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Homework Help: Find speed of quasar from observed and actual wavelength readings

  1. Aug 23, 2010 #1
    1. Astronomers on Earth observe a feature at 442 nm in the spectrum of a distant quasar. However, it is known that this feature corresponds to 581 nm in the rest frame of the quasar. Calculate the speed of the quasar. Enter your answer as a fraction of the speed of light, positive if the quasar is approaching, negative if it's receding.



    2. Relevant equations
    I used 6be9a7bd1b2910eccce5e1c51268ebd4.png


    3. The attempt at a solution
    It was easier for me to think about this in terms of frequency than wavelength, so I converted the wavelengths into frequency by using
    c = speed of light
    L = lambda, wavelength
    f = frequency
    u = speed of source

    c/L = f

    So 3e8/442e-9 = 6.79e14
    3e8/581e-9 = 5.16e14

    So the observed frequency is 6.79e14 Hz and the actual frequency is 5.16e14 Hz.

    fobs=(1-u/c)femitted

    6.79e14 = (1-u/c)(5.16e14)

    1.316 = 1-u/c

    .316=-u/c

    -.316c=u

    Now, this isn't right. I'm using (1-u/c) versus (1+u/c) because the relative speed of the object is negative if they're approaching, and the wavelength is getting blueshifted, so they're moving towards each other. However, that produces a negative answer, and the question states to enter a positive answer if they're moving towards each other.

    Anyone see any mistakes in my work?

    Thanks a lot in advance~
     
  2. jcsd
  3. Aug 23, 2010 #2
    Bump? :( I don't know where else to go with this
     
  4. Aug 24, 2010 #3
    Bump. Due at 10 PM EST tonight
     
  5. Aug 24, 2010 #4
    It's simply that your equation is wrong. The quasar and the observer are moving towards each other, so the ratio should be 1+v/c, not 1-v/c. Your calculation, which is correct, also shows that.
     
  6. Aug 24, 2010 #5
    I tried +.314 and still no luck. I must be making a tiny mistake somewhere, and I just can't find it.
     
  7. Aug 24, 2010 #6
    Suppose that your calculation is correct. We all know that they are approaching each other at +0.314c. But in your formula, the minus sign means that v is the speed at which they are going away from each other. Therefore, that v is negative means that they are approaching each other, which matches the expected phenomenon.
     
  8. Aug 24, 2010 #7
    I believe the +/- discrepancy, but the .314 apparently isn't right either :\
     
  9. Aug 24, 2010 #8
  10. Aug 24, 2010 #9
    Ohhh. That could be it. I'll try that soon, thank you. You should use the relativistic formula for .10c+?
     
  11. Aug 24, 2010 #10
    It depends, but generally, yup :)
     
  12. Aug 24, 2010 #11
    Yes! Thank you. I knew it was something small like that. .267c was the right answer, and using gamma sorted it all out!

    Case closed :P
     
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