I am aware of the command AstronomicalData["Jupiter", "Distance"] from which I can get the distance between the earth and Jupiter. Is there any way to use AstronomicalData to get the distance between the sun and Jupiter?


2 Answers 2


Perhaps naïve:

Norm@AstronomicalData["Jupiter", "Position"]
  edit .... copy/paste error corrected

Checking some consistence

EuclideanDistance @@ (AstronomicalData[#, "Position"] & /@ {"Earth", "Jupiter"}) == 
                                            AstronomicalData["Jupiter", "Distance"]

  • $\begingroup$ @Kuba I can't comment. I've never been in Jupiter. $\endgroup$ Commented Nov 18, 2013 at 6:32
  • $\begingroup$ @Kuba My knowledge of astronomy is only commensurable with my manners: It doesn't suffice to get surprised. $\endgroup$ Commented Nov 18, 2013 at 7:10
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    $\begingroup$ @Kuba Well, I get a 7 now. Eppur si muove; $\endgroup$ Commented Nov 19, 2013 at 0:56

Offered as an alternative to getting the same information and a check on it, one can also get this measurement from a WolframAlpha query:

enter image description here


Of some interest, by these measurements Jupiter appears to have moved quite a ways further from the Sun since belisarius's answer just some 11 hours ago.

67.74204*10^11 vs 7.74232*10^11

WolframAlpha can also give one a plot of Jupiter's position in orbit (although not a particularly satisfying one):

enter image description here

If, Jupiter had crossed its perihelion (the point it passes closest to the sun, which should correspond to the fastest part of its orbit) could it have moved as much as it appears to have moved between belisarius's measure and mine?

Any astronomers or astro physicists about?

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    $\begingroup$ Interesting ... do you think we are solving for two different Jupiters? $\endgroup$ Commented Nov 18, 2013 at 7:13
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    $\begingroup$ @belisarius - Perhaps just slipping glimpses of parallel universes. More (or less given one's point of view) seriously, I ran the Wolfram Alpha query and your code and almost the same time and got essentially identical values (7.743 x^11 meters). I just ran The Wolfram Alpha query again and it comes out 7.744*10^11. Not as big a move, but a move. $\endgroup$
    – Jagra
    Commented Nov 19, 2013 at 0:35
  • $\begingroup$ @belisarius I thought at first you'd forgotten to compensate for the great altitude at which you train... $\endgroup$
    – cormullion
    Commented Nov 22, 2013 at 13:12
  • $\begingroup$ @belisarius Ok, I've deleted old comments $\endgroup$
    – Kuba
    Commented Nov 22, 2013 at 13:35

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