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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. |
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