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The eccentricity constant of solar objects



 
 
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  #21  
Old January 7th 18, 12:08 AM posted to sci.astro
Anders Eklöf
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Posts: 100
Default The eccentricity constant of solar objects

Peter Riedt wrote:

On Friday, January 5, 2018 at 3:20:27 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

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

wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?í? 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.?


Satellites ecc 1-3(a-b)^2/(a+b)^2)
Moon 0.0549 0.99999978624
Io 0.0041 0.99999999999
Europa 0.0090 0.99999999985
Ganymed 0.0013 1.00000000000
Calli 0.0074 0.99999999993
Mimas 0.0202 0.99999999607
Encela 0.0047 0.99999999999
Tethys 0.0200 0.99999999624
Dione 0.0020 1.00000000000
Rhea 0.0010 1.00000000000
Comets
Halley 0.9670 0.85758592313
Encke 0.8470 0.96428781552
Tempel1 0.5190 0.99769836444
Planets
MER 0.2056 0.99995625439
VEN 0.0068 0.99999999995
EAR 0.0167 0.99999999817
MAR 0.0934 0.99999819976
JUP 0.0484 0.99999987116
SAT 0.0542 0.99999979788
URA 0.0472 0.99999988373
NEP 0.0086 0.99999999987
PLU 0.2488 0.99990429852
Asteroid
Pallas 0.2313 0.99992918251


I repeat:

Where does that formula (.5*sqrt(4-3(a-b)^2/(a+b)^2))come from?
What *is* the eccentricity constant (X) in the first place?
Where do your 11-13 digit values for the semi minor axis com from?
How does .5*sqrt(4-3(a-b)^2/(a+b)^2) simplfy to 1-3(a-b)^2/(a+b)^2) ?

--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour
  #22  
Old January 7th 18, 12:08 AM posted to sci.astro
Anders Eklöf
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Posts: 100
Default The eccentricity constant of solar objects

Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....


No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.


Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.


--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour
  #23  
Old January 7th 18, 05:01 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....


No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.


Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.


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



The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):

smajora smina e
MER 57,909,231,029 56,672,064,712 0.2056
VEN 108,209,525,401 108,207,023,568 0.0068
EAR 149,598,319,494 149,577,457,301 0.0167
MAR 227,943,771,564 226,947,353,141 0.0934
JUP 778,342,761,465 777,430,569,626 0.0484
SAT 1,426,714,892,866 1,424,617,764,212 0.0542
URA 2,870,633,540,862 2,867,434,101,795 0.0472
NEP 4,498,393,012,162 4,498,226,658,512 0.0086
PLU 5,906,438,090,764 5,720,709,449,730 0.2488

The two formulas for X differ indeed:

.05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2)
MER 0.999956281 0.999650256
VEN 1.000000000 1.000000000
EAR 0.999999998 0.999999985
MAR 0.999998201 0.999985606
JUP 0.999999871 0.999998969
SAT 0.999999797 0.999998377
URA 0.999999883 0.999999067
NEP 1.000000000 0.999999999
PLU 0.999904311 0.999234522


  #24  
Old January 7th 18, 03:16 PM posted to sci.astro
Martin Brown[_3_]
external usenet poster
 
Posts: 189
Default The eccentricity constant of solar objects

On 03/01/2018 15:18, Peter Riedt wrote:
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik'
StÅ™Ã*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*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.


It isn't a law though it is a mutilated form of eccentricity which is
already defined geometrically from the parameters a,b of an ellipse as:

e^2 = (1-(b/a)^2)

Hence

e^2a^2 = a^2 - b^2

b = a.sqrt(1-e^2)

This can be stuffed into your so called "law" to understand why it maps
all low eccentricity orbits to very approximately 1.

(and insanity check the limiting case of e=1 a parabola is X = -2).

"X" = 1 -3a(1-sqrt(1-e^2))^2/a(1+sqrt(1-e^2))^2

= 1 - 3*(2-e^2 -2sqrt(1-e^2))/(2-e^2+2sqrt(1-e^2))

Taking sqrt(1-e^2) taylor series expansion for small e as
sqrt(1-e^2) ~ 1 - e^2/2 - 1/8e^4 for small e

= 1 - 3*( 2-e^2-2+e^2+1/8e^4)/(4-2e^2)
~ 1 - 3e^4/16/(2-e^2)

So for small eccentricity e it looks like 1 - e^4/32

For the record at larger e it has no merit either.

"X" = e at about e= 0.80481
and
"X"=0 at about e= 0.963433

Neither of these having any physical significance.

--
Regards,
Martin Brown
  #25  
Old January 7th 18, 03:42 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 07/01/2018 v 05:01 Peter Riedt napsal(a):


The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):


If looking at https://www.princeton.edu/~willman/p...y_systems/Sol/
( as you did not provided a particular page ),
the semi major axis are not as accurately known as you provided,
neither the excentricity.

So you are just throwing numbers without much sense in it.


smajora smina e
MER 57,909,231,029 56,672,064,712 0.2056
VEN 108,209,525,401 108,207,023,568 0.0068
EAR 149,598,319,494 149,577,457,301 0.0167
MAR 227,943,771,564 226,947,353,141 0.0934
JUP 778,342,761,465 777,430,569,626 0.0484
SAT 1,426,714,892,866 1,424,617,764,212 0.0542
URA 2,870,633,540,862 2,867,434,101,795 0.0472
NEP 4,498,393,012,162 4,498,226,658,512 0.0086
PLU 5,906,438,090,764 5,720,709,449,730 0.2488

The two formulas for X differ indeed:

.05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2)
MER 0.999956281 0.999650256
VEN 1.000000000 1.000000000
EAR 0.999999998 0.999999985
MAR 0.999998201 0.999985606
JUP 0.999999871 0.999998969
SAT 0.999999797 0.999998377
URA 0.999999883 0.999999067
NEP 1.000000000 0.999999999
PLU 0.999904311 0.999234522

Cannot you see that a nearly constant X parameter value
means the parameter is useless ?

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #26  
Old January 7th 18, 04:17 PM posted to sci.astro
Peter Riedt
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Posts: 83
Default The eccentricity constant of solar objects

On Sunday, January 7, 2018 at 10:16:22 PM UTC+8, Martin Brown wrote:
On 03/01/2018 15:18, Peter Riedt wrote:
On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik'
StÅ™Ã*ž wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅ™Ã*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.


It isn't a law though it is a mutilated form of eccentricity which is
already defined geometrically from the parameters a,b of an ellipse as:

e^2 = (1-(b/a)^2)

Hence

e^2a^2 = a^2 - b^2

b = a.sqrt(1-e^2)

This can be stuffed into your so called "law" to understand why it maps
all low eccentricity orbits to very approximately 1.

(and insanity check the limiting case of e=1 a parabola is X = -2).

"X" = 1 -3a(1-sqrt(1-e^2))^2/a(1+sqrt(1-e^2))^2

= 1 - 3*(2-e^2 -2sqrt(1-e^2))/(2-e^2+2sqrt(1-e^2))

Taking sqrt(1-e^2) taylor series expansion for small e as
sqrt(1-e^2) ~ 1 - e^2/2 - 1/8e^4 for small e

= 1 - 3*( 2-e^2-2+e^2+1/8e^4)/(4-2e^2)
~ 1 - 3e^4/16/(2-e^2)

So for small eccentricity e it looks like 1 - e^4/32

For the record at larger e it has no merit either.

"X" = e at about e= 0.80481
and
"X"=0 at about e= 0.963433

Neither of these having any physical significance.

--
Regards,
Martin Brown


ok, thanks.
  #27  
Old January 7th 18, 09:44 PM posted to sci.astro
Anders Eklöf
external usenet poster
 
Posts: 100
Default The eccentricity constant of solar objects

Peter Riedt wrote:

On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....


No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.


Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.




The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):


So, since you use the eccentricity e to calculate the semi major axis,
what is the point of introducing a new "eccentricity constant" X?

1. What is the geometrical mening of X?
2. In what way is it useful?


smajora smina e
MER 57,909,231,029 56,672,064,712 0.2056
VEN 108,209,525,401 108,207,023,568 0.0068
EAR 149,598,319,494 149,577,457,301 0.0167
MAR 227,943,771,564 226,947,353,141 0.0934
JUP 778,342,761,465 777,430,569,626 0.0484
SAT 1,426,714,892,866 1,424,617,764,212 0.0542
URA 2,870,633,540,862 2,867,434,101,795 0.0472
NEP 4,498,393,012,162 4,498,226,658,512 0.0086
PLU 5,906,438,090,764 5,720,709,449,730 0.2488

The two formulas for X differ indeed:

.05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2)
MER 0.999956281 0.999650256
VEN 1.000000000 1.000000000
EAR 0.999999998 0.999999985
MAR 0.999998201 0.999985606
JUP 0.999999871 0.999998969
SAT 0.999999797 0.999998377
URA 0.999999883 0.999999067
NEP 1.000000000 0.999999999
PLU 0.999904311 0.999234522


So much for you simplification of the formula.
Can you still not see where the error is?
You say my points are valid, and choose to ignore them.

--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour
  #28  
Old January 8th 18, 01:16 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....

No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.

Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.




The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):


So, since you use the eccentricity e to calculate the semi major axis,
what is the point of introducing a new "eccentricity constant" X?

1. What is the geometrical mening of X?
2. In what way is it useful?


smajora smina e
MER 57,909,231,029 56,672,064,712 0.2056
VEN 108,209,525,401 108,207,023,568 0.0068
EAR 149,598,319,494 149,577,457,301 0.0167
MAR 227,943,771,564 226,947,353,141 0.0934
JUP 778,342,761,465 777,430,569,626 0.0484
SAT 1,426,714,892,866 1,424,617,764,212 0.0542
URA 2,870,633,540,862 2,867,434,101,795 0.0472
NEP 4,498,393,012,162 4,498,226,658,512 0.0086
PLU 5,906,438,090,764 5,720,709,449,730 0.2488

The two formulas for X differ indeed:

.05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2)
MER 0.999956281 0.999650256
VEN 1.000000000 1.000000000
EAR 0.999999998 0.999999985
MAR 0.999998201 0.999985606
JUP 0.999999871 0.999998969
SAT 0.999999797 0.999998377
URA 0.999999883 0.999999067
NEP 1.000000000 0.999999999
PLU 0.999904311 0.999234522


So much for you simplification of the formula.
Can you still not see where the error is?
You say my points are valid, and choose to ignore them.

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


X is more useful than SR and GR which cannot calculate any real elements of solar orbits.
  #29  
Old January 8th 18, 09:06 AM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 08/01/2018 v 01:16 Peter Riedt napsal(a):
On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote:




The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):


So, since you use the eccentricity e to calculate the semi major axis,
what is the point of introducing a new "eccentricity constant" X?

1. What is the geometrical mening of X?
2. In what way is it useful?




So much for you simplification of the formula.
Can you still not see where the error is?
You say my points are valid, and choose to ignore them.


X is more useful than SR and GR which cannot calculate any real elements of solar orbits.


An evasive attempt.
X is useless.

BTW, If you cannot calculate it via SR/GR,
it does not mean it cannot be done.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #30  
Old January 12th 18, 03:52 PM posted to sci.astro
Paul B. Andersen[_10_]
external usenet poster
 
Posts: 10
Default The eccentricity constant of solar objects

Den 08.01.2018 01.16, skrev Peter Riedt:
On Monday, January 8, 2018 at 4:44:46 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

On Sunday, January 7, 2018 at 7:08:23 AM UTC+8, Anders Eklöf wrote:
Peter Riedt wrote:

Your points are valid. However, the formula using .5*sqrt(4)... produces
the same result as using 1-3....

No - they don't, except for a circle.

For Mercury 5*sqrt(4-3(a-b)^2/(a+b)^2) gives 0,999956 and
1-3(a-b)^2/(a+b)^2) gives 0.999650.

The difference doesn't look big, but the devation from 1 differs by an
order of magnitude.

The comet Halley produces .85 for X.

Only with 1-3(a-b)^2/(a+b)^2) as you listed.
Just try using .5*sqrt(4-3(a-b)^2/(a+b)^2) instead.

Since I don't have your values for a and b I can't check.




The values for the semi major axis were obtained from Princeton.edu
and the values for the semi minor axis were calculated by me with
the formula semi minor axis = semi major axis * sqrt(1-e^2):


So, since you use the eccentricity e to calculate the semi major axis,
what is the point of introducing a new "eccentricity constant" X?

1. What is the geometrical mening of X?
2. In what way is it useful?


smajora smina e
MER 57,909,231,029 56,672,064,712 0.2056
VEN 108,209,525,401 108,207,023,568 0.0068
EAR 149,598,319,494 149,577,457,301 0.0167
MAR 227,943,771,564 226,947,353,141 0.0934
JUP 778,342,761,465 777,430,569,626 0.0484
SAT 1,426,714,892,866 1,424,617,764,212 0.0542
URA 2,870,633,540,862 2,867,434,101,795 0.0472
NEP 4,498,393,012,162 4,498,226,658,512 0.0086
PLU 5,906,438,090,764 5,720,709,449,730 0.2488

The two formulas for X differ indeed:

.05*sqrt(4-3(a-b)^2/(a+b)^2) 1-3(a-b)^2/(a+b)^2)
MER 0.999956281 0.999650256
VEN 1.000000000 1.000000000
EAR 0.999999998 0.999999985
MAR 0.999998201 0.999985606
JUP 0.999999871 0.999998969
SAT 0.999999797 0.999998377
URA 0.999999883 0.999999067
NEP 1.000000000 0.999999999
PLU 0.999904311 0.999234522


So much for you simplification of the formula.
Can you still not see where the error is?
You say my points are valid, and choose to ignore them.

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


X is more useful than SR and GR which cannot calculate any real elements of solar orbits.


A very strange idea.

X = X(e) = √(4-3⋅(1-√(1-e²))²/(1+√(1-e²))²)/2

e = 0.00 X = 1.0000000000
e = 0.04 X = 0.9999999399
e = 0.08 X = 0.9999990338
e = 0.12 X = 0.9999950691
e = 0.16 X = 0.9999842377
e = 0.20 X = 0.9999609449
e = 0.24 X = 0.9999175202
e = 0.28 X = 0.9998438029
e = 0.32 X = 0.9997265649
e = 0.36 X = 0.9995487110
e = 0.40 X = 0.9992881691
e = 0.44 X = 0.9989163362
e = 0.48 X = 0.9983958716
e = 0.52 X = 0.9976775062
e = 0.56 X = 0.9966953251
e = 0.60 X = 0.9953596037
e = 0.64 X = 0.9935455742
e = 0.68 X = 0.9910751195
e = 0.72 X = 0.9876855065
e = 0.76 X = 0.9829727662
e = 0.80 X = 0.9762812095
e = 0.84 X = 0.9664653233
e = 0.88 X = 0.9512997861
e = 0.92 X = 0.9256679486
e = 0.96 X = 0.8733242883
e = 1.00 X = 0.5000000000

So you have made a function which evaluates to something
very close to 1 for most eccentricities.
What's the point with that?

Wouldn't the function X = 1.0 be equal useful?

What does X tell us about the planetary orbits?

Mercury e = 0.2056 X = 0.9999562811
Venus e = 0.0086 X = 0.9999999999
Earth e = 0.0167 X = 0.9999999982
Mars e = 0.0934 X = 0.9999982007
Jupiter e = 0.0484 X = 0.9999998711
Saturn e = 0.0541 X = 0.9999997986
Uranus e = 0.0472 X = 0.9999998834
Neptun e = 0.0086 X = 0.9999999999
Pluto e = 0.2488 X = 0.9999043107

You said:
"X is more useful than SR and GR which cannot
calculate any real elements of solar orbits."

Can you please explain what elements of solar orbits
you can calculate using X?

--
Paul

https://paulba.no/
 




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