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Stink Fish Pot
06-11-2011, 11:43 PM
If we look at the solar system on edge from the side (assuming you could), and the sun and all planets were in a straight line like you see in science books, would our Solar System actually center on a plane?


If my question isn't clear, let me try again. If I could draw an imaginary line through the center of the sun and continue drawing that line until I got to Mercury, would Mercury be bisected also? Does everything line up on a perfect geometric plane? And does that include the Asteroid Belt and the Ort Cloud? I know Pluto has an eccentric orbit, so it doesn't fit the general model, but in general, does this hold up?

Asympotically fat
06-12-2011, 12:00 AM
If we look at the solar system on edge from the side (assuming you could), and the sun and all planets were in a straight line like you see in science books, would our Solar System actually center on a plane?


If my question isn't clear, let me try again. If I could draw an imaginary line through the center of the sun and continue drawing that line until I got to Mercury, would Mercury be bisected also? Does everything line up on a perfect geometric plane? And does that include the Asteroid Belt and the Ort Cloud? I know Pluto has an eccentric orbit, so it doesn't fit the general model, but in general, does this hold up?

The word your searching for is inclination (http://en.wikipedia.org/wiki/Inclination). You can see from the inclinations quoted in the article that the orbits of the planets do within a few percent lie in a plane. Pluto has a larger inclination than the planets, but is still pretty close to lying in this plane.

For objects in the asteroid belt the inclination is generally less than 30% which means that generally speaking th asteroid belt falls into this plane, but less so than the planets.

I beive the Oort cloud is roughly speaking spherical.

Raygun99
06-12-2011, 12:38 AM
Pluto's actually a pretty clear outlier (http://2.bp.blogspot.com/_jeetJTfzUoU/TSEL0Tq5YII/AAAAAAAAKlQ/gbUgVoEmb80/s1600/NewHorizons-FullTrajectory-SideView-02Jan2011.jpg) relative to the rest of the planets, which is one of the reasons why some believe it to be a captured body or escaped moon.

dtilque
06-12-2011, 04:10 AM
Pluto's actually a pretty clear outlier (http://2.bp.blogspot.com/_jeetJTfzUoU/TSEL0Tq5YII/AAAAAAAAKlQ/gbUgVoEmb80/s1600/NewHorizons-FullTrajectory-SideView-02Jan2011.jpg) relative to the rest of the planets, which is one of the reasons why some believe it to be a captured body or escaped moon.

Who are these people who think it's a captured body or escaped moon? This is not the mainstream astronomical position.

Pluto has a fairly typical orbital inclination for a Kuiper Belt Object, which is one reason it's classed as such rather than considered to be a planet. Eris, which is about the same size as Pluto, has a very large inclination (44 degrees), which is exteme even for a KBO.

However, when considering the inclination (or any other orbital characteristic) of the Solar System as a whole, it's best to just ignore the Kuiper Belt (including Pluto) as well as the asteroids and comets. In terms of mass, these small bodies constitute an unimportant minor clutter in the Solar System.

Alka Seltzer
06-12-2011, 06:09 AM
If we look at the solar system on edge from the side (assuming you could), and the sun and all planets were in a straight line like you see in science books, would our Solar System actually center on a plane?


Yes, all eight planets orbit within 7 degrees of the sun's equator. The reason for this is that the solar system formed from a protoplanetary disc (http://en.wikipedia.org/wiki/Protoplanetary_disc). Smaller objects, such as KBOs and comets, were scattered into more inclined orbits due to gravitational resonances and encounters with the sun or planets. For example, if a comet nucleus passed close to the south celestial pole of Jupiter (hope I am getting my terminology right there) it would be accelerated into a highly elliptical polar orbit.

There is a good article here (http://en.wikipedia.org/wiki/Solar_system_formation) on wikipedia that explains how the solar system formed.

Raygun99
06-12-2011, 11:50 AM
Who are these people who think it's a captured body or escaped moon? This is not the mainstream astronomical position.

I suppose I should have been more precise and said "led some to theorize". It and the rest of the KBOs are oddballs all around.

appleciders
06-12-2011, 12:22 PM
The (rough) plane in which the planets orbit is called the "ecliptic". As the above posters have noted, all of the regular planets fall on or near the ecliptic.

Chronos
06-12-2011, 12:27 PM
In terms of mass, these small bodies constitute an unimportant minor clutter in the Solar System. In terms of mass, everything except the Sun itself constitute minor clutter. In terms of angular momentum, everything except the Sun and Jupiter is minor clutter. Either way, we don't make the cut, either.

Stink Fish Pot
06-12-2011, 12:36 PM
Thank you all for the answers thus far... so now, to push my luck...

To extrapolate this further, can the same be said (more or less) about the stars in the Milky Way galaxy (and assuming all galaxies are formed in a similar way), most if not all of the galaxies in the universe?

Any picture I've ever seen of a galaxy, either a photo or artist's conceptual drawing, shows all stars more or less on the same inclination (thanks, tamop!).

if we pull that out further to look at the galaxies considered in our "local group", this seems to break down. Or does it? Galaxies in our local group seem to be tilted (with respect to our POV) at random angles of inclination, but these galaxies in our local group all rotate around a single spot in space, do they not?

Alka Seltzer
06-12-2011, 02:52 PM
It's true of spiral galaxies (http://en.wikipedia.org/wiki/Spiral_galaxies) (for example, the Milky Way has a diameter of 100,000 light years, but is about 1,000 light years thick), but not elliptical galaxies (http://en.wikipedia.org/wiki/Elliptical_galaxies).

The galaxies in the local group are not rotating around a common point.

ZenBeam
06-12-2011, 03:53 PM
As I recall, the inclination of extra-solar systems is more-or-less random with respect to the Milky Way's orbital plane as a whole. The plane of the ecliptic is about 60 degrees off of the plane of the Milky Way, for example.

ZenBeam
06-12-2011, 04:01 PM
That 60 degrees is an eyeball figure based on where the ecliptic and Milky Way are in the sky on a map I have. I'm sure someone can find a more accurate value.

Yumblie
06-12-2011, 04:16 PM
You can see how the planets more or less line up in a plane from here on Earth. The planets all travel along the same path in the sky as the moon and sun, though at different speeds and sometimes different directions, depending on where they are in their orbit. Last spring there was a neat "alignment" where 3 planets were visible at night forming a dotted line that led from the sun across the ecliptic (I believe they were Venus, Mars, Saturn).

dtilque
06-13-2011, 12:20 AM
That 60 degrees is an eyeball figure based on where the ecliptic and Milky Way are in the sky on a map I have. I'm sure someone can find a more accurate value.

Pretty close. The usual figure cited is 62 degrees, IIRC. Retrograde, that is, which means that it's actually 118 degrees.


I suppose I should have been more precise and said "led some to theorize". It and the rest of the KBOs are oddballs all around.

They're just left over planetisemals from the formation of the Solar System. They missed being incorporated into a planet and got pushed to the outskirts by the migration of Neptune. Nothing really oddball about them as a class, although some have unusual properties.

The only unusual property of Pluto, for example, is its apparent magnitude, which can be as much as 13.6. That's about 100 times brighter than Eris, which has apparent magnitude 18.7 (lower numbers mean brighter). But Eris is several times further away than Pluto. It would be much brighter if you moved it to the same distance as Pluto.

mlees
06-13-2011, 10:36 AM
If there was a planetary alignment, a neutrino released from the sun would not strike all the aligned planets, would it? A difference of one degree in inclination might not sound like much, but it could be several planetary diameters above or below the other planet(s) in relation to the plane of the ecliptic.

My high school math isn't up to this. Mars is at 1.85 degress of inclination, 6800 km in diameter, orbitting at an average distance of 227,939,150 km from the Sun. What is it's "height" above the ecliptic when it is at it's "heighest" point, in km? (I assume it goes from 1.85 degress above, to 1.85 degrees below one half of a Martian year later, then back again. Is that right?)

Earth is at 1.5 degress, Venus at 3.4, Jupiter at 1.3 degrees, Saturn at 2.48. The further out you go from the Sun, the greater the difference these little numbers become.

Bytegeist
06-13-2011, 01:18 PM
If there was a planetary alignment, a neutrino released from the sun would not strike all the aligned planets, would it?

Correct, because I don't think they can ever align quite that well, all at once.

From time to time, Venus and Mercury will transit the Sun (http://en.wikipedia.org/wiki/Transit_of_Mercury) when viewed from Earth (though not both at the same time). On those occasions, a solar neutrino could travel in a straight line through the inner planet and then hit Earth. That kind of alignment only occurs when both planets are sufficiently close to the common line of intersection of their orbital planes. The lines of intersection from the various pairs of orbital planes are not going to be aligned with each other, however.

Meaning that a near-perfect seven-planet transit of the Sun, as seen from Neptune, is impossible, and that's essentially what would need to happen for the neutrino to hit every planet.


A difference of one degree in inclination might not sound like much, but it could be several planetary diameters above or below the other planet(s) in relation to the plane of the ecliptic.

My high school math isn't up to this. Mars is at 1.85 degress of inclination, 6800 km in diameter, orbitting at an average distance of 227,939,150 km from the Sun. What is it's "height" above the ecliptic when it is at it's "heighest" point, in km?

To a good approximation, the ratio of a planet's "height" (h) above the ecliptic to its distance (d) from the Sun is the angle (a) that it makes with the ecliptic, when that angle is expressed in radians. That is, h/d = a; h = ad.

mlees
06-13-2011, 02:08 PM
Correct, because I don't think they can ever align quite that well, all at once.

From time to time, Venus and Mercury will transit the Sun (http://en.wikipedia.org/wiki/Transit_of_Mercury) when viewed from Earth (though not both at the same time). On those occasions, a solar neutrino could travel in a straight line through the inner planet and then hit Earth. That kind of alignment only occurs when both planets are sufficiently close to the common line of intersection of their orbital planes. The lines of intersection from the various pairs of orbital planes are not going to be aligned with each other, however.

Meaning that a near-perfect seven-planet transit of the Sun, as seen from Neptune, is impossible, and that's essentially what would need to happen for the neutrino to hit every planet.


To a good approximation, the ratio of a planet's "height" (h) above the ecliptic to its distance (d) from the Sun is the angle (a) that it makes with the ecliptic, when that angle is expressed in radians. That is, h/d = a; h = ad.

Um... 1.85 degrees = .032375 radians? So .032375 * 227,939,150km = 7,379,530km, which is 1,085 planetary diameters. :eek: There's lot's of space out there.

Even so, it just occured to me that Mars' heighest point above the ecliptic probably isn't the same point on the "clock" as Earth's is. That is to say, when Mars is at it's highest seperation from the ecliptic, and we arbitrarily called that 12 o'clock when viewed from "above", Earth's sister spot could be anywhere on the dial.

What a confusing ballet going on. It's a wonder anything lines up at all (for that "straight as the neutrino flies" shot).

Kinda makes my question look a little dumb. :(

Thanks for the answer, though!

Chronos
06-13-2011, 02:17 PM
Even so, it just occured to me that Mars' heighest point above the ecliptic probably isn't the same point on the "clock" as Earth's is. That is to say, when Mars is at it's highest seperation from the ecliptic, and we arbitrarily called that 12 o'clock when viewed from "above", Earth's sister spot could be anywhere on the dial. Astronomers refer to this angle as the "argument of the ascending node". Though technically, that's the angle of the point at which it goes from south of the ecliptic to north of it, not the point when it's furthest from the ecliptic.

ZenBeam
06-13-2011, 04:25 PM
As a guess, most pairs of planets could align with the Sun the way Venus and Earth do*. Can there ever be three planets aligned so that a neutrino could go through three of them? Say within +/- 1 million years of the present, where we'd have fairly accurate knowledge of the orbits.

* OK smart guy who's going to nitpick: Which can and which can't?

Stink Fish Pot
06-14-2011, 01:17 AM
Another question based on what has been agreed to in this thread.

It is presumed that Uranus was hit by a flying object of such size and speed as to knock it on its side. In order to to this, how was it able to stay within the inclination that the imaginary plane passes through? Can (or has) it be proven mathematically that there was an object that could tip the planet over but not move it out of the planetary plane that the rest of the planets find themselves on?

Or would Uranus move back into place on its own due to the gravitational pull of the sun and other planets?

Chronos
06-14-2011, 02:13 PM
If Uranus' inclination is in fact due to an impact with something else, the impactor was probably itself orbiting more or less in the ecliptic. Compared to the size of the planet, it might have been very off-center (thus causing the inclination change), but compared to the size of the orbit, it wasn't, so the plane of the orbit wouldn't have changed much.

mlees
06-14-2011, 02:20 PM
If Uranus' inclination is in fact due to an impact with something else, the impactor was probably itself orbiting more or less in the ecliptic. Compared to the size of the planet, it might have been very off-center (thus causing the inclination change), but compared to the size of the orbit, it wasn't, so the plane of the orbit wouldn't have changed much.

Does a planet only get it's spin (rotation) from the time it formed out of some primordial cloud of hydrogen?

Wouldn't an object large enough to knock Uranus sideways be big enough to perturb that rotation too? (Does Uranus wobble?)

Chronos
06-14-2011, 03:50 PM
When we say "knock it sideways", we do mean perturbing the rotation. The planets are spheres, so the rotation is the only thing that gives them any sort of preferred axis. Likewise, without anything else to give a preferred axis, you don't get wobbles, either.

mlees
06-14-2011, 04:05 PM
When we say "knock it sideways", we do mean perturbing the rotation. The planets are spheres, so the rotation is the only thing that gives them any sort of preferred axis. Likewise, without anything else to give a preferred axis, you don't get wobbles, either.

I assume Uranus formed from a disk of gas that collapsed due to gravitational attraction, and while doing so, began to have motion. (Spin, or rotation.) Is that correct?

In almost all of the othe planets (I think), this spin is fairly close to the plane of the ecilptic. In the case of Uranus, it is not. Is it likely that Uranus is tilted so far off kilter because of a collision or merger with another object after it's protoplanetary nebula had collapsed to it's present volume? (It's eccentricity and it's inclination appear average.)

I'm asking because I am wondering if Uranus's rotation may have already been "set" in place, nearer to the ecliptic, then got pushed over, or did the rotation just form at that angle?

Whack-a-Mole
06-14-2011, 04:26 PM
I'm asking because I am wondering if Uranus's rotation may have already been "set" in place, nearer to the ecliptic, then got pushed over, or did the rotation just form at that angle?

Not sure anyone knows for sure what happened but I cannot see how Uranus would form out of the protoplanetary disk with a rotation like that. I think the most commonly held belief is that something big smashed into it after it had formed and knocked it on its side (in a manner of speaking).

Chronos
06-14-2011, 04:42 PM
One can imagine scenarios where its angular momentum was skewed drastically some time before it finished coalescing. I don't know enough about the topic to say how likely such scenarios are, though.

mlees
06-14-2011, 04:50 PM
Ok, thanks!

Bytegeist
06-14-2011, 06:05 PM
One can imagine scenarios where its angular momentum was skewed drastically some time before it finished coalescing.

Ok, thanks!

For what it's worth, Uranus's rings and major moons all orbit in the planet's equatorial plane, not its orbital plane — which to planetary astronomers suggests that the "Big Tilt" event, whatever that was exactly, probably happened very early in the planet's history.

mlees
06-14-2011, 10:49 PM
For what it's worth, Uranus's rings and major moons all orbit in the planet's equatorial plane, not its orbital plane — which to planetary astronomers suggests that the "Big Tilt" event, whatever that was exactly, probably happened very early in the planet's history.

I read that, but the implications didn't sink in. Thanks!