Slashdot Mirror
Kuiper Object Discoveries Formally Announced
ewhac writes "The San Francisco Chronicle is reporting that the new Trans-Neptunian objects mentioned in the press earlier this year are being formally announced this week at a planetary conference in Cambridge, England. Bearing the extremely temporary names 'Xena,' 'Santa,' and 'Easterbunny,' the new objects are quite interesting in their own right (Santa is cigar-shaped, rotates end-over-end every four hours, and has a 60-mile-diameter moon). However, even more interesting is the intrigue behind the press conferences revealing Xena earlier this year. It seems that, using the astronomers' own observation logs (publicly available over the Web) and some key details inadvertently revealed in earlier announcements, someone was planning on 'discovering' the objects first and claiming credit. This was why the scientists 'pre-announced' the existence of Xena back in July, to establish priority. The conference in Cambridge represents the first formal, scientific disclosure of the objects."
(Santa is cigar-shaped, rotates end-over-end every four hours, and has a 60-mile-diameter moon)
Ack! It's the cheesy alien probe from Star Trek IV!
I don't have access to the necessary data and my training in this area is thin, but one possibility that springs to mind is that the object has too much angular momentum. I'm sure every slashdotter knows that the Earth is slightly flattened by its rotation. As you add more angular momentum you normally expect the object to just flatten more and more as it spins faster and faster. It turns out that after a certain point the body will be more stable as a tumbling elongated shape than a fast spinning disc. Continue to increase the angular momentum and the body will ultimately separate in to two.
Now, this won't result in a perfect cigar shape - especially the high length to width ratio and straight sides - so another theory may be necessary, depending on the data. This is what sprang in to my mind when they mentioned it, though.
It's not that the new planet is brighter than Pluto, it's that it's brighter than a snowball at the same size as Pluto and the same distance as the new planet.
At the distances the planets are from us, both objects look like specks. They will probably be larger on something like Gemini but there won't be a difference in sizes due to the distance.
In addition, we already know how far away the objects are due to measurements of it's position. E.g. once you get a few observations you can plug that information into Kepler's equations and get an orbit from it. Once we know that the object is further away from pluto and still brighter, we can figure out that the object has to be either larger than pluto or a perfect mirror. One of the pages gives various size estimates based on reflectivity.
"When you sit with a nice girl for two hours, it seems like two minutes. When you sit on a hot stove for two minutes, it
Given:
From this we can calculate the brightness of a perfect mirror the size of Pluto if it were in the new object's orbit.
From observations we know that the object is almost as bright as a Pluto-sized mirror would be at this distance.
Thus, the smallest the object can be is 97% the size of Pluto. Since the object cannot be a perfect mirror, it is bigger than Pluto.
Likewise, the reflectivity of other substances can be tried. If the object is made of snow (90% reflectivity) it will be 2% larger Pluto, and if the object has the same composition as Pluto it will be 25% larger than Pluto.
Show me on the doll where his noodly appendage touched you.
Dont be dense.
This planet is billions of km away, and only a few 1000km in diameter.
Its size when viewed from the earth is MUCH lower than the seeing from the athmosphere. In fact its so small that even the spitzer space telescope couldnt resolve it as anything more than a point.
So you have a pointsource.
brightness of the point= (light from planet)/(distance from earth)^2
light from planet=light recieved from sun*albedo
light recieved from sun= constant*(area of planet disc)/(distance from sun)^2
-> brightness oft the point= albedo*solar constant*(radius of planet)^2*pi/(distance from sun*distance from earth)^2
You know the solar constant, you know the distances, and you know that the albedo cannot bigger than 1 (perfect lambertian reflection).
If you just meassure the light recieved from the point, you have only albedo and radius left, which allows a minimum size estimate)
HI O WISE PRINCE. WHT TOOK U SO DAM LONG?
Diameter = 1329/sqrt(p) * 10^(-0.2*H),
where p = albedo and H = absolute magnitude (-1.2 in the case of 2003 UB313)
Albedo (reflectivity) can be between 0 (no light is reflected) and 1 (all light is reflected). Pluto's albedo is 0.6.
So, if 2003 UB313 has the same reflectivity as Pluto, it would be about 3000 diameter. If it is somewhat brighter (albedo = 0.8) its diameter would still be 2600 km. If p = 1.0 it would be about the size of Pluto. In the unlikely case that it is very dark, it would be far larger than Pluto.