apodman wrote:neufer wrote:(If you do find out, be sure to let us know.)
I was hoping to post a link that would beg the question and motivate someone else to answer it for me. I know that turnabout is fair play, but I did get my bid in first.
I'll try to answer. Grain of salt and all that!
First, why should a ring form at all? Ring particle collisions are not perfectly elastic, meaning that they conserve angular (and linear) momentum and do not conserve kinetic energy. Collisions reduce the relative velocity of the colliding bodies, so a collision between two bodies whose orbits are tilted relative to each other will decrease their tilt angle. Over the course of many collisions, a puffed up ring system will relax into a disk.
Why should the ring form in the equatorial plane? As stated in the link you provided (
http://www.astronomycafe.net/qadir/q58.html), orbits inclined with respect to Saturn's equatorial plane
precess (change orientation in space) due to the planet's oblateness. The extra mass around Saturn's equator pulls inclined bodies back toward the equatorial plane, so they cross the plane slightly sooner than they would in a perfectly spherical potential. This results in precession of the line of nodes; it rotates in the opposite direction of orbital motion. (The pericentre also precesses, but that's only relevant to eccentric orbits.) Here's a diagram of nodal precession due to an exterior planet. Different source of equatorial mass, same idea:
Omega: longitude of ascending node (see
Keplerian Elements)
black: unperturbed orbits
green: inner planet's trajectory under the perturbing influence of the massive outer planet
Getting to my point: the nodal precession frequency is a function of distance from the planet. Imagine that Saturn's rings are instantly tilted to some nonzero inclination with respect to Saturn's equator. Each ring orbit undergoes nodal precession, but due to the rings' extent in semimajor axis, each particle orbit precesses at a slightly different rate. The initial rings eventually smear out, and we end up with a puffed up disk that settles into a thin ring (equatorial this time) through collisions, as described above.
Digression: in a spherically symmetric potential,
inelastic collisions conserve angular momentum, and do not conserve kinetic energy. That's basic mechanics. Now, nodal precession of an orbit is the same as precession of the angular momentum vector, which projects out of the orbital plane according to the
right-hand rule. Mentally tilt Saturn's rings again. Initially, the projection of the rings' total angular momentum vector onto the equatorial plane is nonzero. In a few nodal precession timescales, the equatorial component of angular momentum will average out to zero. This
non-conservation of total angular momentum is a consequence of
Noether's theorem, which links symmetries to conservation laws. In particular, rotational symmetry implies conservation of angular momentum. The laws of motion are
not symmetric in the polar angle of an oblate planet's potential, so angular momentum is not conserved in any tilted plane.
(This is really pedantic, but that link begot, not begged, the question 
http://en.wikipedia.org/wiki/Begging_th ... uial_usage )
apodman wrote:Living where we do, we get to see more than our share of Washington, D.C. postcards and such. It is very popular to show a long shot down the mall including the Washington Monument and the Capitol Building. They usually include the moon, which is "photoshopped" in more often than not. Sometimes you can tell by the impossible angle of the lit side. Other times it appears impossibly large, and I end up calculating how far back the camera would have to be for it to be real. The question you answered about the apparent size of Titan reminded me of that.
A telephoto lens can make the moon appear impossibly large. For example:
http://apod.nasa.gov/apod/ap081212.html
...But photoshopped postcards wouldn't surprise me.
[quote="apodman"][quote="neufer"](If you do find out, be sure to let us know.)[/quote]
I was hoping to post a link that would beg the question and motivate someone else to answer it for me. I know that turnabout is fair play, but I did get my bid in first.[/quote]
I'll try to answer. Grain of salt and all that!
First, why should a ring form at all? Ring particle collisions are not perfectly elastic, meaning that they conserve angular (and linear) momentum and do not conserve kinetic energy. Collisions reduce the relative velocity of the colliding bodies, so a collision between two bodies whose orbits are tilted relative to each other will decrease their tilt angle. Over the course of many collisions, a puffed up ring system will relax into a disk.
Why should the ring form in the equatorial plane? As stated in the link you provided (http://www.astronomycafe.net/qadir/q58.html), orbits inclined with respect to Saturn's equatorial plane [url=http://en.wikipedia.org/wiki/Precession#Of_the_Earth.27s_axis]precess[/url] (change orientation in space) due to the planet's oblateness. The extra mass around Saturn's equator pulls inclined bodies back toward the equatorial plane, so they cross the plane slightly sooner than they would in a perfectly spherical potential. This results in precession of the line of nodes; it rotates in the opposite direction of orbital motion. (The pericentre also precesses, but that's only relevant to eccentric orbits.) Here's a diagram of nodal precession due to an exterior planet. Different source of equatorial mass, same idea:
[img]http://www.utsc.utoronto.ca/~05leimbi/precession.png[/img]
Omega: longitude of ascending node (see [url=http://en.wikipedia.org/wiki/Orbital_elements#Keplerian_elements]Keplerian Elements[/url])
black: unperturbed orbits
green: inner planet's trajectory under the perturbing influence of the massive outer planet
Getting to my point: the nodal precession frequency is a function of distance from the planet. Imagine that Saturn's rings are instantly tilted to some nonzero inclination with respect to Saturn's equator. Each ring orbit undergoes nodal precession, but due to the rings' extent in semimajor axis, each particle orbit precesses at a slightly different rate. The initial rings eventually smear out, and we end up with a puffed up disk that settles into a thin ring (equatorial this time) through collisions, as described above.
Digression: in a spherically symmetric potential, [url=http://en.wikipedia.org/wiki/Inelastic_collision]inelastic collisions[/url] conserve angular momentum, and do not conserve kinetic energy. That's basic mechanics. Now, nodal precession of an orbit is the same as precession of the angular momentum vector, which projects out of the orbital plane according to the [url=http://en.wikipedia.org/wiki/Right-hand_rule#Direction_associated_with_a_rotation]right-hand rule[/url]. Mentally tilt Saturn's rings again. Initially, the projection of the rings' total angular momentum vector onto the equatorial plane is nonzero. In a few nodal precession timescales, the equatorial component of angular momentum will average out to zero. This [i]non[/i]-conservation of total angular momentum is a consequence of [url=http://en.wikipedia.org/wiki/Noether%27s_theorem]Noether's theorem[/url], which links symmetries to conservation laws. In particular, rotational symmetry implies conservation of angular momentum. The laws of motion are [i]not[/i] symmetric in the polar angle of an oblate planet's potential, so angular momentum is not conserved in any tilted plane.
[size=85](This is really pedantic, but that link [i]begot[/i], not [i]begged[/i], the question :P http://en.wikipedia.org/wiki/Begging_the_question#Colloquial_usage )[/size]
[quote="apodman"]Living where we do, we get to see more than our share of Washington, D.C. postcards and such. It is very popular to show a long shot down the mall including the Washington Monument and the Capitol Building. They usually include the moon, which is "photoshopped" in more often than not. Sometimes you can tell by the impossible angle of the lit side. Other times it appears impossibly large, and I end up calculating how far back the camera would have to be for it to be real. The question you answered about the apparent size of Titan reminded me of that.[/quote]
A telephoto lens can make the moon appear impossibly large. For example: http://apod.nasa.gov/apod/ap081212.html
...But photoshopped postcards wouldn't surprise me.