# Find speed of quasar from observed and actual wavelength readings

1. Aug 23, 2010

### Sarial

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

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?

2. Aug 23, 2010

### Sarial

Bump? :( I don't know where else to go with this

3. Aug 24, 2010

### Sarial

Bump. Due at 10 PM EST tonight

4. Aug 24, 2010

### hikaru1221

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.

5. Aug 24, 2010

### Sarial

I tried +.314 and still no luck. I must be making a tiny mistake somewhere, and I just can't find it.

6. Aug 24, 2010

### hikaru1221

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.

7. Aug 24, 2010

### Sarial

I believe the +/- discrepancy, but the .314 apparently isn't right either :\

8. Aug 24, 2010

### hikaru1221

9. Aug 24, 2010

### Sarial

Ohhh. That could be it. I'll try that soon, thank you. You should use the relativistic formula for .10c+?

10. Aug 24, 2010

### hikaru1221

It depends, but generally, yup :)

11. Aug 24, 2010

### Sarial

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