# #760 EK Physics Doppler Effect

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#### [email protected]

##### Member
15+ Year Member

I was not too sure how to even begin solving this problem and the EK explanation was not too helpful thus I was wondering if anyone could help me out.

"760. An interstellar gas circles the core of earth's galaxy. If the wavelength of the light reflecting off the gas coming toward the earth is 499 nm, and the wavelength of light reflecting off the gas moving away from earth is 501 nm, what is the speed of the gas?
A. 4.2xl0^4 m/s
B. 1.2xl0^5 m/s
C. 6.0xl0^5 m/s
D. 1.5xl0^11 m/s

760. C is correct. The relative velocities of the gas moving away from the earth and the gas moving toward the earth are equal and opposite; therefore the change in wavelength must be the same for both but in opposite directions, and the wavelength of light coming from earth must be 500 nm. Now use the Doppler approximation: vic= (change in wavelength)/ (wavelength of source).. v/(3x10^8) = 1/500. This velocity must be divided by 2 because the change in the distance of the path traveled by the light is decreased or increased by a factor twice the velocity of the gas. To visualize this, notice that the box below travels toward the man at 1 rn/s. However, the path of light going to the box and back to the man changes from 20m to 18m in a second, or 2 rn/s. This division by 2 is required here because the source is also the observer, and the wave is making a round trip."

Did EK take the average of the two wavelengths? I think it is accounting for the relative velocities of the gas. The gas is slowing down and speeding up the light waves, right? Shouldn't the change in wavelength be 2 m/s instead.

In addition, why are they saying that the velocity should be divided with two? I don't get their box analogy. In addition, if I solve for the equation v/(3x10^8) = 1/500. I get v= 6.0xl0^5 m/s without dividing by 2 unless my math is wrong somehow.

Anyways, if you have any ideas please let me know. I want to finish this section of my MCAT studying and move on. Thank you for all your help.

Verónica

#### BryanNextStep

##### Full Member
(1) Did EK take the average of the two wavelengths? I think it is accounting for the relative velocities of the gas.

(2) The gas is slowing down and speeding up the light waves, right?

(3) In addition, why are they saying that the velocity should be divided with two? I don't get their box analogy. In addition, if I solve for the equation v/(3x10^8) = 1/500. I get v= 6.0xl0^5 m/s without dividing by 2 unless my math is wrong somehow.

Hey there Verónica!

(1) Sounds like they're describing two different situations. In one, you've got the Doppler effect from something coming at you - which gives you a wavelength of 499. Then you've got something else going away from you - which gives you a wavelength of 501.

It's kind of like when an ambulance drives by - when it's coming towards you, the pitch is higher by a certain amount. Then, when the ambulance is driving away from you, it goes down by the same amount. The actual pitch emitted by the ambulance is going to be at a point between that high pitch and low pitch.

So in this case the actual wavelength being emitted is 500nm. While the cloud is approaching us, the light gets blue-shifted to 499 nm. When the gas cloud is going away, it gets red-shifted to 501nm.

(2) Pretty sure it's got nothing to do with the velocity of light. Remember, the speed of light depends on the medium you're moving through. In the case, the light ray bounces off the object, goes through the vacuum of space, and hits the telescope.

(3) I'm with you here - I'm not understanding how their explanation relates to this question. You're right that the equation gives you the right answer without any "dividing by 2".

The thing is, the situation they're describing with the box is a good example, but it just doesn't seem relevant to this question.

A case where it would be relevant would be something like a bat's sonar.

Imagine a bat flying at a wall. The bat emits a sonar squeak while he's flying towards the wall. The wall "hears" the sonar pitch, but the frequency is higher because the bat is flying towards the wall.

Then, the sound bounces off the wall, so now the wall becomes the "source" of the sound. The sonar squeak goes back to the bat, but because the bat is *still* flying towards the wall, the pitch goes up even more.

Thus, the doppler shift in this case would be twice as much as you'd normally expect - it gets shifted up when the wall "hears" the bat squeak, and then it gets shifted up when the bat "hears" it emitted from the wall.

Again, that has nothing to do with the light example, but I think maybe that's the phenomenon they were trying to describe with "the box".

Good luck #### [email protected]

##### Member
15+ Year Member
Okay one more question... why did EK plug in 1 as the change in wavelength in the equation: v/c= (change in wavelength)/ (wavelength of source). v/(3x10^8) = 1/500.

Thank you for all your help by the way. I am starting to understand. The question threw me off because it was different from all the other questions.

Best,

Verónica