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The eccentricity constant of solar objects



 
 
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  #11  
Old January 4th 18, 09:53 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 04/01/2018 v 10:36 Peter Riedt napsal(a):
On Thursday, January 4, 2018 at 3:39:00 PM UTC+8, Libor 'Poutnik' StÅ™Ã*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' StÅ™Ã*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.


sqrt is not necessary for my formula but irreplaceable in Pythagoras


sqrt is necessary for determining the position of the ellipse focus.
That is what the ellipse eccentricity means.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #12  
Old January 4th 18, 09:59 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 04/01/2018 v 20:20 Anders Eklöf napsal(a):

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?


Peter generally is not bothered
by the insane numerical resolution of low accuracy physical data.


--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #13  
Old January 6th 18, 01:23 AM posted to sci.astro
Peter Riedt
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Posts: 83
Default The eccentricity constant of solar objects

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?Ã*?

wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?Ã*? 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


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

All the planets yield at least 0.999 for X. The comet Halley with e close too .99 produces 8.5 for X.

Thank you for your comments.
  #14  
Old January 6th 18, 01:31 AM posted to sci.astro
Peter Riedt
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Posts: 83
Default The eccentricity constant of solar objects

On Saturday, January 6, 2018 at 8:23:49 AM UTC+8, 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?Ã*?

wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?Ã*? 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


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

All the planets yield at least 0.999 for X. The comet Halley with e close too .99 produces 8.5 for X.

Thank you for your comments.


Correction: The comet Halley produces .85 for X.
  #15  
Old January 6th 18, 01:52 AM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

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?Ã*?

wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' St?Ã*? 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


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
  #16  
Old January 6th 18, 12:46 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 06/01/2018 v 01:23 Peter Riedt napsal(a):

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

All the planets yield at least 0.999 for X. The comet Halley with e close too .99 produces 8.5 for X.

Thank you for your comments.


What is the geometrical or physical meaning of X ?

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #17  
Old January 6th 18, 12:51 PM posted to sci.astro
Libor 'Poutnik' Stříž
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Posts: 48
Default The eccentricity constant of solar objects

Dne 06/01/2018 v 01:52 Peter Riedt napsal(a):

Satellites ecc 1-3(a-b)^2/(a+b)^2)
Moon 0.0549 0.99999978624
Io 0.0041 0.99999999999


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


You can see X is quite useless
as it is *very* insensitive to the low-medium ecc ellipse geometry.

Furthermore, it has no clear geometrical meaning.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #18  
Old January 6th 18, 12:55 PM posted to sci.astro
Peter Riedt
external usenet poster
 
Posts: 83
Default The eccentricity constant of solar objects

On Saturday, January 6, 2018 at 7:46:56 PM UTC+8, Libor 'Poutnik' StÅ™Ã*ž wrote:
Dne 06/01/2018 v 01:23 Peter Riedt napsal(a):

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

All the planets yield at least 0.999 for X. The comet Halley with e close too .99 produces 8.5 for X.

Thank you for your comments.


What is the geometrical or physical meaning of X ?

--
Poutnik ( The Pilgrim, Der Wanderer )

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


The eccentricity of solar objects is not random but conforms to the same rule for all orbiting objects (interaction of the aether with matter).
  #19  
Old January 6th 18, 02:32 PM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 06/01/2018 v 12:55 Peter Riedt napsal(a):
On Saturday, January 6, 2018 at 7:46:56 PM UTC+8, Libor 'Poutnik' StÅ™Ã*ž wrote:
Dne 06/01/2018 v 01:23 Peter Riedt napsal(a):

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

All the planets yield at least 0.999 for X. The comet Halley with e close too .99 produces 8.5 for X.

Thank you for your comments.


What is the geometrical or physical meaning of X ?

--
Poutnik ( The Pilgrim, Der Wanderer )

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


The eccentricity of solar objects is not random but conforms to the same rule for all orbiting objects (interaction of the aether with matter).


There is no rule for eccentricity value.
they can have any value from 0 to 1.

Sure, the objects with e too close to 0
do not exist for very long, as they collide with the Sun.



--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.
  #20  
Old January 6th 18, 02:34 PM posted to sci.astro
Libor 'Poutnik' Stříž
external usenet poster
 
Posts: 48
Default The eccentricity constant of solar objects

Dne 06/01/2018 v 14:32 Libor 'Poutnik' StÅ™Ã*ž napsal(a):
Dne 06/01/2018 v 12:55 Peter Riedt napsal(a):

The eccentricity of solar objects is not random but conforms to the same rule for all orbiting objects (interaction of the aether with matter).


There is no rule for eccentricity value.
they can have any value from 0 to 1.

Sure, the objects with e too close to 0
do not exist for very long, as they collide with the Sun.


Errata: ".. too close to 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.
 




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