<p>This is a supplemental answer to @RussellBorogove's <a href="https://space.stackexchange.com/a/25856/12102">excellent and intuitive answer</a>. </p>

<p>Starting from the <a href="https://en.wikipedia.org/wiki/Vis-viva_equation" rel="nofollow noreferrer">vis-viva equation</a> one of the few equations I can actually remember:</p>

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

<p>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 <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> and is about 3.793E+07 km^3/s^2. (beware of meters vs kilometers).</p>

<p>The circumference is $C = 2 \pi a$, so </p>

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

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

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

<hr>

<p>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 <em>will not scale precisely as $a^{3/2}.$</em></p>

<p>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 <a href="https://en.wikipedia.org/wiki/Shell_theorem" rel="nofollow noreferrer">Shell Theorem</a> first proven by Sir Isaac Newton.</p>

<p>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 <a href="https://space.stackexchange.com/q/25868/12102">Equation for orbital period around oblate bodies, based on J2?</a></p>

CC BY-SA 3.0
2018-03-06T04:10:49.827Z
@ChrisB.Behrens bingo! :-)
uhoh
CC BY-SA 3.0
2018-03-06T03:30:42.803Z
2018-03-06T10:28:23.650Z
uhoh
<p>Hours to days.</p>

<p>The orbital period is proportional to the 3/2 power of the orbital radius, and the orbital period of the moon <a href="https://en.wikipedia.org/wiki/Methone_(moon)" rel="noreferrer">Methone</a> (and thus of the particles in its associated ring arc) is very close to 24 hours. Working from the <a href="https://en.wikipedia.org/wiki/Rings_of_Saturn#Major_subdivisions" rel="noreferrer">orbital radii given in this table</a>, I get: </p>

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

CC BY-SA 3.0
2018-03-05T16:44:20.443Z
2018-03-05T16:44:20.443Z
Russell Borogove
<p>I don't mean necessarily a granular chart of all of the different rings at different distances, but is it days? Weeks?</p>

<orbit><saturn>
What are the periods of Saturn's rings?
Chris B. Behrens
# What are the periods of Saturn's rings?

(Asked by: username_0 on 2018-03-05)

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

**Tags:** orbit, saturn

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## 2 answers

### Answer by username_2 on 2018-03-05. Score: 11 | Accepted answer

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](https://en.wikipedia.org/wiki/Methone_(moon)) (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](https://en.wikipedia.org/wiki/Rings_of_Saturn#Major_subdivisions), 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


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### Answer by username_1 on 2018-03-06. Score: 4

This is a supplemental answer to @RussellBorogove's [excellent and intuitive answer](https://space.stackexchange.com/a/25856/12102).

Starting from the [vis-viva equation](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/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?](https://space.stackexchange.com/q/25868/12102)

> **Comments:**
> **username_1**: @ChrisB.Behrens bingo! :-)  

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