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Orbital period of a star in the Milky Way?

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  1. Apr 22, 2013 #1
    So here's a question I'm struggling with:
    The rotation speed of the sun around the Milky Way center is 220 km/s and it takes the sun around 200 million years to orbit once around center of the galaxy.
    Given that the rotation curve is relatively flat (i.e. the rotation speed stays the same as function of distance), how long does it take to make one full orbit for a star that is two times further from Milky Way center than the Sun?

    I've tried many things and can't seem to find an answer that fits. I used the formula for orbital period (T): T = 2∏√(r^3/GM) where r is the distance of the star from the center of the milky way (supposedly a supermassive black hole), G is the gravitational constant (6.67*10^-11 m^3/kgs^2), and M is the mass of the supermassive black hole, Sagittarius A (8.2*10^36 kg), around which all stars in our galaxy rotate.
    If r is 56,000 ly (twice the distance from our Sun to the center of the galaxy which is 28,000 ly) which is 5.298*10^20 m, this all calculates to be 3.27*10^18 seconds. This is 1.036222602924*10^-7 years... which seems WAY too small compared to the given orbital period of our Sun around the center of the galaxy, which is 200 million years. Where did I go wrong?
     
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  3. Apr 22, 2013 #2
    Think dark matter.

    The orbital period of stars further from the galactic center should be slower the further you go out. However the orbital periods are roughly the same regardless of distance from the center.

    This lead to the search for the missing mass or dark matter. Essentialy you don't know the correct mass for your calculations. Even if dark matter was not a factor. Your only using Sagittarious A. Theres far more mass than just the BH at the center of the galaxy to account for.
     
    Last edited: Apr 22, 2013
  4. Apr 22, 2013 #3
    While not the best of quality these links are they do show the breakdown between Keplers Laws and the rotational period of a star and our galaxy

    http://physics.uoregon.edu/~jimbrau/astr123/notes/chapter23.html [Broken]

    see the section on keplers laws.

    This one has a decent graph.

    https://www.e-education.psu.edu/astro801/content/l8_p8.html
     
    Last edited by a moderator: May 6, 2017
  5. Apr 22, 2013 #4
    So are you saying that the orbital period of a star twice as far away from the galactic center as the sun would also be roughly 200 million years?

    Because Kepler's law that says the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. So if the semi-major axis, a, is twice that of our Sun's, it would be P ∝ √(2a^3) which is about 2.83 times the orbital period of the Sun or 566 million years.
     
  6. Apr 22, 2013 #5
    No I am saying that Keplers laws breaks down on galactic scales due to the dark matter halo. Keplers laws predict a slower rate than observation due to dark matter.
    I misimplied that it would be the same but it is faster than what Keplers kaws would predict. See graph second link above.
     
  7. Apr 22, 2013 #6

    marcus

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    I would think it would be twice whatever the sun's period is. If sun's is 200 My then roughly speakibg the other star's would be 400 My.

    If sun's period is 225 My which I recall from many years back in astro class, then the other stars would be roughly 450 My.

    The approximate flatness of the rotation curve means that the SPEED does not fall off as rapidly as you'd expect. But the circumference or length of orbit scales up with the radius, so is twice as long. As a very rough approximation, imagine that the speed stays constant out that far. That is the crude assumption I'm using.
     
  8. Apr 22, 2013 #7
    Finally found the link to an article that will help. For further references google milky way rotation curves.

    Anyways here is one such reference.
    http://ircamera.as.arizona.edu/astr_250/Lectures/Lec_22sml.htm
    O
    according to the graph a star twice the distance from galactic center is roughly 230 My

    here is a technical paper showing how its calculated however it involves the mass density distribution due the gas, dark matter and star planets etc. Each type of galaxy will have its own forms of a power law index.
     
    Last edited: Apr 22, 2013
  9. Apr 23, 2013 #8

    marcus

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    ?

    There is a graph in those lecture_22 notes that does not show the star's period is 230 My.
    It shows the star's speed is roughly 230 km/s

    That is roughly the same as the sun's 220 km/s

    But the length of its orbit is twice that of the sun's, so its orbital period will be roughly twice.

    I can't figure out what you are doing, unless it could be not reading the vertical axis of the graph correctly.

    ?
     
  10. Apr 23, 2013 #9
    Lol your right I jumped ahead one shouldn't be in a hurry working from a phone oops.lol.

    Anyways I should have typed radial velocity of 230 km/s you know the radius rest is simple math involving its circumference. Sorry misread it and thought it had done so on the graph

    looks like your memory above worked Marcus lol
     
  11. Apr 23, 2013 #10

    Chalnoth

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    Just for a sense of scale, Sagittarius A is only around 0.1%-0.2% the mass of the stars in the Milky Way.
     
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