5
$\begingroup$

I'm trying to find data on the heliocentric velocities of planets in our solar system and was directed to Mathematica's AstronomicalData in a previous topic. Unfortunately I cannot find any x,y and z velocity components and was hoping someone might be able to point me in the right direction.

$\endgroup$
4
$\begingroup$

An approximation (speed is in m/s and Position in m):

AstronomicalData["Mars", {"Speed", {2019, 3, 1, 0, 0, 0}}] Normalize @@
   Differences[ AstronomicalData[ "Mars", {"Position", {2019, 3, 1, 0, 0, #}}] & /@ {0, 1}]

(*
  {-22373.4, 8780.75, 734.403}
*)
$\endgroup$
  • 1
    $\begingroup$ +1 Consider a central difference instead: it should be more accurate. It makes a substantial difference, by the way: it affects the third sig. fig. in your example, because it reflects the net planetary acceleration during the course of 12 hours. $\endgroup$ – whuber May 24 '13 at 22:18
  • $\begingroup$ @whuber Perhaps I'm wrong, but I think I'm taking one second differences, not one day. Even within 10 seconds, there aren't appreciable speed changes (as expected, of course) $\endgroup$ – Dr. belisarius May 25 '13 at 4:44
  • 1
    $\begingroup$ Central difference: {1, -1}.(AstronomicalData["Mars", {"Position", DatePlus[{2019, 3, 3}, {#, "Second"}]}] & /@ {1, -1}/2). Normalize and multiply with the speed afterwards. :) $\endgroup$ – J. M. will be back soon May 25 '13 at 9:48
  • $\begingroup$ Would something like this not work as well? v = ((AstronomicalData[ "Mars", {"Position", {2013, 1, 1, 1, 1, 1.01}}] - AstronomicalData["Mars", {"Position", {2013, 1, 1, 1, 1,1}}]))/0.01 $\endgroup$ – InquisitiveInquirer May 25 '13 at 12:08
  • 1
    $\begingroup$ @J.M. Normalize[{1, -1}.(AstronomicalData[ "Mars", {"Position", DatePlus[{2019, 3, 1}, {#, "Second"}]}] & /@ {1, -1}/ 2)] AstronomicalData["Mars", {"Speed", {2019, 3, 1, 0, 0, 0}}] gives the same result as mine. That's why I don't understand whuber's comment. I'm forgetting something, surely $\endgroup$ – Dr. belisarius May 25 '13 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.