Find Tangential Velocity of Star: "k" Conversion Factor Explained

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Homework Help Overview

The discussion revolves around finding the tangential velocity of a star using its proper motion and distance. The original poster presents data including a distance of 32 parsecs and a proper motion of 0.24 arc seconds per year, along with a formula for tangential velocity.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate tangential velocity using a provided formula but expresses uncertainty about their method and final result. They also reference an alternative explanation involving a conversion factor "k" for proper motion.

Discussion Status

Participants are engaged in clarifying the conversion of proper motion from arc seconds to radians and the implications of using the correct conversion factor. Some participants have pointed out potential errors in the original poster's calculations and offered guidance on unit conversion.

Contextual Notes

There is a noted confusion regarding the conversion of units, specifically from arc seconds to radians and the time frame involved. The original poster's reliance on an online conversion tool is questioned, and there is a suggestion to revisit fundamental unit conversion principles.

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


[/B]
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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JoAstro said:
μ= 0.24 = 1.16×10-16rad/s
This is not correct. Note that μ is given in arcseconds per year. Also, make sure you are converting arcseconds to radians correctly.
 
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? :)
 
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.
 
Definitely need that refresh!

Thank you :)
 

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