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In years, what is the time it takes our sun to orbit around our solar system's barycenter?

Glorfindel
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sunSis
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1 Answers1

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As you can see from the diagrams below of the motion of the Solar System's barycenter relative to the Sun (courtesy of Wikipedia), the Sun's motion with respect to the system's barycenter is not a simple closed curve, and it doesn't have a simple period.

Solar system barycentre 1945-1995

(Click the above image for a SVG version).

Solar System barycenter 2000-2051

It is primarily dominated by Jupiter's period (11.862 years) because Jupiter has most of the non-solar mass of the Solar System. (On a related note, Jupiter has most of the angular momentum of the solar system). Jupiter's mass is approximately 0.0009543 solar masses, so the Sun is almost 1048 times more massive than Jupiter.

According to Wikipedia

To calculate the actual motion of the Sun, you only need to consider the motions of the four giant planets (Jupiter, Saturn, Uranus, Neptune). The contributions of all other planets, dwarf planets, etc. are negligible.

Here is a Sage / Python script which displays an interactive 3D plot of the above diagrams, using data from the JPL Horizons on-line solar system data and ephemeris computation service. Horizons can produce data covering a wide range of dates, please see the Horizons documentation for details.

FWIW, here's a plot of the Sun's speed relative to the barycentre.

Sun speed relative to barycentre

PM 2Ring
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  • In years (see OP) – sunSis Oct 16 '18 at 01:08
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    @sunSis It varies quite a bit, but it's roughly 12 years. – PM 2Ring Oct 16 '18 at 01:16
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    @SunSis - The correct answer is in this answer: "the Sun's motion with respect to the system's barycenter is not a simple closed curve, and it doesn't have a simple period." By insisting on a simple period, you are demanding an answer that doesn't exist. – David Hammen Oct 16 '18 at 10:30
  • Thank you for your edit here! Seeing my mistakes that you fixed made me laugh out loud in a library :o – uhoh Nov 02 '18 at 11:09
  • Could the motion of the sun w.r.t. the solar system barycenter be considered as quasiperiodic, e.g., if we consider the orbital timescale to be much smaller than the timescale over which the orbital angular momentum changes appreciably? – Daddy Kropotkin May 03 '21 at 15:53
  • @DaddyKropotkin Sure. Although the Sun's motion wrt the barycentre doesn't form a simple closed loop, it's actually fairly well-behaved, because the periods of Saturn, and Uranus are roughly 2.5× and 7× the period of Jupiter. My Sage script uses cubic Bezier curves to plot the path. The parameter ss controls the step size (in years) used for the Bezier curves. Currently, ss=1 so every year point is used in the curve calculation. If you set ss=3, only every 3rd year point will be used, but the curve will still go fairly close to the in-between year points. – PM 2Ring May 03 '21 at 16:19
  • Very neat! Are you using cubic Bezier curves because you are considering four points in the plane, i.e., Saturn, Uranus, Jupiter, and the Sun? Or is there another reason? Do the gravitational dynamics reduce to the Bernstein polynomials? Like, this isn't just computer graphics, right? You're implementing Newtonian gravity? – Daddy Kropotkin May 03 '21 at 18:25
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    @DaddyKropotkin I'm not calculating the data, I'm pulling it directly from the JPL Development Ephemeris via their Horizons interface. I'm using that 3D position & velocity data (of the Sun, relative to the Solar System barycentre) to calculate the Bezier control points, which Sage then uses to plot the curves. Although the curves are in 3D, at this scale it's not easy to see that the curves aren't perfectly planar. The Z coordinate variation is ~1% of the X & Y variation. – PM 2Ring May 03 '21 at 19:56
  • That's awesome! Thanks for explaining. – Daddy Kropotkin May 03 '21 at 21:44
  • Also see https://astronomy.stackexchange.com/a/44903/16685 – PM 2Ring Dec 20 '22 at 13:45