# How would one calculate the synodic period of the Earth and an elliptical orbit?

For example, when would someone with a telescope be able to see Starman and his Roadster (when will the Tesla roadster's elliptical orbit cross ours again, and how would that be calculated?)

For two circular orbits, I understand this is given by $$\frac{1}{S} = \frac{1}{T_2} - \frac{1}{T_1}$$ .

## 1 Answer

tl;dr:

The synodic period would be the same. For this object it will be about:

$$\left( \frac{1}{365.25} - \frac{1}{557} \right)^{-1} \approx 1061 \ days.$$

You can confirm that by reading this new ArXiv preprint about Roadster: The random walk of cars and their collision probabilities with planets. In the 2nd paragraph of Section 3 it says:

The bodies reach the same orbital longitude on their synodic timescale of ∼ 2.8 yrs.

There's a slight difference, since the Roadster's orbit continues to be refined, and the paper is about evolution over long timescales.

However, I think what you really want to know are the times of closest approach when (at least) one orbit is significantly eccentric, and those will not be periodic or come at regular intervals.

Well the Roadster is quite far away now and moving farther rapidly. See this answer for some details and some nice images from medium-sized tracking telescopes. It's already dimmer than +22 magnitude as seen from Earth, and as far as I know there won't be any really close approaches to Earth for a long time. But to answer your question, the period is about 557 days. You can see this for example if you go to JPL's Horizons and look at the osculating elements. If you look a few weeks in the future when the Roadster is sufficiently far away from Earth not to be perturbed, the period around the Sun flattens out to 557 days.  above x2: Top: Distance data from JPL's Horizons plotted using Python from here. Bottom: Estimated apparent magnitude using distances from Horizons plus math shown in the text of this answer. Your milage may vary, but probably within +/-2 magnitudes of "official" predictions when they come out.

A really cool GIF can be seen in the Space.com article Observatory Spots Elon Musk's Tesla Roadster Zooming Through Space (Video). It is over 7 MB so I can't add it here. However, here are ten frames from somewhere in the middle. Still, you should go see the whole thing.

GIF: DEMIOS image below, from here as also tweeted by Jonathan McDowell. The little dot near the center moving to the right and up is Roadster in reflected sunlight, probably mostly from the white FH 2nd stage still attached.

GIF: • Why is it not periodic, does it not follow an elliptical orbit with that period of 557 days? Are there a lot of perturbations? Lastly that synodic period equation assumes circular orbits, does it really still apply to elliptical ones as well? – Rithwik Sudharsan Feb 12 '18 at 2:54
• @hawaii12 Both Earth's and Roadster's elliptical orbits are periodic. But a plot of the distance between the two will only be periodic if the two orbits are circular. You can always calculate a synodic period using your equation, but the more that one or both orbits are elliptical, it is less useful it is because the idea of "appears in the same position" is less and less correct. It's the usefulness of the synodic period that requires nearly-circular orbits, not the definition. There won't be a different equation for elliptical orbits. – uhoh Feb 12 '18 at 5:03
• So, the more elliptical the orbits are, the farther the synodic period will be from a true intersection of the two objects? Is there any place I can go to see why the distances between two elliptical orbits is not periodic; this is the visual I am seeing (the distance seems to be periodic). – Rithwik Sudharsan Feb 13 '18 at 6:05
• @hawaii12 Oh, your visual is of two bodies orbiting around a common center of mass. That's totally different than two bodies orbiting around a third body such as the Sun in the case of your question. That little green dot in the visual linked in your comment is the location of the center of mass of the two, but there isn't anything there physically. They are orbiting around each other. Mathematically you can see it as a single orbit. The smaller body's orbit is exactly 3.6 times larger than the heavier body, they both have identical eccentricity, and they both lie in a single plane. – uhoh Feb 13 '18 at 11:53
• Your graphic is the same as the others shown in Wikipedia's article about the two body problem. In these figures, there's a red dot in the middle, but there is no physical body or mass at that point. But with Earth and Roadster both orbiting around a third, things are very different, so Earth and Roadster's orbits around the Sun are mostly independent. – uhoh Feb 13 '18 at 11:59