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Tangential velocity

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data

    I am trying to find the tangential velocity of a star but I am confused with the whole procedure.

    2. Relevant equations

    They give me the following data:
    ⋅Distance: 32parsecs
    ⋅μ: 0.24 arc seconds per year

    I have a section explaining proper motion that states the following formula:

    Vt= μ⋅d
    where Vt is the tangential velocity expressed in SI units, μ the proper motion in arc seconds per year expressed in radiants per second, and d the distance expressed in SI units.


    3. The attempt at a solution

    d= 32pc = 9.87×1017metres
    μ= 0.24 = 1.16×10-16rad/s
    (I used google to convert these values)

    Thus,

    Vt = μ⋅d = (1.16×10-16 rad/s) × (9.87×1017 m.)
    Vt = 114.49 m/s

    But I am not 100% sure with my method and therefore my final result.

    Then, I decided to read a little further and I stumbled upon a website that explained the tangential velocity in a more "complete" or rather different way. It said:

    ***To get the tangential velocity, you need to first measure the angular velocity of the star across the sky (d[PLAIN]http://www.astronomynotes.com/starprop/theta.gif/[I]dt[/I]). [Broken] This is how many degrees on the sky the star moves in a given amount of time and is called the proper motion by astronomers. If you determine the star's distance from its trigonometric parallax or the inverse square law method, you can convert the angular velocity (proper motion) to tangential velocity in physical units such as kilometers/second. The tangential velocity = k × the star's distance × the proper motion, where k is a conversion factor that will take care of the conversion from arc seconds and parsecs and years to kilometers/second. Using the Pythagorean theorem for right triangles, you find that the star's total velocity = Sqrt[(radial velocity)2 + (tangential velocity)2].

    My question is, what is that "k" they mention being the conversion factor?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 26, 2016 #2

    TSny

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    This is not correct. Note that μ is given in arcseconds per year. Also, make sure you are converting arcseconds to radians correctly.
     
  4. Mar 26, 2016 #3
    TSny, thanks for pointing that out.

    I used google as I don't know a way to go at it. May I ask if you know how to convert these values? :)
     
  5. Mar 26, 2016 #4

    TSny

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    Homework Helper
    Gold Member

    You must have misread the 10-16 factor when using Google.

    To do the conversion yourself, you will need to know how many radians are in one arcsecond (or vice versa) and you will need to know how many seconds are in a year. Then use a standard systematic method to convert arcseconds/year to radians/second. If you need a refresher on converting units, try https://www.mathsisfun.com/measure/unit-conversion-method.html or search for some other sites.
     
  6. Mar 26, 2016 #5
    Definitely need that refresh!

    Thank you :)
     
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