Answer
(a) Both arms show a red shift.
(b) $8.149\times 10^{14}Hz$
(c) $8.114\times 10^{14}Hz$
Work Step by Step
(a) We know that the spiral arm of the galaxy and spiral galaxy are moving away from the Earth. Thus, the net motion is away and both arms(spiral) show the red shift.
(b) We know that
$f^{\prime}=f(1-\frac{u}{c})$
We plug in the known values to obtain:
$f^{\prime}=(8.230\times 10^{14}Hz)(1-\frac{3.600\times 10^6m/s-6.400\times 10^5m/s}{3\times 10^8m/s})$
$f^{\prime}=8.149\times 10^{14}Hz$
(c) As $f^{\prime}=f(1-\frac{u}{c})$
We plug in the known values to obtain:
$f^{\prime}=(8.230\times 10^{14}Hz)(1-\frac{3.600\times 10^6m/s+6.400\times 10^5m/s}{3.00\times 10^8m/s})$
$f^{\prime}=8.114\times 10^{14}Hz$
