How fast must this star be moving?

[Joyeuse]

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!

 

[Curious3141]

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 $\frac{f_s}{f_o} = \sqrt{\frac{1+\beta}{1-\beta}}$

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).

 

[Joyeuse]

You need to use the Relativistic Doppler Effect formula.

It's given by $\frac{f_s}{f_o} = \sqrt{\frac{1+\beta}{1-\beta}}$

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)?

 

[Curious3141]

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.

 

[asteriatic]

I would first clarify the given information and assumptions. Is the star assumed to be emitting a monochromatic (single frequency) light? Is the observer (Earth) assumed to be stationary? Also, it is important to define what is meant by "11% higher frequency" - is it 11% higher than the original frequency, or 11% higher than the observed frequency? These details will affect the approach to the problem.

Assuming that the star is emitting a monochromatic light and the observer is stationary, the equation used in the attempt at solution is correct. However, there may be a calculation error or approximation error that is causing the answer to be slightly off. I would suggest double-checking the calculations and considering any possible sources of error.

If the Earth's speed is assumed to be significant, it can be taken into account by using the relative velocity formula (v = v_s + v_obs), where v_obs is the observer's velocity relative to the star. This would require additional information such as the Earth's speed and direction of motion relative to the star.

In terms of reflection, if the star is assumed to be a point source, then there would be no reflection. However, if the star is assumed to have some sort of surface or atmosphere that could cause scattering or reflection of light, then this could affect the observed frequency. In this case, additional information about the star's properties would be needed to accurately calculate the observed frequency.

Overall, the approach and equation used seem to be correct, but the answer may be slightly off due to calculation or approximation errors. Additional information about the assumptions and given parameters would be needed to provide a more accurate answer.