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
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