# Are there any moons that exhibit a non-convex curvature in their orbit about the Sun?

In my answer at https://physics.stackexchange.com/a/266444/59023, it took me several iterations before David Hammen and others could help me understand why the Earth's moon's orbit about the Sun is always convex (David got this correct the first time). Part of my initial issue was the ratio of orbital periods of the moon about Earth (a little less than 28 days) versus the Earth-moon system about the Sun (a little more than 365 days), i.e., about a ~8% ratio.

This got me wondering if the other solar system moons still had convex orbits despite very different orbital period ratios, e.g., Io has a ~0.04% ratio or a 200 times faster orbital period around it's planetary host body.

So would a moon like Io (or a moon orbiting any of the other outer planets) have a non-convex curvature at some point in its orbit about the sun?

• I answered a related question over on Astronomy about a year ago. If I understand your question correctly, any moon listed there would qualify. However, there might also be moons that answer this question but don't answer the old Astronomy question. Commented Oct 14, 2021 at 13:55
• Ah okay, so it is a matter of orbital speed of the moon relative to the orbit of the planet-moon system about the sun? Commented Oct 14, 2021 at 14:02
• The previous condition was for moons that actually reverse direction relative to the sun, i.e., make self-intersecting "loops" in their path. If an epitrochoid does that, it definitely changes the sign of its curvature; but there are also epitrochoids that change the sign of their curvature while not intersecting themselves. Commented Oct 14, 2021 at 14:05
• Note also that not all epitrochoids change the sign of their curvature; if you make $d$ small enough in the parametric equation for a epitrochoid, you'll get a curve that is always convex. So the literal answer to the question you ask in the title is "yes, the orbits of all moons can be approximated by epitrochoids (except maybe those of Uranus)". But that's not really what you're asking; it might be worth rephrasing the title. Commented Oct 14, 2021 at 14:23
• @MichaelSeifert - Another good point, yes perhaps I should... Commented Oct 14, 2021 at 14:32

We can the path of any moon as an epicycloid: we treat the planet as though it moves in uniform circular motion with radius $$r_p$$, and the moon as though it moves in uniform circular motion about the planet with radius $$r_m$$. The net acceleration of the moon will be the vector sum of the planet's acceleration towards the sun (with magnitude $$r_p \omega_p^2$$) and the moon's acceleration towards the planet (with magnitude $$r_m \omega_m^2$$). Here, $$\omega_p$$ is the angular velocity of the planet about the sun, and $$\omega_m$$ is the angular velocity of the moon about the planet.
Since the planet's acceleration always points towards the sun, the only way for the moon to have a net acceleration vector pointing away from the sun is for its acceleration vector towards the planet to be of greater magnitude: $$r_m \omega_m^2 > r_p \omega_p^2,$$ or in terms of the respective periods $$T_p$$ and $$T_m$$, $$\frac{r_m}{T_m^2} > \frac{r_p}{T_p^2}.$$ Under these approximations, any moon for which this is satisfied will have some point of "concavity" in its orbit when it is between the Sun and its parent planet.