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New Moon System Around Uranus
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.'"
"Moon System Around Uranus"? Why even bother thinking of something funny? :/
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?
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
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".
----
Squirrel
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*
"A door is what a dog is perpetually on the wrong side of" - Ogden Nash
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?
Turkeyphant
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.
...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.
www.rexguo.com - Technologist + Designer
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 somehow? I'm kind of confused. Can anyone explain this?
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?
InfoSec that matters, when it counts.
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)
An infinite number of monkeys will eventually come up with the complete works of
There are entirly too many moons. In the case of Uranus we will soon need more female Shakspearian characters to name them after.
"Most interesting how often you humans seem to obtain that which you do not want" -Spock
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, ideal for initial value problems, yields exact functional solutions, involves simple algebra [yes, that's right: simple algebra solutions to almost any set of partial differential equations] and turns out doubling precision for every iteration.
Oh, yes: there is a version out for Maple, too.
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, and asteroid motion. It works.
[PS: That last link has code for you code monkeys]
Correct Horse Battery Staple: 72 bits of entropy. Enter "Correct H" into google. When it generates the phrase, that's
The story is also on space.com. they also have a article showing how to find Uranus 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..
"You lied to me! There is a Swansea!"
It's not a moon system, these are Haemorroids.
"The obvious mathematical breakthrough would be development of an easy way to factor large prime numbers." Bill Gates,
I downloaded the "incredibly easy, conceptually simple" description in the PDF file.
Now I remember why I switched to computer science. *sigh*
GCHQ Quantum Insert installed. If only our tongues were made of glass, how much more careful we would be when we speak
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?
GCHQ Quantum Insert installed. If only our tongues were made of glass, how much more careful we would be when we speak
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
Correct Horse Battery Staple: 72 bits of entropy. Enter "Correct H" into google. When it generates the phrase, that's
The time on the Hubble is FREE from STSI (Space Telescope Science Institute, in Baltimore MD) NASA pays them to operate HST, as a scientist you just have to submit a good experiment to STSI, get approved and wait in line for years for your observation. If the HST is down or croaks during your observation window, too bad, you go back to the end of the line and wait. Hubble will soon be upgraded (SM4) to have better "eyes" when the Shuttle is back up. I have worked some on the SM4 software, its a LOT of very very old C code on an old 286 processor, plus some one of a kind processors from 25 yrs ago that no one else but NASA uses. Several of the instruments have NO software. But it gets the job done. If Uranas has the two Moons, I would have thought HST would have found them by now as they DO look at the planets and not just the neato nifty nebulas and stuff sveral Sagans ("billions and billions") of light years away
JWST (James Web Space Telescope) is just now reaching the requirements definition phase, expect 7-10 yrs before launch. It will be stationed at L-4 or L-2 (the Earth Sun Lagrange Points where the gravity pull of the Earth and Sun are equal, so the object kinda just sits there with very litte station keeping needed) which is a LONG way from Earth, so the JWST better work right the FIRST time cuz there ain't no way to get there to put glasses on it!
Any other questions?
...and rename it to Urektum to stop all those stupid jokes. Oh, wait...
Kjella
Live today, because you never know what tomorrow brings
.. if they had just found a way to use the word Klingon in it.
"Derp de derp."