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Space Science

New Moon System Around Uranus 247

An anonymous reader writes "Astronomers have discovered two of the smallest moons yet found around Uranus. The new moons, uncovered by NASA's Hubble Space Telescope, are about 8 to 10 miles across (12 to 16 km) -- about the size of San Francisco. The two moons are so faint they eluded detection by the Voyager 2 spacecraft, which discovered 10 small satellites when it flew by the gas giant planet in 1986. The newly detected moons are orbiting even closer to the planet than the five major Uranian satellites, which are several hundred miles wide. The two new satellites are the first inner moons of Uranus discovered from an Earth-based telescope in more than 50 years. "It's a testament to how much our Earth-based instruments have improved in 20 plus years that we can now see such faint objects 1.7 billion miles (2.8 billion km) away," says Mark Showalter, a senior research associate at Stanford University. 'The inner swarm of 13 satellites is unlike any other system of planetary moons,' says co-investigator Jack Lissauer. 'The larger moons must be gravitationally perturbing the smaller moons. The region is so crowded that these moons could be gravitationally unstable. So, we are trying to understand how the moons can coexist with each other.'"
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New Moon System Around Uranus

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  • Too easy (Score:5, Funny)

    by mgebbers ( 252737 ) on Saturday September 27, 2003 @04:41AM (#7070965)
    "Moon System Around Uranus"? Why even bother thinking of something funny? :/
  • by Anonymous Coward on Saturday September 27, 2003 @04:42AM (#7070966)
    Although the new moons are large in terms I can understand, they seem very, very small when comparing them to planets.

    Is there a definition of a moon? Must something be X miles/kilomters in size or volume?
    • > Is there a definition of a moon?

      Let's start from planet. Planet is a body orbiting a sun, which do not emit light on its own.
      A moon is a body orbiting a planet. It may be just a bigger rock, like Phobos and Deimos around Mars or a big one like our Moon. But of course it must be smaller than a planet.

      This is not a formal definition of course :-)

      Regards

      • But of course it must be smaller than a planet.

        Really?

        Pluto is smaller than seven of the solar system's moon according to here [arizona.edu]( Moon, Io, Europa, Ganymede, Callisto, Titan and Triton) and its a planet.

        • Misunderstanding. English is not my first language.
          I mean comparing moon to the planet it is orbiting. Not planets among them. Pluto probably could have been some big planet's moon. But it is orbiting Sun, not any other planet, so it is a planet too.
          Well, it isn't very precise, because there is a lot of smaller rocks orbiting Sun, which are not called planets but planetoids or asteroids (sorry, I don't know proper English terminology). And astronomers are not sure, if Pluto should be treated as a planet o
          • A moon is a natural satellite revolving around a planet. There definition of a Planet is still somewhat ambigious. This is because we have not been able to observe other planet systems with great detail and thus have not been able to formulate a good definition of Planet.
        • My understanding is that, given its orbit and size, whether Pluto is a planet is debatable. It exists as a planet more for historic reasons. There are those who argue Pluto ought to be considered a Kuiper Belt object and not a planet. If so it is certainly among the largest. And, of course, debates about what is or isn't a planet at a certain point seem silly.

          One should further point out that Pluto and Charon (it's "moon") are typically referred to as a planet system since Charon is so big relative to

        • Pluto is smaller than seven of the solar system's moon according to here( Moon, Io, Europa, Ganymede, Callisto, Titan and Triton) and its a planet.

          Pluto is small than many moons but it's still a planet, if only because it has a Sailor Senshi assigned to it.

      • Let's start from planet. Planet is a body orbiting a sun, which do not emit light on its own. A moon is a body orbiting a planet. It may be just a bigger rock, like Phobos and Deimos around Mars or a big one like our Moon. But of course it must be smaller than a planet. This is not a formal definition of course :-)

        Sorry, it is not quite that simple. A planet does indeed orbit the sun, but it must also have a certain size. Minor rocks orbiting the sun are asteroids. There is currently some controversy o
        • I thought the difference between a planet-moon combination and a planet-planet combination was that winth in a planet moon combination the centre of gravity lay within the body of the larger body, whereas a planet-planet combination has it centre of gravity outside both bodies.
          • >whereas a planet-planet combination has it centre of gravity outside both bodies

            So that means our moon and earth is actually a planet-planet combination then.

            I do believe the centre of gravity between the 2 planets lay outside of both the earth and the moon.
            • even if the Moon is a very big moon for a planet our size, it's still a moon. the centre of gravity is within earth, 4700km from earth's centre.
              http://www1.tpgi.com.au/users/mpaine/scie nce.html# moontide
    • i wonder this myself. you would think it obvious that they should just be referred to as rocks. a fly around my head is not a moon, nor something to because of interest or awe. maybe a moon should be X% in size of what it orbits to be a moon. Or perhaps someone could just better define "rock floating in space" so we dont have to keep inflating the number of objects we can call moons around planets.
      • Your fly is not a moon because you are not a planet, and the fly is not obriting. It's "fly"ing.

        Why bother trying to make a definition on something that we really have no idea how to define. Every new star or system we find doesn't usually have much in common with our own, and we'd have to constantly redefine it all the time anyway.

        Basically, you can call it a moon or a rock or whatever you like, at the end of the day it's a whopping great big chunk of something and probably doesn't care what you call i

    • IANAA (I am not an astronomer), but as far as I can remember, a moon is defined as a celectrial body that orbits another "Planet" and whose "center of gravity" of the pair, lies within the volume of the planet.

      If you follow this definition then the moon (Luna) is not a moon but a planet. So this definition is not widely used. It makes sense to me but I have always thought of Luna as to big to be a moon relative to the earth.

      Oh well...
  • about the size of San Francisco

    Bah, that's 2D. How many VW Bugs is it?
  • by Anonymous Coward on Saturday September 27, 2003 @04:42AM (#7070971)
    After VW-Beetles, Libraries of Congress we have
    now the unit of "San Francisco" to describe huge
    masses. So how many Kilo-VW-Beetles is one
    San Francisco?

    johnboy
  • *sighs* (Score:4, Funny)

    by Pompatus ( 642396 ) on Saturday September 27, 2003 @04:42AM (#7070972) Journal
    New Moon System Around Uranus

    I saw that headline on slashdot and immediately thought to myself, "this is definately one of those times to read the article and NOT the comments".
    • I saw that headline on slashdot and immediately thought to myself, "this is definately one of those times to read the article and NOT the comments".

      Strangely, I was thinking exactly the opposite.
  • by Rosco P. Coltrane ( 209368 ) on Saturday September 27, 2003 @04:51AM (#7070994)
    This article is a trap. it's not an article about planets, it's the yearly Slashdot teenagers counter : they post an article with "Uranus" in it, wait a bit, then count the number of people who made witty rectal comments.

    Seriously though, is it not possible to read an article about Uranus without seeing all those "uranus *lol* *giggle* *pffft!*" posts ?

    *sigh*
    • Seriously though, is it not possible to read an article about Uranus without seeing all those "uranus *lol* *giggle* *pffft!*" posts ?

      Of course it's possible. I'm amazed at the discovery of this new moon system, and duly amazed that the two new moons evaded detection for so many years. I imagine it speaks to the quality of NASA's latest technologURANUS! lol *giggle* *pffft!*

      OK, maybe it's not possible, but you have to admit, it's a bit funnier than all those jokes about NO CARRI^ath0

    • Seriously though, is it not possible to read an article about Uranus without seeing all those "uranus *lol* *giggle* *pffft!*" posts ?

      Don't worry. By the year 2620 scientists will rename Uranus to end that joke once and for all.

      Of course, renaming it Urectum won't help much...
    • by metamatic ( 202216 ) on Saturday September 27, 2003 @08:33AM (#7071584) Homepage Journal
      Yes, there's nothing funny about Uranus. let's forget the childish humor and take a serious, scholarly look at Uranus. To many people it's just a giant cloud of gas where the sun doesn't shine, but those of us who are enthusiastic about Uranus know that it has many secrets.

      Surprising as it may seem, we don't have all that many photographs of Uranus. Yes, the Pioneers sent back pictures of Uranus, lots of them. But there are very few images that are high enough resolution and quality to show the faint rings around Uranus. Perhaps the excitement around these new moons will give us the excuse we need to take another long, hard look at Uranus.

      Even if you have no idea how to find Uranus, you can still appreciate its unusual configuration. Scientists still don't understand why Uranus is tilted sideways. Also, while we know what's near the surface, we still aren't sure of the exact chemical mixture deep inside Uranus. Are the moons stable, or are they spiraling into Uranus?

      With so much to learn, we must hope that NASA will probe the depths of Uranus soon. Yes, there are many technical issues that will need to be resolved, and problems to be faced--but we put men on the moon, and I'm sure that given sufficient motivation, NASA's engineers can lick Uranus too.

      Oh, and yes, the size comparisons are silly, but can you think of a more sensible unit of size than San Francisco for an object in the vicinity of Uranus?
  • "Moons are unstable" (Score:5, Interesting)

    by turkeyphant ( 648612 ) on Saturday September 27, 2003 @04:51AM (#7070995) Homepage Journal

    If these moons are gravitationally astatic, stochastic motion could account for their motion in this deterministic system. We all know how complication three-body motion is, so with the number of objects affected by various gravitational fields out there, it would be incredibly hard to predict any movement at all. I wish them good luck in trying to precisely "understand how the moons can coexist with each other".

    Is it not possible that these moons are so unstable that they will have relatively short lifespans? Might they soon end up crashing into the planet's surface or interact together and get flung off out of the solar system?

    • by MickLinux ( 579158 ) on Saturday September 27, 2003 @06:37AM (#7071239) Journal
      We all know how complication three-body motion is, so with the number of objects affected by various gravitational fields out there, it would be incredibly hard to predict any movement at all.

      What they said was correct at one time. It is no longer correct.

      It actually isn't all that hard to predict their motion. There's a new mathematical tool, the Parker-Sochacki solution to the Picard Iteration, that has made great strides in the ability to predict this.

      What's even better, this solution method is incredibly easy, conceptually simple [jmu.edu], ideal for initial value problems, yields exact functional solutions [jmu.edu], involves simple algebra [yes, that's right: simple algebra solutions to almost any set of partial differential equations] [jmu.edu] and turns out doubling precision for every iteration.

      Oh, yes: there is a version out for Maple, too. [jmu.edu]

      The solution that it turns out is a MacLauren series [functionally equal to the Taylor Series] dependant on as many variables as you need. However, for this you'd have everything dependent on time.

      Also, this method *has* been used to predict planetary, moon [aps.org], and asteroid motion. [aps.org] It works.

      [PS: That last link has code for you code monkeys]

      • by metamatic ( 202216 ) on Saturday September 27, 2003 @08:15AM (#7071519) Homepage Journal
        I downloaded the "incredibly easy, conceptually simple" description in the PDF file.

        Now I remember why I switched to computer science. *sigh*
        • by MickLinux ( 579158 ) on Saturday September 27, 2003 @09:53AM (#7071900) Journal
          Okay, first of all, when you read that thing, know that IVP = "Initial Value Problem"; "ODE"=Ordinary Differential Equation.

          Now, here goes: Suppose you have a function that is a taylor series: y=a + bt + ct^2 + dt^3...

          Alternatively, I could write that y=cy(0)+cy(1)t^1 + cy(2)t^2 + cy(3)t^3... where cy(n) is the coefficient a,b,c,d...I'm going to switch back and forth a little, for convenience' sake.

          Now, at time t=0, what is the value of y? y=a. Suppose y measures position. At time t=0, what is the value of b? b is the initial velocity. That is because y'=b+2ct+3ct^2+4ct^3..., and at t=0, everything except b drops out. But the time derivitive of y is velocity. So b is equal to the initial velocity.

          That's the concept of the Picard iteration: it's incredibly easy to deal with differentials if you have a Taylor series.

          Let's stop here, and instead of calling the coefficients a,b,c,d... let's name them as mathematicians do: cy(0),cy(1),cy(2),cy(3)... That is, coefficient for y #0, #1, #2, and so on.

          Now, suppose I have three Taylor functions, f,g,and h, and I know two of them, and I have an equation f=g+h. How do I solve for g, for example, knowing f and h? Well, this one's easy from algebra. Each coefficient can be calculated from the relationship cf(n)=cg(n)+ch(n). So that one's easy. So is subtraction, same method, different sign.

          Now multiplication is harder, and division is incredibly hard, and so that's kindof where Picard stopped. So it didn't seem all that useful to him. But Parker and Sochacki got it past that.

          If we had f(t)=g(t)*h(t), well, for these to be functionally equivalent, then the coefficients for the g*h entries on the right should be equal to the coefficient for f, same power of t. So...

          power of t=0 (that's the a coefficient for each):

          cf(0) = cg(0) * ch(0).

          Everything else has a nonzero power of t, so that one's easy.

          power of t=1 (that's the b coefficient for f, but that's the a coefficient of g times the b coefficient of h, plus the b coefficient of g times the a coefficient of h):

          cf(1) = cg(0)*ch(1) + ch(0)*cg(1)

          That's the next one.

          Power of t=2:

          cf(2) = cg(0)*ch(2)+cg(1)*ch(1)+cg(2)*ch(0).

          Here, we're beginning to get a pattern.

          cf(n) = SUM (i=0...n) cg(i)ch(n-i)

          So if we have all the values 0...n for g and h, then we can calculate value n for f, as well.

          DIVISION

          Okay, up through this, picard got. He couldn't get division. However, Parker and Sochacki posited that you could take the differential of f=g/h, to get:

          f' = d [g*h^-1]/dt = (g'h - gh')/(h^2)

          so

          f'*h*h = g'*h - g*h'

          Now, if we have coefficients for g and h through n, and we want the f' coefficient #n, then we need to look at the coefficients that accompany t^(n-1), because on the left we have f', and if we know

          f=cf(0) + cf(1)t + cf(2)t^2 + ... + cf(n-1)t^(n-1)+ [unknown]cf(n)t^n

          then

          f'=cf(1)+2cf(2)t + ... (n-1)cf(n)t^(n-1)

          where cf(n) again is unknown.

          Looking at the rest of the left hand side f'*h*h, we note that since we have h through point n, we have the coefficients of h*h through point n as well. So calling k=h*h, we have

          f'k = g'h-h'g

          where it's the coefficients of the (n-1) powers of t that are of interest.

          However, when you multiply that lefthand side out, you quickly see that there is only one coefficient in k that multiplies against the (n-1) power of t, and all other values are known! So dropping our interest in all other powers of t, and just dealing with the coefficient of interest:

          SUM(i=0...n-1) cf'(i)k(n-1-i) = SUM(q=0...n-1)cg(n-1-q)ch'(q) - SUM (r=0...n-1)cg'(r)ch(n-1-r)

          but cg'(r) = (r+1)cg(r) so

          I'm going to stop it here, because I'm doing this in my head, and rather than give you a wrong answer, I'm going to say it should be really obvious if you take the

          • Mr. MickLinux, may I be excused? My brain is full.
          • What throws me off all the time is the use of different letters to represent certain things, but what they represent isn't usually explained. Does "c" always mean the same thing to everybody in all problems? (isn't it the speed of light?) And y, and b, etc...

            Is there some generally understood meaning for certain letters if not specified? Such as a programmer could use variables "i" and "s" and people would probably assume that they are an integer and a string, respectively.

            BTW, your post was very infor
            • f,g, h, and k were functions.

              i,q, and r were iteration variables.

              In this, also, a,b,c, and d were coefficients. An alternative naming convention for the same coefficients, but generalized to functions f,g, and h were

              a for function f is "cf(0)". That is, [c]oefficient for [f] number zero.

              a for function g is "cg(0)". That is, [c]oefficient for [g] number 0.

              b for function f is "cf(1)" that is, [c]oefficient for [f] number 1.

              and so on.

    • ...for a few thousand years than for a few billion.
  • Several Moons (Score:3, Insightful)

    by NeoGeo64 ( 672698 ) on Saturday September 27, 2003 @04:51AM (#7070996) Homepage Journal
    Uranus is distinguished by the fact that it spins on its side.

    Uranus is a gas giant with no solid surface. Like the other gas planets, Uranus has bands of clouds that blow around rapidly.

    Uranus is sometimes just barely visible with the unaided eye on a very clear night; it is fairly easy to spot with binoculars (if you know exactly where to look). There are several Web sites that show the current position of Uranus.

    Sorry guys, I couldn't help but post some immature humor. ;)
    • Uranus is not the only planet that spins on its side. It's equtorial inclination to orbit is 98 degrees. Pluto(in my opinion not worth of being called planet) has an angle of 122 degrees qutorial inclination to orbit. Venus is bizzare with an equtorial inclination to orbit of 177.4 degrees. Another ineresting fact about Uranus is that it was accidentally discovered in 1781 by the British astronomer Sir William Herschel and was originally named the Georgium Sidus (Star of George) Here are other facts ab
  • "They were over the moon with joy with the discovery"

    Sorry I will get my coat

    Rus
  • Why is NASA... (Score:5, Interesting)

    by rexguo ( 555504 ) on Saturday September 27, 2003 @04:58AM (#7071022) Homepage
    ...so eager to take Hubble down, when it's still contributing so much to astrophysics? The new space telescope isn't even ready for launch yet, and who knows if it will work at first go? I'd rather have Hubble as backup until the new one is working smoothly and flawlessly before even thinking about bringing it down. Capitalism and politics just don't mix well with science.
    • Yeah, those damn rocket scientists have no idea what they're doing.

    • Well, here's a conspiracy theory: ;)

      It costs money to run the Hubble space telescope. Taking it down will save money.

      Creating a new one will cost a lot of money... if you create a new one.

      My *conspiracy theory* is that they fill a 500 gallon drum up with various junkyard waste, spray paint "Hubble II" on the side, and have the rocket carrying it blow up half way to orbit.

      Thereby saving hundreds of millions of dollars which can be funnelled to the war in Iraq.
    • The real problem is that the Webb scope (NGST) does not replace HST. It is not a visible light instrument. NASA does not have the funds to keep HST going. No real surprise, it's expensive.

      We keep putting up very expensive things: HST, the Shuttle, ISS - it seems that a few expensive things are easier to sell than a bunch of cheap things.

      It costs about $250 million per service flight. Instead of servicing Hubble, send up a new scope every five years or so. How much scope can you get for $250 million?

  • by Anonymous Coward
    Perhaps people finally just launched hilary rosen into space?
  • The new moons, uncovered by NASA's Hubble Space Telescope, are about 8 to 10 miles across (12 to 16 km) -- about the size of San Francisco.

    The two new satellites are the first inner moons of Uranus discovered from an Earth-based telescope in more than 50 years. "It's a testament to how much our Earth-based instruments have improved in 20 plus years that we can now see such faint objects 1.7 billion miles (2.8 billion km) away," says Mark Showalter.

    Is Hubble considered an Earth-based telescope someho
    • Well, since it is made up of components and materials FROM earth then it could be considered earth based for the sake of the artical. However, I think it should not be, it is in orbit is it not? Dfn of an object in orbet is? (http://space.about.com/library/glossary/bldeforbi t.htm)
    • ...first inner moons of Uranus discovered from an Earth-based telescope in more than 50 years.

      Is Hubble considered an Earth-based telescope somehow? I'm kind of confused. Can anyone explain this?

      I think this is referring to that fact that Hubble is bound to the Earth ie. not mobile and flying through the solar system like the Voyager probe etc.

      • So a telescope on the Moon would also be Earth-based? That must be what the conspiracy theorists mean when they say that the US hasn't been to the Moon...
        • So a telescope on the Moon would also be Earth-based? That must be what the conspiracy theorists mean when they say that the US hasn't been to the Moon...

          Can you provide some evidence for the moon landing that does not rely on the honesty of a government that has a strong history of lying to suit its purposes?

  • Last I checked... (Score:3, Informative)

    by andreMA ( 643885 ) on Saturday September 27, 2003 @05:11AM (#7071063)
    Hubble was an orbital instrument, not "Earth-based"
    "It's a testament to how much our Earth-based instruments have improved in 20 plus years that we can now see such faint objects 1.7 billion miles (2.8 billion km) away," says Mark Showalter, a senior research associate at Stanford University.
    Or did it re-enter the atmosphere and I missed it?
    • ...land-based telescopes would be looking through a dirty window (the atmosphere), orbiting earth is like sticking your nose out the door, while the alternative is to send a probe which would be to walk over and examine it. The point is, we're observing this "from home", which is Earth. So I don't see a big problem with that statement.

      Kjella
  • Pardon my ignorance, but I thought that Jupiter and Saturn were the only true Gas Giants, and Uranus wasn't (irrespective of composition). Then again, I am not an Astronomer, so I could be wrong here. This seems pretty impressive for the Hubble, who knows whether it would have found them earlier if the optics had worked from day one?
  • It seems that discovering moons is all tthe rage now...

    Late-Working NASA Scientists Discover Moons Over My Hammy! [augustana.edu]

    It appears that the onion no longer has it in it's archives. Bummer.

  • Those are not moons those are...
  • San Francisco? Must be like that metric/english unit Mars screwup - I thought all orbiting objects and ice flows were measured in "Rhode Islands".
  • no no those weren't moons, those were klingons. capt. kirk will be there shortly to pick them up.
  • Possibilities (Score:4, Interesting)

    by SkArcher ( 676201 ) on Saturday September 27, 2003 @05:41AM (#7071132) Journal
    The small moons are probably fragments knocked off one of the larger moons and may be in descending spiral orbits. If they are permanent, then I would suspect that they would be in some sort of synchonicity with the larger moons, using the gravitation to help maintain a stable orbit (if they were not synchroeous they would be unstable, IIRC)
  • There are entirly too many moons. In the case of Uranus we will soon need more female Shakspearian characters to name them after.
  • by adeyadey ( 678765 ) on Saturday September 27, 2003 @07:11AM (#7071328) Journal

    The story is also on space.com [space.com]. they also have a article showing how to find Uranus [space.com] in the sky - it is quite close to Mars at the moment. Perhaps we should start calling it the 6th planet at /. just to avoid tedious jokes..

    • Or we could just wait until 2620, when Uranus will be renamed to end that stupid joke once and for all.
    • Here is a trivia question: How many planets are visible without a telescope? Most people will answer "five" (Mercury, Venus, Mars, Jupiter and Saturn). But if you answered "six," congratulations, you can go to the head of the class!

      -http://www.space.com/scienceastronomy/uranus_moon s_030925.html

      Perhaps we should start calling it the 6th planet at /. just to avoid tedious jokes

      The other comments almost have it right. The space.com article isn't counting Earth as a planet visible without a telesceo

    • Perhaps we should start calling it the 6th planet at /. just to avoid tedious jokes..

      Yeah, THAT'D be a good idea. There aren't enough factual inconsistencies on slashdot to start.

      The planets are:

      Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto

      (though Neptune and Pluto switch sometimes)

      Making Uranus the 7th planet
  • by Sri Lumpa ( 147664 ) on Saturday September 27, 2003 @07:18AM (#7071349) Homepage

    It's not a moon system, these are Haemorroids.
  • by cca93014 ( 466820 )
    The new moons, uncovered by NASA's Hubble Space Telescope, are about 8 to 10 miles across (12 to 16 km) -- about the size of San Francisco.


    Aha! But are they polyamorous out of work web designers?

  • Dang frat boys

    That's because I wear boxers now

    You oughta see my wife's

    Just what Uranus needs another moon

    This is what I spent 87B USD on?

  • That's no moon.

    [Fired. -Ed]
  • Hubble is hardly Earth-based... tell that to the astronauts that routinely have to service it at amazing price and effort!
  • by NanoGator ( 522640 ) on Sunday September 28, 2003 @02:21AM (#7076120) Homepage Journal
    .. if they had just found a way to use the word Klingon in it.

Every nonzero finite dimensional inner product space has an orthonormal basis. It makes sense, when you don't think about it.

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