Blackhole at the center of each galaxy, think again

Discussion in 'Astronomy, Exobiology, & Cosmology' started by Anomalous, Jun 7, 2005.

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  1. Anomalous Banned Banned

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    If there a massive blackhole at the center of each galaxy then; near the center of the galaxy, stars should be revolving the BH at very high speeds, hence we should be able to note the revolutions in months, days or even hours.

    And we should also be able to see some of the stars being hidden behind the blackhole. If all these thing happen then yes there must be a blackhole at the center of each galaxy.
     
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  3. Novacane Registered Senior Member

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    O.K. What's next?

    Novacane
     
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  5. Anomalous Banned Banned

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    You missed it. You were supposed to tell me if that is what is observed, ie. if you have any such information.
     
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  7. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    I am far from an expert in these questons, but it is my understanding that one can not see any stars near the postulated BH because there is so much dust there also. Perhaps radio astonomy can see something. The stars that are observable and relatively centrally located in our galaxy (and many others) do have high orbital speeds - They are in orbit about a collection of mass that is much greater than the observed dust and presumbed stellar density, but this does not prove that this collection of mass is a small (zero) volume of space that the standard theory of BHs predicts.
     
  8. Dinosaur Rational Skeptic Valued Senior Member

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    I have read about observations of at least a few galaxies with motion near the core that strongly supports the existence of a black hole. I also think that quasars are best explained by black holes at their core.

    I do not know if it is believed that there are black holes in many or most galaxies. I am reasonably sure that there is observational evidence for black holes at the center of only a small percentage of the observed galaxies. The belief that lots (?percentage) of galaxies have central black holes is a theory based on the observation of a few such galaxies.
     
  9. superluminal I am MalcomR Valued Senior Member

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  10. Pete It's not rocket surgery Registered Senior Member

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    I have an unreliable recollection of exactly such an observation in our galaxy.

    I really don't think so. The black hole isn't large enough, and stars aren't packed closely enough for an alignment to happen at all often. If an alignment did occur and was observerd, it wouldn't necessary hide the star, either. I'm not sure the black hole is larger than a star to begin with, and there's also lensing to consider.
     
  11. superluminal I am MalcomR Valued Senior Member

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    Pete,

    Not unreliable.
     
  12. Pete It's not rocket surgery Registered Senior Member

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    Thanks super!

    The article I was thinking of was actually this one in Nature:
    A star in a 15.2-year orbit around the supermassive black hole at the centre of the Milky Way

    Note the orbit time - 15 years. More than what Anomalous expected... but still impressive. It's about half the period of Saturn's orbit around the Sun. The orbit radius is about 17 light hours, or about 250 au. That's damn close in interstellar terms (further than the Kuiper belt, closer than the Oort cloud).

    I wonder how close to the hole a star would have to be to have a period of hours? And how stable would that orbit be?
     
    Last edited: Jun 9, 2005
  13. superluminal I am MalcomR Valued Senior Member

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    Pete,

    If we assume a 1M solar mass hole, and a 1 solar mass star, then we can use

    Vorb = sqrt(GM/r)

    G = 6.67300e-11
    M = 1.98892e36 kilograms

    with r = 2e11 meters (200 million km)

    Vorb = 2.58e7m/s yielding a period of 13.6 hours.

    (EDIT: c = 300e6m/s. The orbit above is almost 0.1c!!! [0.086c])

    The centripetal acceleration experienced by the star is

    a = v^2/r = 2.58e7^2/2e11 = 3328m/s^2 = 340g's

    Will this tear the star apart by tidal forces? Probably.

    EDIT: Actually, the tidal acceleration (differential acceleration) across a star of 1e6 km in this orbit looks like approx. 33m/s or about 3.3g. So the star will probably just be streched out a bit.
     
    Last edited: Jun 9, 2005
  14. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Pete: You are very knowledgable and almost always very correct. What also impresses me is that you are willing to grind thru the numbers. (I have grown much too lazy for that in retirement.) Thus, I am surprized that you are asking this question.

    Surely you know, as Newton showed long ago, that the square of the orbit period, P, is proportional to the cube of the orbit radius. This remains true for elliptical orbits but the "radius" is the semi-major axis SMA. Or mathematically if the circle's R = SMA, then the periods are equal. That is, the periods of an eliptical orbit that could be just circumscribed inside a circular orbit are identical. Note the "circumscribable eliptic orbit" crosses the circular as the sun (or gravity center to be more general) is at one of the focii of the eliptical orbit, not the center of the elipse. (To show that it is "circumscribable" one would conceptually rigidly slide the elipse to put the sun at the center of the elipse. (Some earlier versions of this post were not clear and implied that the eliptical orbit WAS circumscribed inside the circular one)

    The distance traveled around such an eliptic orbit inscribed inside a circular one is of course less than the circumference of the circular one. This lesser travel distance in the eliptic orbit is exactly commenstated by the fact that in the region near apogee, the object in the eliptic orbit has lower velocity. The higher velocity it has near perigee can not "buy back" the time lost when it was traveling slow - this is why both orbits have exactly the same period despite their different travel distances. (If I am remembering my orbital mechnics with error, please someone correct me.)

    Thus, (to partially answer your question - give radius for a one year period orbit):

    15x15/(250x250x250) must equal 1x1/(R^3) where R is the radius of circular orbit (or the semi major axis of an eliptical one).

    Hence R = 250 / 225^1/3. The cube root of 15X15 OR 225 is a little more than 6, I'll call it 6.12 - thus R is approximately 250/6.12 or 40.8AU

    Only out of great respect for you do I do this much calculation!

    I let you do it again with hours. We know now, assuming a year is 8766 hours, that

    8766x8766/(40.8x40.8x40.8) = t x t/(r^3) where t is the number of hours you want to select to calculate the corresponding orbit radius (I made it r, not R, as it will be less that the 40.8AU I calculated for the one year orbit.)

    As far as "is it stable?" that depends on a lot of things. I bet that the one year orbit is, if it is roughly circular and the star is not too big, I.e. sun like. You will need to explore (mainly) the gravity and gravity gradient of the black hole at the star surface to answer this question, which is more complex.

    As you bring the BH in closer to the star, the gravity gradient of the BH will distort the star from a sphere (just as moon makes tides on Earth). When the BH's gravity at the top of the stellar tidal bulge is greater than the star's gravitation hold on that part of the star, that part of the star will no longer be bound to the star's core. It will initialy try to orbit the BH but there will be a lot of internal momentum transfer between varrious pieces of the star being ripped away from tha star.

    Net effect is that some parts stolen from the star will be both trying to fall into the BH and to orbit it. (I.e. moving in a death spiral towards the BH.) As these pieces compress in this spiral and collide, they will get very much hotter than they were on the stellar surface. That is, a quasar is forming.

    I find it a little hard to accept, but believe that this great compression and collisional heating produces such copious radiation that a significant fraction of the mass being "eaten by the BH" will be converted into radiant energy (mainly gamma rays by E=Mc^2) before it disappears inside the event horizon of the black hole. The fraction of the mass converted in to radiation by this quasar process is much greater than the mass fraction converted by fusion in the star. I don't know it for fact, and it surely varries with how long the fusion process has been going on and size/ separation of BH and star when the quasar process starts, but I bet that in some cases, more mass is converted into radiation by the quasar process than the star converted into radiation in its entire lifetime!
     
    Last edited by a moderator: Jun 10, 2005
  15. Anomalous Banned Banned

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    Great Points Pete.

    Thats an interesting question, It may be possible that the reason why there are no stars at hours orbit may be because as U said , it may be too unstable and the star would be eaten by the BH.

    It is said that if U fall in a BH then the difference of G between your head and feet will be so great that U will be streched into a rope. The stars themselves have too powerful G so its quite possible that with a small step closer the star will be torn apart
     
  16. Dinosaur Rational Skeptic Valued Senior Member

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    We normally think about black holes containing a lot of mass.

    If an Earth Mass were compressed into a black hole, the gravitational field 4000 miles (about 6400 km) from the center would be the same as Earth surface gravity. Around such a black hole there would surely be some distance at which an object would orbit in an hour. There would be no fierce tidal effects on small objects until you got fairly close to the event horizon of such a light weight black hole.

    For any size black hole, there is a distance at which a ray of light would orbit in a circle. Inside that distance, I do not think that there are any stable orbits. Many years ago, I read an article in Scientific American which stated that inside this critical distance, rotational effects added to the gravitational force due to the black hole. I think they were saying that once inside this critical distance, an object falling directly toward the center of the black hole would fall faster if you applied a force orthogonal to the velocity vector.
     
  17. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    Everything you state is essentially correct, but one part may be missleading. At the event horizon of a big massive black hole, the "tidal forces" as you call it (more correctly the "gravitational gradient") is much less that the gravity gradient at the event horizon of a small mass BH. The gradient falls off as the cube of the distance from the black hole. The small BH's EH is closer to the BH and thus the gradient has not decresed as much (the cubic fall off is more important factor than the mass diff.) If you were at the event horizon of a really big BH, there would be essentially no tidal force acting on your body to rip it apart. That is, you would disappear from our universe before you were streched spaghetti like, but you would be "spagetti like" before you disappeared, if the BH is smaller (but obviously not too small). Near a very small BH, there is very hot "Hawking radiation" - your spaghettii would be cooked before streched much! :bugeye:
     
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  18. superluminal I am MalcomR Valued Senior Member

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    Another interesting thing about black holes is that time as seen at the EH from far away from the hole comes to a stop. Of course from the perspective of the guy at the EH nothing is wrong and he proceeds on to destruction.
     
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  20. superluminal I am MalcomR Valued Senior Member

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    Hi!

    Nope. It's just a bit of theoretical playtime with one BH and one normal star.
     
  21. Pete It's not rocket surgery Registered Senior Member

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    I think you could pretty happily ignore other stars for most purposes. Like you can mostly ignore Jupiter when figuring Earth's orbit around the Sun - Jupiter's effect is only significant over many orbits.
     
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  23. Pete It's not rocket surgery Registered Senior Member

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    If they're close enough, sure.

    How big do you picture the centre of the galaxy to be?
     
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