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.


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

  • 2
    $\begingroup$ It would probably help to post the equations for those that don't have the book, or at least a reference to page/chapter. $\endgroup$ Oct 7, 2022 at 14:18
  • $\begingroup$ sidereal, synodic, draconitic, anomalistic, tropical... $\endgroup$
    – uhoh
    Oct 7, 2022 at 16:07
  • $\begingroup$ @uhoh - what are you trying to tell us? $\endgroup$
    – Peter
    Oct 7, 2022 at 17:30
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Oct 7, 2022 at 19:22
  • $\begingroup$ In reality, the choice is arbitrary. But, D is the count of days of the Gregorian calendar, which is based on the tropical year. $\endgroup$ Oct 7, 2022 at 20:20


You must log in to answer this question.

Browse other questions tagged .