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Non-Spherical Stars

An anonymous reader writes "Now that the large interferometers are coming on line, the stars are no longer dots. Achernar (Alpha Eridani), is a huge ellipsoid whose polar radius (due to fast spinning) is 50% smaller than the equatorial one!"

14 of 70 comments (clear)

  1. Press release here: by molo · · Score: 5, Informative

    More details at the press release:

    http://www.eso.org/outreach/press-rel/pr-2003/pr-1 4-03.html

    Including more technical drawings.

    -molo

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  2. Energy output by VendingMenace · · Score: 4, Interesting

    50% smaller? Wow, this must be spinning incredibly fast. With so must mass being displaced from where it would be in a sphere, it must effect the pressure inside the star. As such, i wonder how much this effects the fusion within the star. Since fusion is driven by the compressional forces of the suns mass, the effective reduced mass must reduce the energy output of this star. RIght?

    Perhps i don't really kow what i am talking about.

  3. Cosmic Rugby by follower_of_christ · · Score: 3, Funny

    Hey God!

    I found your rubgy ball!

  4. Amateur Astronomy by Anonymous Coward · · Score: 3, Funny

    Nicole Kidman or Gwyneth Paltrow are the flattest stars that can be seen with the naked eye or possibly binoculars.

  5. I crush you... by FroMan · · Score: 5, Funny

    You know when you take your index finger and thumb and look at something pretty far away. Then you squish them till they touch.

    I think someone was doing that at the end of the telescope.

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  6. This has been done before. by mph · · Score: 4, Informative

    The oblateness of Altair was measured using the Palomar Testbed Interferometer (PTI) in 1999-2000.

  7. our sun, the planets by kamukwam · · Score: 4, Insightful

    I really don't think the fact that the star isn't a perfect sphere is surprising. The fact that we can measure it is a breakthrough. If we look at the sun, we can see that isn't a perfect sphere. It's not very much an ellipsoid either, but you could imagine a star (much younger) that spins very fast and is more like an ellipsoid. Even Jupiter (and also the earth!) are somewhat flattened.

  8. The Very Large Telescope Interferometer by jwachter · · Score: 5, Informative

    This site describes the telescopes that comprise the interferometer used to make the observations:
    http://www.eso.org/projects/vlti/

    Quote:
    The Very Large Telescope Interferometer (VLTI) consists in the coherent combination of the four VLT Unit Telescopes and of several moveable 1.8m Auxiliary Telescopes. Once fully operational, the VLTI will provide both a high sensitivity as well as milli-arcsec angular resolution provided by baselines of up to 200m length.

  9. Who writes these articles? Or am I iggernint? by Atario · · Score: 3, Interesting

    Due to its daily rotation, the solid Earth is slightly flattened...

    Solid Earth? Only the surface (and part of the core) is solid, right? The rest is [Dr. Evil] liquid hot magma.

    The observed flattening cannot be reproduced by the "Roche-model" that implies solid-body rotation and mass concentration at the center of the star.

    I thought stars were pretty much all plasma, which is to say, a fluid. Why, therefore, should stars obey any "solid-body" rule at all?

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  10. Re:Who writes these articles? Or am I iggernint? by mwtown · · Score: 3, Informative

    Actually, I think they mean "solid-body" as "cohesive object" in this case.

    While I'm getting technical, Plasma can't be considered a fluid either, as it's not a liquid, it's a different state of matter altogether.

  11. Theoretical maximum for common stellar materials? by Raindance · · Score: 3, Interesting

    Any physics buffs know what the largest theoretical ratio would be between a star's polar radius and equatorial radius, for the stuff that stars are made out of? Is the ratio for this star anywhere close to that?

    I'd imagine one can only attain this through centrifugal force, which necessarily puts structural stress on the star, and past a certain amount of structural stress stars should disintegrate.

  12. Re:Theoretical maximum for common stellar material by CheshireCatCO · · Score: 4, Interesting

    I believe a star has zero tensile strength* (it's just a fluid), so once you're spinning too fast for gravity to hold you together, it's bye-bye time.

    The better question is this: how did that star form? If it was spinning too fast to hold together, how did it accrete matter with that much angular momentum at all?

    * Barring magnetic fields, mind you. But you'd need an ass-kicking field to hold a star together very long, I would think.

  13. Re:Who writes these articles? Or am I iggernint? by X-rated+Ouroboros · · Score: 4, Informative

    How is a plasma not a fluid?
    because it's plasma!

    This exchange is about on par with "How is a liquid not a fluid?" "Because it's a liquid."
    "Fluid" is not a state of matter, no one's claiming it's a state of matter, saying plasma can't be a fluid because plasma is the 4th state of matter is a category error. Liquid is the second state of matter. Gas is the third state of matter. Both are fluids.

    A fluid is any substance which undergoes continuous deformation when subjected to a shear stress. The problem we're probably having is that the obvious sources for the shear stresses in the couse of, say, water being poured from a cup (normal force of the side of the cup vs gravity) are paralleled for the case of plasma by electromagnetic feilds. It just don't grok intuitively but, plasma behaves like a fluid... ergo, it is a fluid.

    --
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  14. yes, it does affect luminosity of the star by Doctor+Fishboy · · Score: 3, Informative

    In the more recent surveys of bright stars in a cluster, they've seen that faster rotating stars (seen indirectly by the rotational broadening of spectral lines of the star) of the same spectral type have a wider scatter of observed brightnesses. The explanation for this is that:

    (i) Faster rotating stars are brighter at their poles than their equators (because of centripetal force slightly expanding the distance of the equator from the core of the star), and:

    (ii) The spin axes of stars are randomly oriented with respect to telescopes on Earth, so:

    (iii) For a large sample of fast rotating stars, you sample all the brightnesses from the equator to the poles, hence a large scatter in measured brightness. You can assume that all stars are effectively at the same distance if they are in a distant cluster.

    Hope that's reasonably clear,

    Dr Fish