Saturn's rings

1. Dec 13, 2007

natski

Dear all,

Can only the outter planet form rings? If Saturn were moved very close to the Sun, would it be able to maintain its ring system? I personally don't think so due to the fact the high tmperatures would vaporise a lot of the ice in the ring system but I would like to everyone else's viewpoint....

Natski

2. Dec 13, 2007

JasonRox

Also, the Sun's gravitational pull might ruin it too if it gets too close.

3. Dec 13, 2007

Contrapositive

If you moved Saturn to say, Mercury's orbit, yes it would probably keep it's rings, the rocky stuff anyway. However, I think if you move it really, really close to the sun then the sun's gravity is going to start disturbing the rings.

The reason you only see the outer planets with rings is because they are so massive they pull lots of stuff into orbit around them. I believe the popular theory speculate that moons move in to close the planet and break apart forming the rings.

4. Dec 13, 2007

natski

So one would perhaps estimate that when the gravitational force of the Sun is equal to the gravitational force of the planet, for the position of the rings, the rings will be disrupted...

GM*/a^2 = GMp/r^2
a = SQRT[ (M*/Mp) r^2 ]

where r is radius of rings and a is semi-major axis of orbit....

If r = 6.839E7 m (about half of Saturn's ring's radius)
and M*/Mj=1047.5

=> a = 0.015 AU roughly

Natski

5. Dec 13, 2007

natski

Presumeably then, if Saturn did somehow migrate inwards, it would lose its moon system before it lost its ring system...?

6. Dec 13, 2007

topherfox

Yes, as planets move inward, the ratio of the gravity from the sun to the planet increases, so either they get off course, or they orbit farther out.

7. Dec 13, 2007

Janus

Staff Emeritus
What you want to use here is the Hill Sphere. This gives the maximum radius at which a satellite can orbit a palnet without being torn away from by the Sun. (If you apply your method to the Earth-Moon system you would determine that the Moon can't orbit he Earth at the distance that it does.)

The radius of the Hill sphere can be found by:

$$\frac{r^3}{a^3}= \frac{m}{3M}$$

Solving for a for an r of 6.839e7 we get about 0.01 AU.