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 (k_o)  to its orbital angular speed, W.

Or:   c =  k_o / 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 (k_o) 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 k_o   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: c =  k_o / W .  Thus the epicyclic frequency:

k_o = 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 k_o

 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 _L =  W  - n k_o / m   =   W  - k_o / 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