How fast must this star be moving?

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Homework Statement



How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 11.0% higher than the frequency of the light it is emitting?


Homework Equations



(f_L)= (f_s)*(v+v_L)/(v+v_s)

The Attempt at a Solution



Okay, so, if the frequency of light we receive is 11% higher than emitted, then that means that (f_L)/(f_s) = 1.11, right? I know that for the frequency of the star, which is the source in this case, to be less than the frequency we receive, it has to be moving toward us. So, the sign on v_s is negative. The earth's speed is really, really small relative to the speed of the star, so the listener (the Earth) is effectively 0 in this case, right? So, I had:

1.11 = v/(v-v_s)

I calculated v_s to be 0.099*c, where c is the speed of light (3.00*10^8 m/s). I also used the speed of light for v in the above calculations. The thing is, it tells me that my answer is close, but not close enough to count as correct.

So, my thoughts were: a) do we have to take the earth's speed into account (and if so, how?), or b) is there some facet that I'm missing (like, is there some sort of reflection of the wave that I have to take into account)? Thank you!
 

Answers and Replies

  • #2
Curious3141
Homework Helper
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Homework Statement



How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 11.0% higher than the frequency of the light it is emitting?


Homework Equations



(f_L)= (f_s)*(v+v_L)/(v+v_s)

The Attempt at a Solution



Okay, so, if the frequency of light we receive is 11% higher than emitted, then that means that (f_L)/(f_s) = 1.11, right? I know that for the frequency of the star, which is the source in this case, to be less than the frequency we receive, it has to be moving toward us. So, the sign on v_s is negative. The earth's speed is really, really small relative to the speed of the star, so the listener (the Earth) is effectively 0 in this case, right? So, I had:

1.11 = v/(v-v_s)

I calculated v_s to be 0.099*c, where c is the speed of light (3.00*10^8 m/s). I also used the speed of light for v in the above calculations. The thing is, it tells me that my answer is close, but not close enough to count as correct.

So, my thoughts were: a) do we have to take the earth's speed into account (and if so, how?), or b) is there some facet that I'm missing (like, is there some sort of reflection of the wave that I have to take into account)? Thank you!

You need to use the Relativistic Doppler Effect formula.

It's given by [itex]\frac{f_s}{f_o} = \sqrt{\frac{1+\beta}{1-\beta}}[/itex]

where s is source, o is observer, β is the ratio of the relative velocity between source and observer to c, the speed of light. In this equation, source and observer are assumed to be moving away from one another. If you solve it and get a negative value for β, you'll know they were moving toward each other.

In Special Relativity, what matters is the relative velocity between two bodies (source and observer, in this case), so there's no point in distinguishing source and observer movement as in the Classical Doppler formula.

Applying this will give you an answer that's distinct, but close (less than 5% error) to what you got with the Classical Doppler application.

This site will tell you more about it: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

(BTW, this is more correctly called the Longitudinal Relativistic Doppler Effect, as it applies when the source and observer are approaching or receding in a direct line to each other. There's another effect called the Transverse Relativistic Doppler Effect that applies to light emissions when the source and observer are moving at right angles to each other. There's no Classical analogue to the latter).
 
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  • #3
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You need to use the Relativistic Doppler Effect formula.

It's given by [itex]\frac{f_s}{f_o} = \sqrt{\frac{1+\beta}{1-\beta}}[/itex]

where s is source, o is observer, β is the ratio of the relative velocity between source and observer to c, the speed of light. In this equation, source and observer are assumed to be moving away from one another. If you solve it and get a negative value for β, you'll know they were moving toward each other.

In Special Relativity, what matters is the relative velocity between two bodies (source and observer, in this case), so there's no point in distinguishing source and observer movement as in the Classical Doppler formula.

Applying this will give you an answer that's distinct, but close (less than 5% error) to what you got with the Classical Doppler application.

This site will tell you more about it: http://en.wikipedia.org/wiki/Relativistic_Doppler_effect

(BTW, this is more correctly called the Longitudinal Relativistic Doppler Effect, as it applies when the source and observer are approaching or receding in a direct line to each other. There's another effect called the Transverse Relativistic Doppler Effect that applies to light emissions when the source and observer are moving at right angles to each other. There's no Classical analogue to the latter).

Ah, so, since the frequency of the observer is 11% greater than the frequency of the source, (f_s)/(f_o) = 1/1.11, correct? Then, just solve for beta and report the absolute value of it (wikipedia is telling me that beta is the velocity of the observer/speed of light)?
 
  • #4
Curious3141
Homework Helper
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Ah, so, since the frequency of the observer is 11% greater than the frequency of the source, (f_s)/(f_o) = 1/1.11, correct? Then, just solve for beta and report the absolute value of it (wikipedia is telling me that beta is the velocity of the observer/speed of light)?

Yes.
 

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