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JoAstro
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Homework Statement
I am trying to find the tangential velocity of a star but I am confused with the whole procedure.
Homework Equations
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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.
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]). 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?
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