesting.

.5*sqrt(4-3(r-r)^2/(r+r)^2) = .5*sqrt(4) = 1

Similarly, any ellipse similar enough to a circle
like those of planets has this constant 1.0,
if rounded to 1 decimal place.


Actually, the eccentricity of a circle is 0 (zero).
Poutnik should maybe have pointed that out to you...


I have improved the formula to read 1-3(a-b)^2/(a+b)^2).


How does .5*sqrt(4-3(a-b)^2/(a+b)^2) simplfy to 1-3(a-b)^2/(a+b)^2) ?

Let's do the algebra. Oh bummer!

..5*sqrt(4) = 1 !!! Trust me, you'd fail any high school math test with
that one (pun not intended).

Orbits are subject to the Law of X.


What's the Law of X.?

--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour
 




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