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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 …..57,909,175,000 .…56,671,636,475 VEN …108,208,930,000 ...108,206,447,840 EAR …149,597,890,000 ...149,577,002,324 MAR …227,936,640,000 ...226,939,989,085 JUP …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 ……….a-b …………..a+b X= MER …1237538525 .....114,580,811,475 1.0 VEN ……..2482160 .....216,415,377,840 1.0 EAR ……20887676 .....299,174,892,324 1.0 MAR …..996650915 .....454,876,629,085 1.0 JUP …..911996157 ..1,555,912,023,843 1.0 SAT …2093320195 ..2,851,357,479,805 1.0 URA …3195437522 ..5,738,748,962,478 1.0 NEP …..165802278 ..8,996,339,997,722 1.0 PLU 185738436432 11,627,021,563,568 1.0 |
#2
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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
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The eccentricity constant of solar objects
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. |
#4
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The eccentricity constant of solar objects
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. |
#5
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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*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. 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
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The eccentricity constant of solar objects
Dne 03/01/2018 v 21:00 Libor 'Poutnik' StÅ™Ã*ž 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*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. 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
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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 |
#8
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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. |
#9
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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. |
#10
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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 |
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