Consider a spiral galaxy that is moving directly away from Earth with a speed V
ID: 2261178 • Letter: C
Question
Consider a spiral galaxy that is moving directly away from Earth with a speed V = 3.600 ? 106 m/s at its center, as shown in the figure. The galaxy is also rotating about its center, so that points in its spiral arms are moving with a speed v = 6.800 x 105 m/s relative to the center. Light with a frequency of 7.070 x 1014 Hz is emitted in both arms of the galaxy, and this light is detected by astronomers on Earth.
(a) What frequency is detected by the astronomers observing the upper arm? (Enter your answer to 4 significant figures.) Hz
(b) What frequency is detected by the astronomers observing the lower arm? (Enter your answer to 4 significant figures.) Hz
Explanation / Answer
The velocities given are much less than the speed of light, so ordinary doppler equations apply.
For the arms moving toward the earth i.e Upper Arm
velocity of the emitters is 3.600*10^5 - 6.800*10^5 = -3.2*10^5 m/s.
From the doppler formula :
v/c =(fe - fo)/fo
fo*v/c = fe - fo
fo*(1 + v/c) = fe
fo = fe/(1 + v/c)
a)
v/c = -3.2*10^5 / 2.998*10^8 = -1.067*10^-3
fo = fe /(1 - 1.067*10^-3)
= fe * 1.00107
= 7.070*10^14 * 1.00107
fo = 7.0776*10^14 Hz
b)
For the other arm (lower arm)
v = 3.600*10^5 + 6.800*10^5 = 10.4*10^5 m/s
v/c = 10.4*10^5 / 2.998*10^8 = 3.469*10^-3
fo = fe/(1 + .003469)
= 7.070*10^14 / (1.003469)
= 7.0456*10^14 Hz