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The eccentricity constant of solar objects



 
 
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  #31  
Old January 12th 18, 10:48 PM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
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/


A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant.
Ads
  #32  
Old January 13th 18, 11:07 AM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 12/01/2018 v 23:48 Peter Riedt napsal(a):
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:


[...]

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?


A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant.


Only because you made your eccentricity formula
to be near independent on the orbit shape.


--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #33  
Old January 13th 18, 11:15 AM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 12/01/2018 v 23:48 Peter Riedt napsal(a):

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


How do you then explain, how could GR be able to predict
the unexplained extra precession of the Mercury orbit,
the famous success of the GR.


--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #34  
Old January 13th 18, 11:54 AM posted to sci.astro
Paul B. Andersen[_10_]
external usenet poster
 
Posts: 7
Default The eccentricity constant of solar objects

Den 12.01.2018 23.48, skrev Peter Riedt:
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
Den 08.01.2018 01.16, skrev Peter Riedt:

X = .5*sqrt(4-3(a-b)^2/(a+b)^2)


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/


A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant.


Come again?

The eccentricities of the orbits a
Mercury e = 0.2056
Venus e = 0.0086
Earth e = 0.0167
Mars e = 0.0934
Jupiter e = 0.0484
Saturn e = 0.0541
Uranus e = 0.0472
Neptun e = 0.0086
Pluto e = 0.2488

They are very different, and not "close to the same constant".

The values for your X, with same precision as above, a
Mercury X = 1.0000
Venus X = 1.0000
Earth X = 1.0000
Mars X = 1.0000
Jupiter X = 1.0000
Saturn X = 1.0000
Uranus X = 1.0000
Neptun X = 1.0000
Pluto X = 0.9999

So despite the fact that the eccentricities of the planets
varies a lot, your X is close to 1 for all of them.

So I repeat the questions:
What can the X tell you about the orbits?
Wouldn't the function X = 1.0 be equal useful?

--
Paul

https://paulba.no/
  #35  
Old January 14th 18, 02:47 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Saturday, January 13, 2018 at 7:54:50 PM UTC+8, Paul B. Andersen wrote:
Den 12.01.2018 23.48, skrev Peter Riedt:
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
Den 08.01.2018 01.16, skrev Peter Riedt:

X = .5*sqrt(4-3(a-b)^2/(a+b)^2)


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/


A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant..


Come again?

The eccentricities of the orbits a
Mercury e = 0.2056
Venus e = 0.0086
Earth e = 0.0167
Mars e = 0.0934
Jupiter e = 0.0484
Saturn e = 0.0541
Uranus e = 0.0472
Neptun e = 0.0086
Pluto e = 0.2488

They are very different, and not "close to the same constant".

The values for your X, with same precision as above, a
Mercury X = 1.0000
Venus X = 1.0000
Earth X = 1.0000
Mars X = 1.0000
Jupiter X = 1.0000
Saturn X = 1.0000
Uranus X = 1.0000
Neptun X = 1.0000
Pluto X = 0.9999

So despite the fact that the eccentricities of the planets
varies a lot, your X is close to 1 for all of them.

So I repeat the questions:
What can the X tell you about the orbits?
Wouldn't the function X = 1.0 be equal useful?

--
Paul

https://paulba.no/


In terms of X, the orbits are indistinguishable.
  #36  
Old January 14th 18, 08:52 AM posted to sci.astro
Anders Eklf
external usenet poster
 
Posts: 95
Default The eccentricity constant of solar objects

Peter Riedt wrote:

On Saturday, January 13, 2018 at 7:54:50 PM UTC+8, Paul B. Andersen wrote:
Den 12.01.2018 23.48, skrev Peter Riedt:
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
Den 08.01.2018 01.16, skrev Peter Riedt:

X = .5*sqrt(4-3(a-b)^2/(a+b)^2)


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-e?))?/(1+√(1-e?))?)/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/

A very good analysis by you. X may not be used to calculate solar
orbits but it shows the various eccentricities of orbits are close to
a constant.


Come again?

The eccentricities of the orbits a
Mercury e = 0.2056
Venus e = 0.0086
Earth e = 0.0167
Mars e = 0.0934
Jupiter e = 0.0484
Saturn e = 0.0541
Uranus e = 0.0472
Neptun e = 0.0086
Pluto e = 0.2488

They are very different, and not "close to the same constant".

The values for your X, with same precision as above, a
Mercury X = 1.0000
Venus X = 1.0000
Earth X = 1.0000
Mars X = 1.0000
Jupiter X = 1.0000
Saturn X = 1.0000
Uranus X = 1.0000
Neptun X = 1.0000
Pluto X = 0.9999

So despite the fact that the eccentricities of the planets
varies a lot, your X is close to 1 for all of them.

So I repeat the questions:
What can the X tell you about the orbits?
Wouldn't the function X = 1.0 be equal useful?

--
Paul

https://paulba.no/


In terms of X, the orbits are indistinguishable.


Which, as quite a few of us have tried to show you, renders X utterly
meaningless, as it doesn't describe reality.

--
I recommend Macs to my friends, and Windows machines
to those whom I don't mind billing by the hour
  #37  
Old January 14th 18, 10:03 AM posted to sci.astro
Paul B. Andersen[_10_]
external usenet poster
 
Posts: 7
Default The eccentricity constant of solar objects

Den 14.01.2018 03.47, skrev Peter Riedt:
On Saturday, January 13, 2018 at 7:54:50 PM UTC+8, Paul B. Andersen wrote:
Den 12.01.2018 23.48, skrev Peter Riedt:
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
Den 08.01.2018 01.16, skrev Peter Riedt:

X = .5*sqrt(4-3(a-b)^2/(a+b)^2)


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/

A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant.


Come again?

The eccentricities of the orbits a
Mercury e = 0.2056
Venus e = 0.0086
Earth e = 0.0167
Mars e = 0.0934
Jupiter e = 0.0484
Saturn e = 0.0541
Uranus e = 0.0472
Neptun e = 0.0086
Pluto e = 0.2488

They are very different, and not "close to the same constant".

The values for your X, with same precision as above, a
Mercury X = 1.0000
Venus X = 1.0000
Earth X = 1.0000
Mars X = 1.0000
Jupiter X = 1.0000
Saturn X = 1.0000
Uranus X = 1.0000
Neptun X = 1.0000
Pluto X = 0.9999

So despite the fact that the eccentricities of the planets
varies a lot, your X is close to 1 for all of them.

So I repeat the questions:
What can the X tell you about the orbits?
Wouldn't the function X = 1.0 be equal useful?

--
Paul

https://paulba.no/


In terms of X, the orbits are indistinguishable.


So the answers to my questions a

Q: What can the X tell you about the orbits?
A: Nothing.

Q: Wouldn't the function X = 1.0 be equal useful?
A: Yes.


--
Paul

https://paulba.no/
  #38  
Old January 14th 18, 10:51 PM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Sunday, January 14, 2018 at 6:03:55 PM UTC+8, Paul B. Andersen wrote:
Den 14.01.2018 03.47, skrev Peter Riedt:
On Saturday, January 13, 2018 at 7:54:50 PM UTC+8, Paul B. Andersen wrote:
Den 12.01.2018 23.48, skrev Peter Riedt:
On Friday, January 12, 2018 at 10:52:32 PM UTC+8, Paul B. Andersen wrote:
Den 08.01.2018 01.16, skrev Peter Riedt:

X = .5*sqrt(4-3(a-b)^2/(a+b)^2)


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/

A very good analysis by you. X may not be used to calculate solar orbits but it shows the various eccentricities of orbits are close to a constant.

  #39  
Old January 15th 18, 06:08 AM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 14/01/2018 v 23:51 Peter Riedt napsal(a):
On Sunday, January 14, 2018 at 6:03:55 PM UTC+8, Paul B. Andersen wrote:

So the answers to my questions a

Q: What can the X tell you about the orbits?
A: Nothing.

Q: Wouldn't the function X = 1.0 be equal useful?
A: Yes.


Q: What can the X tell you about the orbits?
A: In terms of X, planetary orbits are the same as agreed by you.


What makes the parameter X useless.
It is useless even as a constant, because it is not a constant.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #40  
Old January 15th 18, 08:13 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Monday, January 15, 2018 at 2:08:42 PM UTC+8, Libor 'Poutnik' Stř*ž wrote:
Dne 14/01/2018 v 23:51 Peter Riedt napsal(a):
On Sunday, January 14, 2018 at 6:03:55 PM UTC+8, Paul B. Andersen wrote:

So the answers to my questions a

Q: What can the X tell you about the orbits?
A: Nothing.

Q: Wouldn't the function X = 1.0 be equal useful?
A: Yes.


Q: What can the X tell you about the orbits?
A: In terms of X, planetary orbits are the same as agreed by you.


What makes the parameter X useless.
It is useless even as a constant, because it is not a constant.

--
Poutnik ( The Pilgrim, Der Wanderer )

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


X is a constant of 4 decimals for 8 out of 9 planets. It is a lot
for such a diverse group of entities. It joins them together in a common bond of extraordinary magnitude.
 




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