2

I'm reading Practical Astronomy With Your Calculator Or Spreadsheet by Duffett-Smith and Zwart. When calculating the positions (in equatorial coordinates) of the Sun and planets the authors use the tropical orbital period T in days (for the Sun this is the Earth's tropical orbital period). Their method involves finding the mean angular speed of the orbiting body using 360/T degrees/day. What's the reason for using tropical rather than sidereal orbital periods? Thanks.

EDIT

Here's a photo of the calculation. Why are they using 365.242191 (Earth tropical year)? Why not 365.256 (Earth sidereal year)? They give the orbital periods of the planets in tropical years. For example, Earth's is 0.999996 tropical years.

enter image description here

Peter
  • 1,033
  • 6
  • 17
  • 2
    It would probably help to post the equations for those that don't have the book, or at least a reference to page/chapter. – Greg Miller Oct 07 '22 at 14:18
  • sidereal, synodic, draconitic, anomalistic, tropical... – uhoh Oct 07 '22 at 16:07
  • @uhoh - what are you trying to tell us? – Peter Oct 07 '22 at 17:30
  • @Peter I always get them mixed up and refer to that table to sort it out. It seems that tropical relates to the observing body's axis, and the question is about the observing body's equatorial coordinates which are defined not completely by the celestial sphere, but in part by the observing body's axis. I can't answer your question but I think this may be the path down which an answer may be found. – uhoh Oct 07 '22 at 19:22
  • In reality, the choice is arbitrary. But, D is the count of days of the Gregorian calendar, which is based on the tropical year. – Greg Miller Oct 07 '22 at 20:20
  • Those calculations seem... strange. See https://en.wikipedia.org/wiki/True_anomaly & the related articles about mean & eccentric anomaly. And https://en.wikipedia.org/wiki/Kepler%27s_equation – PM 2Ring Oct 08 '22 at 02:07

0 Answers0