What river is found in the western hemisphere
2 answers:
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Answer:
A. z (galaxy 1) = -0.00515, z (galaxy 2) = -0.01707, z (galaxy 3) = -0.04008
B. v (galaxy 1) = -0.00517c, v (galaxy 2) = -0.01707c, v (galaxy 3) = -0.0401c
C. d (galaxy 1) = -21 Mpc,d (galaxy 2) = -69.4 Mpc, d (galaxy 3) = -163 Mpc
Explanation:
We begin by listing out the parameters we were given:
λ (obsv) = 656.3 nm, λ (emit 1) = 659.7 nm, λ (emit 2) = 667.7 nm,
λ (emit 3) = 683.7 nm
A) Using the Redshift formula, we have:
z = [λ (obsv) - λ (emit)] ÷ λ (emit)
For galaxy 1:
z = [λ (obsv) - λ (emit 1)] ÷ λ (emit 1)
z = (656.3 - 659.7) ÷ 659.7 = -0.00515
z = -<u>0.00515</u>
For galaxy 2:
z = [λ (obsv) - λ (emit 2)] ÷ λ (emit 2)
z = (656.3 - 667.7) ÷ 667.7 = -0.01707
z = -<u>0.01707</u>
For galaxy 3:
z = [λ (obsv) - λ (emit 3)] ÷ λ (emit 3)
z = (656.3 - 683.7) ÷ 683.7 = -0.04008
z = -<u>0.04008</u>
B) Using the Doppler formula, we have:
(Δλ ÷ λ) = v ÷ c
v = c * (Δλ ÷ λ)
but, z = (Δλ ÷ λ)
⇒ v = c * z
speed of light (c) = 3 x m/s
For galaxy 1:
v = c * z
Substitute z into the equation calculated from A) above
v = 3 x * (-0.00515)
v = -1.55 x m/s
v = -<u>0.00517c</u>
For galaxy 2:
v = c * z
Substitute z into the equation calculated from A) above
v = 3 x * (-0.01707)
v = -5.12 x m/s
v = -<u>0.01707c</u>
For galaxy 3:
v = c * z
Substitute z into the equation calculated from A) above
v = 3 x * (-0.04008)
v = -12.03 x m/s
v = -<u>0.0401c</u>
N.B: the negative value of velocity connotes that the galaxies are moving away from us (not towards us)
C) Using Hubbleʹs law, we have:
v = H · d
where:
v = velocity of a galaxy (km/s), d = distance (Mpc),
H = Hubble's constant (km/s/Mpc) = 73.8 km/sec
d = v ÷ H
We use the velocities calculated in B) above
For galaxy 1:
v = -1.55 x km/s
d = -1.55 x ÷ 73.8
d = -<u>21 Mpc </u>
For galaxy 2:
v = -5.12 x km/s
d = -5.12 x ÷ 73.8
d = -<u>69.4 Mpc </u>
For galaxy 3:
v = -12.03 x km/s
d = -12.03 x ÷ 73.8
d = -<u>163 Mpc</u>
N.B: distance cannot be a negative value