# Calculate the line of sight velocity dispersion of the cluster in km/sec

"Calcium H and K lines have rest frame wavelengths of 3968.5 A (angstrom). Spectroscopy of 5 stars in a cluster revealed that the Ca H line was detected at
3970.1 A
3966.3 A
3967.5 A
3971.0 A
3968.2 A
in each of these five stars. Calculate the line of sight velocity dispersion of the cluster in km/sec. The rms of the proper motions of these 5 stars was measured to be 1.3 arcseconds / year. Calculate the distance to the star cluster.

Last edited:

Pengwuino
Gold Member

I have no idea how to do this. My dad said if I can find the answer online he will buy me a new car. He is trying to prove to me that people online are wasting their lives. Anyone that can help please?

...... what?

People online do waste their lives but... what?

Except for physics forums.

Delta2
Homework Helper
Gold Member

Your dad is teaching astronomy or physics? This is a problem that has to do with the doppler shift effect of the light the stars emmit as they move closer or further from earth.

I could help if this was a simple case with stars moving in the line that connects them with earth but this isnt the case , a star movemnt doesnt necessarily follow this line.

Ok i ll try to use the formula from wikipedia $$f_i=(1-\frac{v_i}{c})f_0$$ where $$v_i$$ is the velocity of the star i relative to earth and i=1...5. Solving for $$v_i$$ we have

$$v_i=(1-\frac{f_i}{f_0})c=(1-\frac{\lambda_0}{\lambda_i})c$$

We know that $$\lambda_0=3968.5A$$ and also the values for the $$\lambda_i$$ so we can now compute $$v_i$$.

$$v_1=120,9Km/s, v_2=-166,4Km/s, v_3=-75,6Km/s, v_4=188,8km/s, v_5=-22,6Km/s$$. The mean value of these 5 velocities is 9,02km/s and the dispersion is 179,78km/s.

Got no clue how to answer the second question for the distance.

Last edited:

Thanks :) Anyone else know the second part?

Delta2
Homework Helper
Gold Member

btw, i am not sure at all how velocity dispersion is defined. if it is $$\sqrt{ \sum_{i=1}^{5}\frac{(v_i-m)^2}{5}}$$ where m is the mean value then it is 129,4Km/s.

Last edited:

1".3arc = (1.3/3600) degrees = (1.3/3600) X (TT / 180) Radians
= proper motion / distance.
=> distance = proper motion / 6.302577854 X 10^-6 Radians
= 158,665.2356 X proper motion.

Derive the rms value of the five (divide by 5 to get mean square) that gives standard deviation. And to get the 'proper motion' put that in this equation to get 'Distance'.

Someone else gave me this. I am sooo confused lol

Redbelly98
Staff Emeritus