I got the Orbital Elements of Jupiter around the Sun, which describe how it orbits relative to the "fix" Sun.

Jupiter Semi Major Axis (AU): 5.20336301

Eccentricity: 0.04839266

Inclination to Ecliptic Plane (deg): 1.30530

Longitude of the Ascending Node (deg): 100.55615

Longitude of Perihelion (deg): 14.75385

Mean Anomaly at J2000.0 (deg): 19.65053

I want to convert this One-Body Problem back into the Two-Body Problem to find out where the Sun is gonna be at a given time (i couldnt find the Orbital Elemens of Sun around Sun-Jupiter Barycenter).

enter image description here

So the Semi-Major Axis depends on the Mass Ratio. The Eccentricity and Inclination should be equal for both, just like Longitude of the Ascending Node. The Longitude of Perihelion should be the one from Jupiter+180°, doesnt it?

The main Problem is the Mean Anomaly at J2000.0. The two Ellipses are equal, besides of the Size. Thus, the True Anomaly at a given time should be the same for Each Coordinate System. Then the eccentric Anomaly will be the same too. Does that mean the Mean Anomaly for J2000.0 will be also the same for both?

Thanks for reading and sunny greetings!!

enter image description here

  • 2
    $\begingroup$ This is an interesting general mathematical question: given an elliptical orbit, find the individual orbits around the barycenter. $\endgroup$
    – user21
    Commented Nov 19, 2018 at 19:43
  • 1
    $\begingroup$ Please be careful: the Sun revolves around the barycenter. Rotation always refers to spin along an internal axis $\endgroup$ Commented Nov 19, 2018 at 20:08

1 Answer 1


The barycenter of two point masses M1 and m2 -- or their 'center of gravity' (which to avoid any doubt is not in itself a center of gravitational attraction) -- lies on the straight line joining them, and divides the distance between them in the proportion m2:M1 -- in classical/Newtonian terms. That is, the barycenter is separated from center M1 by the fraction m2/(M1+m2) of the total distance M1 to m2, and by the complementary fraction M1/(M1+m2) from the center of m2.

In a two-body system, the positions and motions of each body relative to their barycenter are similar in the way that two triangles can be similar: the corresponding angles for each are the same, the distances are in a constant proportion. (See also https://en.wikipedia.org/wiki/Barycenter and https://en.wikipedia.org/wiki/Two-body_problem .)

So at least part of the answer to the question about the Sun and Jupiter -- imagined as a 2-body system -- is that the angles including the mean anomaly are indeed the same for each.

However, this only applies approximately to the figures quoted in the question. These are not 'one-body' or 'two-body' orbit figures, they are heliocentric approximate mean elements for Jupiter in the n-body solar system as at J2000 (i.e. 2000 Jan 1.5), from a derivation applied to data over the interval 1800-2050. The figures appear in a GSFC/NASA Jupiter fact-sheet, and also on p.316 of 'Explanatory Supplement to the Astronomical Almanac' (1992, ed. K P Seidelmann et al.), where the error in Jupiter's position computed from the elements is stated as up to 300" in right ascension.

Mean elements such as those quoted in the question can be derived from the time-varying osculating elements obtained over a long period. Such elements may be obtained for the chosen period in different ways, for example

(a) by deriving them from results of a long-term combined numerical integration of the orbits of a set of solar-system bodies -- usually involving all of the major planets along with either some of the minor planets as further individual bodies and/or an approximate allowance made for the effects of those not included individually; or

(b) by deriving them from an analytical theory (a method used for example to obtain the results in J L Simon et al., 'Precession formulae and mean elements for the Moon and the planets', Astronomy & Astrophysics 282 (1994) 663-683).

In either of these cases, subtraction of the periodic terms leaves the mean elements as non-periodic polynomials as a function of time-interval from a chosen epoch associated with the data.

A third and approximate approach, which seems to have been the one adopted for the figures quoted in the question, involves fitting the osculating elements to a chosen polynomial function over a chosen interval. The website http://www.met.rdg.ac.uk/~ross/Astronomy/Planets.html suggests that the figures here resulted from a 250-year least squares fit of data from the JPL DE200 numerically integrated planetary ephemeris to a Keplerian orbit where each element was allowed to vary linearly with time. The mean elements for a date such as J2000 would then represent the value of the linear function at the chosen date.

If it is wished to take the approximate mean elements for Jupiter at J2000 to represent 1-body elements on that date, then that is perhaps about as close an approximation of the kind as can be made: and the angles will then be similar as already described, and the barycentric distances will be obtained by the fractional split already mentioned.

A more precise way of tracking the relation between the Sun and the solar-system barycenter is outlined in a 1983 paper from JPL on solar-system numerical integration by XX Newhall et al.). This takes into account all of the planets and their masses, and the result is a complex path that has perhaps never been represented by any trigonometric series, but only numerically and diagrammatically -- e.g. at https://en.wikipedia.org/wiki/File:Solar_system_barycenter.svg, thus:

from Wikimedia

  • $\begingroup$ Thanks for the good explanation. Tomorrow there is the Sun over here and I am gonna test if it works. Okay, so this is the mean orbital elements won from measuring over a long period. Still, I can take this element and make a model. I think getting the "counter-Orbital-Elements" for the Sun should at least be a good first Approximation into this complex Problem. Wikipedia en.wikipedia.org/wiki/Orbital_elements: "...When viewed from a non-inertial frame centred on one of the bodies, only the trajectory of the opposite body is apparent; Keplerian elements describe these non-inertial $\endgroup$ Commented Nov 22, 2018 at 10:56
  • $\begingroup$ trajectories. " This means the Orbital Elements describe an object orbiting relative to a second, fixed object. So it should be correct to get the "absolute" orbit elements like I did. Anyways Jupiter seems to raise the accuracy of the Sun position in my algorithm, now I would like to build Saturn and Neptune. Do you have any idea how to start with it? I don't think I can add the Sun-Saturn orbit geometrically to the first one. We will see, I will report tomorrow. $\endgroup$ Commented Nov 22, 2018 at 10:56
  • $\begingroup$ what an awesome diagram! $\endgroup$
    – Fattie
    Commented Nov 25, 2018 at 15:01

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