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Posers and Puzzles

Posers and Puzzles

  1. Subscriber sonhouse
    Fast and Curious
    10 Dec '09 13:43
    Just found by a new optical technique, Alcor itself is about 80 LY away and masses about 2X that of our sun, which is about 2E30 Kg. So 4E30 Kg for Alcor. The newly discovered red dwarf is about 1/4th the mass of the sun so it is about 5 E29 Kg. The article in physics.org mentions it orbits around Alcor in a 90 year orbit.
    Anyone up on celestial navigation? How can you calculate the separation of the two stars given those three figures, Alcor A mass 4E30 Kg, Alcor B mass 5E29 Kg and its orbital period is 90 years. Any takers?
  2. Standard member wolfgang59
    Infidel
    10 Dec '09 18:19
    There must have been some estimation of the size of the orbit surely?
  3. Subscriber sonhouse
    Fast and Curious
    11 Dec '09 13:38 / 6 edits
    Originally posted by wolfgang59
    There must have been some estimation of the size of the orbit surely?
    I'm thinking those three numbers can lead to a solution but they did not mention the size of the orbit, radius or diameter. I think a solution would relate the orbital period to the escape velocity of Alcor A. In any orbital the escape velocity is 1.414 (sqr root of 2) times the orbital velocity, but not sure how to apply that in this case. If the orbit is say, at 16 billion Km (ten billion miles) you get one answer but if it is at 1.6 billion Km you get another. I think we can relate it to our suns' orbitals, since the mass of the sun is half that of Alcor, it should be easy to equate what a 90 year orbit would be in our solar system and then double the mass and recompute for Alcor. Seems kind of cheating though.

    I just did a bit of google and found the orbital period of Uranus is 84.1 years, close to Alcor's newly found companion. I guess I could work it out from there.
    So Uranus orbits at 6729.3 meters/sec, escape velocity of Uranus from Sol is 1.414 times that or 9515.3 meters/second. The sun is about 2E30 Kg. That's a start anyway.

    One Wiki piece says the orbital period is 84.323326 years or 30,799.095 days which works out to 2,661,041,808 seconds. Uranus maxes out at 3 E9 Km and mins at 2.7 E9 Km, so I used a # halfway between to get a halfasssed assessment of the average distance, used 2.85 E9 Km as distance. Sort of making an attempt at simplifying to a circular orbit.

    The formula for orbital velocity is V=(GM/r)^0.5 G=grav. constant, M=mass, r=radius. So solving for R, = GM/V^2. So The mass of the dwarf doesn't matter, just the mass of the primary. Seems a bit of a conundrum, if I knew the radius I could figure the velocity, if I don't know the radius, I don't know the velocity.
  4. Subscriber AThousandYoung
    It's about respect
    17 Dec '09 10:58
    Is this evidence that the Dipper is really a three dimensional Dipper? It has a near and a far corner right?

    Which star in the Dipper are we talking about? Handle or Dipper?
  5. Subscriber sonhouse
    Fast and Curious
    17 Dec '09 12:24
    Originally posted by AThousandYoung
    Is this evidence that the Dipper is really a three dimensional Dipper? It has a near and a far corner right?

    Which star in the Dipper are we talking about? Handle or Dipper?
    Here is a link to a piece about this star, its in the bend of the handle:

    http://www.sciencecodex.com/a_faint_star_orbiting_the_big_dippers_alcor_discovered

    So what about my original question? Do we have to resort to studying our own sun and the planets in it to determine the distance to the new star? It seems to be around the distance to Uranus but would like to calc the actual distance.
  6. Subscriber AThousandYoung
    It's about respect
    17 Dec '09 13:29
    Newton's Theorem of Revolving Orbits might solve this but I'm too lazy to try to work it out at this time.
  7. 19 Dec '09 21:42
    The accelleration from gravity needs to equal the centrifugal accelleration, so (assuming we can consider the larger star stationary):

    G * m1 / r^2 = v^2 / r

    We know G and m1 (mass of the LARGER star), but not v and r. We know the revolution time, though (let's call it T), so we express v in terms of r and T (assuming circular orbit):

    v * T = 2 * pi * r -> v = 2 * pi * r / T

    Substituting that for v in the first equation we get:

    G * m1 / r^2 = 4 * pi^2 * t^2 * r

    which we rearrange into:

    r^3 = G * m1 * t^2 / (4 * pi^2)

    setting T at 90 years and m1 at 4E30 kg I get r = 3.8E12 m which is about 25 AU (distance between sun and earth).

    I'm sorry if this is hard to read, I'm not sure how to do equations in text. I'm pretty sure the result is correct, though.
  8. Subscriber sonhouse
    Fast and Curious
    21 Dec '09 20:57
    Originally posted by zzyw
    The accelleration from gravity needs to equal the centrifugal accelleration, so (assuming we can consider the larger star stationary):

    G * m1 / r^2 = v^2 / r

    We know G and m1 (mass of the LARGER star), but not v and r. We know the revolution time, though (let's call it T), so we express v in terms of r and T (assuming circular orbit):

    v * T = 2 * pi * ...[text shortened]... ad, I'm not sure how to do equations in text. I'm pretty sure the result is correct, though.
    Sounds good but the distance between Earth and the sun is 1 AU by definition. 25 AU is about 2.3 billion miles or 3.7 billion kilometers. Sounds about right. Wasn't sure if the orbital period and the two masses were enough data to get the radius. Thanks. I'll go over your math and learn something, eh!
    It really sucks to want to do math when all you have is regular typing.