Timeline for Earth's orbit and historical sun-earth distance
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 28, 2017 at 15:18 | comment | added | user967 | My mistake. At the end of the results, it says "LT = One-way down-leg Newtonian light-time (day)", so it's actually in light days, not light days, not light seconds. I've corrected the post. | |
| Jul 28, 2017 at 15:18 | history | edited | user967 | CC BY-SA 3.0 |
seconds -> days
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| Jul 28, 2017 at 10:47 | comment | added | user1141649 | OK, what I am gonna do is that I export all days from 1600 to 2400 AD. I will calculate the diagonal of the cube box coordinates (d=sqrt(xx*+yy+z*z)) to get distance and diagonal of the cube box velocity vectors to get speed. Then I will save the distances and speed to file and I will add it to my program so it will be able to get the data needed to calculate exact radius. | |
| Jul 28, 2017 at 10:43 | vote | accept | user1141649 | ||
| Jul 28, 2017 at 10:18 | comment | added | user1141649 | I tried to look for the coordinates of day 1849-Jul-28 and I got 5,87E-03 light seconds which should be 1,76E+03km. Compared with my program where R=1,01677040986377 AU ... My resultis 1,52107E+11 . 1.76E+03km is 1760 km - too small number... My calculation from coords. d=sqrt(xx + yy + zz) = 6,08E-01*6,08E-01 + -8,15E-01-8,15E-01 + -2,17E-04*-2,17E-04 = 1,02E+00ld. Roughly 155000 km. | |
| Jul 27, 2017 at 18:20 | history | answered | user967 | CC BY-SA 3.0 |