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

In summary, the tangential velocity of a star can be found by measuring its angular velocity and then converting it to kilometers per second. The conversion factor used is k.
  • #1
JoAstro
14
0

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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  • #2
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.
 
  • #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? :)
 
  • #4
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.
 
  • #5
Definitely need that refresh!

Thank you :)
 

What is tangential velocity and why is it important in astronomy?

Tangential velocity refers to the speed at which an object moves in a circular or elliptical path. In astronomy, it is important because it helps us understand the motion of celestial objects, such as stars, planets, and galaxies.

How is "k" conversion factor used in finding the tangential velocity of a star?

The "k" conversion factor is used to convert between different units of measurement, such as kilometers per second and astronomical units per year. It is used in the formula for tangential velocity, which is v = k * r, where v is the tangential velocity, k is the conversion factor, and r is the distance from the center of the circular or elliptical path.

What are some common values of "k" conversion factor used in astronomy?

The most commonly used values of "k" conversion factor in astronomy are 1 km/s per AU/yr (kilometers per second per astronomical unit per year) for objects within our solar system, and 1 AU/yr (astronomical units per year) for objects outside of our solar system.

Can "k" conversion factor be used for all types of celestial objects?

Yes, "k" conversion factor can be used for all types of celestial objects as long as their distance from the center of their circular or elliptical path is known. However, different conversion factors may be used depending on the units of measurement and the specific object being studied.

How does the tangential velocity of a star affect its orbit and overall motion?

The tangential velocity of a star affects its orbit and overall motion by determining the shape, size, and speed of its path around the center of mass. A higher tangential velocity can result in a more elliptical orbit, while a lower tangential velocity can result in a more circular orbit. The tangential velocity also plays a role in the stability and dynamics of a star's orbit within a larger system, such as a galaxy.

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