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.'"

Comments

Several Moons

[NeoGeo64]

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. ;)

Okay, here's the concept in plain English

[MickLinux]

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

Re:What Makes a Moon a Moon?

[bpd1069]

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...