9
$\begingroup$

From my understanding, a ring can form around a planet when a moon gets too close to its Roche limit, and gets ripped appart by the planet's gravity pull. That makes sense to me, but I don't understand why Saturn has both moons and rings at the same place.

I know the F ring is supposedly created by Enceladus' rejections, but I don't get why Pandora and Prometheus are not ripped appart as well as the ancient objects that formed the ring they are in.

I couldn't find specific explanations for this. Does it have any link to the moon's density?

$\endgroup$
1
  • $\begingroup$ Jupiter has both rings and moons too. The rings are just dust particles tho. $\endgroup$
    – ott--
    Commented Jul 30, 2016 at 10:38

2 Answers 2

5
$\begingroup$

You're right that density is the important thing here. The Roche limit is the distance from the main body $d$ such that $$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$ where $_M$ denotes the main body and $_m$ denotes the satellite.

As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half times the Roche limit from Saturn. Therefore, they're in no danger of being ripped apart any time soon.

$\endgroup$
1
  • $\begingroup$ Thank you very much for clarifying, this is the answer I needed. Also, since my question has been edited : pardon my english. French user here :p $\endgroup$
    – Nico
    Commented Dec 2, 2015 at 7:57
-4
$\begingroup$

This answer is just misleading. The rings are not formed by the moons. They are clearly formed by a force. And I don't think its gravity. Gravity forms spheres or spherical shells if as you suggest its the roche limit.

FYI saturns rings are about 30 feet thick and extend for millions of miles. Gravity does not confine things to thin discs. There is an magnetic force at work here. And it has been demonstrated by plasma physicists.

$\endgroup$
1
  • 3
    $\begingroup$ Several statements here need justification. $\endgroup$
    – James K
    Commented Jul 30, 2016 at 10:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .