Tuesday, May 28, 2024

Solutions to Galaxy Density Wave Problem

 The Problem again:

The number c of epicycle oscillations per orbit about a galactic center, is given by the ratio of the star’s epicycle frequency (ko)  to its orbital angular speed, W.

Or:   c =  ko / W  

a) Compute W for the Sun given that the Sun’s period around the galactic center is: 2 x 10 8 yrs.
  
b) Then find the Sun’s epicyclic frequency (ko) if we know c = 1.35. (Note: if the ratio c is integral, or a non-decimal number, we say the orbit is “closed”. If non-integral it is not closed). Because of differential rotation rates and the local angular velocity (W L) differing from the inertial value, the solar region is open.


c) Compute the local angular velocity (W L) for the Sun if:  

m(W  -  W L) = n ko   where m = 2 and n = 1

Solutions:

a) Assume that the Sun moves in a nearly circular orbit about the galactic center with the radius of orbit R = 104 parsecs. It then makes an entire orbital revolution in 2 x 108 years. We need to obtain this period in seconds:

T(s) = (2 x 10 8 yrs.) (365.25 days/yr.) (86,400 s/ day)

T(s) = 6.3 x 10 15 s

 Then: W  = (2 π rad)/ T(s) = 2 π rad/ 6.3 x 10 15 s  

W  » 10 -15 rad s-1


(b) We know c = 1.35 and we have: cko / W .  Thus the epicyclic frequency:

ko = 1.35  W    = 1.35 (10 -15 rad s-1) = 1.35 x 10 -15 rad s-1

 (c) Re-arrange the given expression:

m(W  W L) = n ko

 By deliberately choosing to have the Sun complete n orbits in the rotating frame of reference (while executing m epicycle oscillations) we can write:

W LW  - n ko / m   =   W  - ko / 2

Therefore:

W L =  10 -15 rad s-1   - ½(1.35 x 10 -15 rad s-1 ) =

3.2 x 10 -16 rad s-1


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