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CHAPTER 45 - 5

# CHAPTER 45 - 5 - whichgivesd=1.4 10 ly 26 v=Hd 6 3 8 3.5 10...

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which gives  d  =       1.4 × 10 8  ly . 26. We find the distance from Hubble’s law: v  =  Hd ; 3.5 × 10 6  m/s = [(70 × 10 3  m/s/Mpc)/(10 6  pc/Mpc)(3.26 ly/pc)] d which gives  d  =       1.6 × 10 8  ly . 27. We estimate the speed of the galaxy from Hubble’s law: v  =  Hd  = [(70 × 10 3  m/s/Mpc)/(3.26 ly/pc)](12 × 10 3  Mly) = 2.6 × 10 8  m/s =       0.86 c . 28. ( a ) We find the receding speed of the galaxy from v  =  Hd  = [(70 × 10 3  m/s/Mpc)/(3.26 ly/pc)](1.0 Mly) = 2.15 × 10 4  m/s = (7.2 × 10 –5 ) c . For the Doppler shift of the wavelength we have λ / λ 0  =  { [1 + ( v / c )]/[1 – ( v / c )] } 1/2 ; λ /(656 nm) = [(1 + 7.2 × 10 –5 )/(1 – 7.2 × 10 –5 )] 1/2 , which gives  λ =      656 nm . ( b ) We find the receding speed of the galaxy from v  =  Hd  = [(70 × 10 3  m/s/Mpc)/(3.26 ly/pc)](1.0 × 10 2  Mly) = 2.15 × 10 6  m/s = (7.2 × 10 –3 ) c . For the Doppler shift of the wavelength we have λ / λ 0  =  { [1 + ( v / c )]/[1 – ( v / c )] } 1/2 ; λ /(656 nm) = [(1 + 7.2

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CHAPTER 45 - 5 - whichgivesd=1.4 10 ly 26 v=Hd 6 3 8 3.5 10...

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