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## The eccentricity constant of solar objects

#1 January 3rd 18, 03:21 AM posted to sci.astro
 Peter Riedt external usenet poster Posts: 83 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 â¦..57,909,175,000 .â¦56,671,636,475
VEN â¦108,208,930,000 ...108,206,447,840
EAR â¦149,597,890,000 ...149,577,002,324
MAR â¦227,936,640,000 ...226,939,989,085
JUP â¦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

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

#2 January 3rd 18, 08:12 AM posted to sci.astro
 Libor 'Poutnik' StÅÃ­Å¾ external usenet poster Posts: 48 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 January 3rd 18, 08:23 AM posted to sci.astro
 Libor 'Poutnik' StÅÃ­Å¾ external usenet poster Posts: 48 The eccentricity constant of solar objects

Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅÃ*Å¾ 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 January 3rd 18, 03:18 PM posted to sci.astro
 Peter Riedt external usenet poster Posts: 83 The eccentricity constant of solar objects

On Wednesday, January 3, 2018 at 4:23:40 PM UTC+8, Libor 'Poutnik' StÅÃ*Å¾ wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅÃ*Å¾ 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 January 3rd 18, 08:00 PM posted to sci.astro
 Libor 'Poutnik' StÅÃ­Å¾ external usenet poster Posts: 48 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' StÅÃ*Å¾ wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅÃ*Å¾ 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 January 3rd 18, 08:19 PM posted to sci.astro
 Libor 'Poutnik' StÅÃ­Å¾ external usenet poster Posts: 48 The eccentricity constant of solar objects

Dne 03/01/2018 v 21:00 Libor 'Poutnik' StÅÃ*Å¾ 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' StÅÃ*Å¾ wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅÃ*Å¾ 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 January 3rd 18, 11:35 PM posted to sci.astro
 Peter Riedt external usenet poster Posts: 83 The eccentricity constant of solar objects

On Thursday, January 4, 2018 at 4:19:46 AM UTC+8, Libor 'Poutnik' StÅÃ*Å¾ wrote:
Dne 03/01/2018 v 21:00 Libor 'Poutnik' StÅÃ*Å¾ 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' StÅÃ*Å¾ wrote:
Dne 03/01/2018 v 09:12 Libor 'Poutnik' StÅÃ*Å¾ 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 January 4th 18, 07:38 AM posted to sci.astro
 Libor 'Poutnik' StÅÃ­Å¾ external usenet poster Posts: 48 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' StÅÃ*Å¾ 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 January 4th 18, 09:36 AM posted to sci.astro
 Peter Riedt external usenet poster Posts: 83 The eccentricity constant of solar objects

On Thursday, January 4, 2018 at 3:39:00 PM UTC+8, Libor 'Poutnik' StÅÃ*Å¾ 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*Å¾ 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 January 4th 18, 07:20 PM posted to sci.astro
 Anders Eklöf external usenet poster Posts: 98 The eccentricity constant of solar objects

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.?

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

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