I don't mean necessarily a granular chart of all of the different rings at different distances, but is it days? Weeks?

  • 1
    $\begingroup$ A very granular chart of many rings is to be found in Wikipedia here. The article is in german, the english version misses the nice chart. $\endgroup$
    – Uwe
    Mar 5, 2018 at 16:39

2 Answers 2


Hours to days.

The orbital period is proportional to the 3/2 power of the orbital radius, and the orbital period of the moon Methone (and thus of the particles in its associated ring arc) is very close to 24 hours. Working from the orbital radii given in this table, I get:

  • D ring: ~5 hours
  • C ring: ~6-7 hours
  • B ring: ~8-10 hours
  • A ring: ~12-14 hours
  • F ring: ~14.5 hours
  • Janus/Epimetheus ring: ~16-17 hours
  • G ring: ~19-20 hours
  • Methone & Anthe arcs: ~24 hours
  • Pallene ring: ~27 hours
  • E ring: ~21 hours to ~90 hours
  • Phoebe ring: 90-500 days

This is a supplemental answer to @RussellBorogove's excellent and intuitive answer.

Starting from the vis-viva equation one of the few equations I can actually remember:

$$v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right)$$

You can set $r=a$ for a circular orbit, giving $v=\sqrt{GM/a}$ where the product of the gravitational constant times Saturn's mass $GM$ is called the standard gravitational parameter and is about 3.793E+07 km^3/s^2. (beware of meters vs kilometers).

The circumference is $C = 2 \pi a$, so

$$ T = \frac{C}{v} = 2 \pi a \sqrt{a/GM} = \frac{2 \pi a^{3/2}}{\sqrt{GM}}$$

which shows the 3/2 power scaling mentioned in @RB's answer.

For the Methone Ring with a distance of 194,230 km, this gives 24.3 hours, same as the other answer of course.

One caveat is that for more accurate calculations, it's important to remember that for strongly oblate planets, the radial variation of the gravity field does not vary strictly as $1/r^2$ but will deviate significantly at shorter distances, and so the period will not scale precisely as $a^{3/2}.$

For a spherically symmetric body, no matter what the radial profile you can put all of the mass at the center point instead of integrating throughout the sphere, as long as you are outside of it. This is called the Shell Theorem first proven by Sir Isaac Newton.

But the oblate figure is only cylindrically symmetric, not spherically symmetric, so you actually do have to integrate the force from every point within the planet. Most of this will already be captured in the $J_2$ term, and so I've just asked Equation for orbital period around oblate bodies, based on J2?

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    $\begingroup$ Teach a man to fish, and all. $\endgroup$
    – Chris B. Behrens
    Mar 6, 2018 at 4:07
  • $\begingroup$ @ChrisB.Behrens bingo! :-) $\endgroup$
    – uhoh
    Mar 6, 2018 at 4:10

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