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

[ladybug1829]

Please Help :)

"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.

 

[Pengwuino]

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?

 

[Kevin_Axion]

Except for physics forums.

 

[Delta2]

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 isn't the case , a star movemnt doesn't 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.

 

[ladybug1829]

Thanks :) Anyone else know the second part?

 

[Delta2]

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.

 

[ladybug1829]

1".3arc = (1.3/3600) degrees = (1.3/3600) X (TT / 180) Radians
= 6.302577854 X 10^-6 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]

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?

Physics Forums is not the place to have other people do your homework for you. Probably best to look elsewhere.