Ek1001 physics #760

[JingleChips]

There's another post about this question, but I think the explanation given by the other member is wrong. Can anyone give input? (http://forums.studentdoctor.net/showthread.php?p=13654142)

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

EK Physics says the answer is C, and that you have to divide the velocity by 2 because of the reflection. However, they don't divide by 2 when arriving at their answer.

Furthermore, I don't think the correct answer is even listed as an answer choice. Please correct me if I'm wrong.

We know the source wavelength is 500nm emitted from the earth because of information given in the problem. The wavelength shift APPEARS to be 1nm, but since really it is a double shift (a shift to 500.5nm or 499.5nm from the gas cloud acting as a moving observer, and then a shift to 501nm or 499nm from the gas cloud acting as a moving source when it reflects the light), the wavelength change we should use is .5nm.
Using EK's approximation formula (v/c=delta lambda / source lambda) solve for v, getting 3.0x10^5 m/s. This is analogous to using the double shift of 1nm, and dividing the calculated velocity by 2.

This is correct, right? and the EK answer choices are all wrong? If not, can somebody tell me the difference between this problem and other "double doppler shift" problems such as a stationary police radar gun targeting a moving car (http://hyperphysics.phy-astr.gsu.edu/hbase/sound/radar.html#c4) ?

---

Members don't see this ad.

 

[IlyaR]

I don't understand why they would divide by 2.

The way that I did this problem is assuming that the light reflected off the STATIONARY gas would be the same 500nm. Since the light is now coming from the gas (which is now the source), I used the doppler approximation and got 1/500=v/(3.0x10^8), which gives C as the answer.

EK screws up here and there. (Not to mention the tons of typos)

 

[JingleChips]

I don't understand why they would divide by 2.

The way that I did this problem is assuming that the light reflected off the STATIONARY gas would be the same 500nm. Since the light is now coming from the gas (which is now the source), I used the doppler approximation and got 1/500=v/(3.0x10^8), which gives C as the answer.

EK screws up here and there. (Not to mention the tons of typos)

I divided by two for the same reason you divide by two when a stationary radar gun is used to detect the speed of a moving car ( http://hyperphysics.phy-astr.gsu.edu/hbase/sound/radar.html#c4 ). Does this not apply in this situation? The earth (radar gun) is stationary, and the gas cloud (vehicle) is moving, both situations rely on reflection.

From that website,

The Doppler shift for relatively low velocity sources such as those encountered by police RADAR is given by

delta f / f = v/c

but in this case there are two shifts: one because the wave incident on the moving car is Doppler shifted and an additional shift because the reflection is from a moving object. The frequency shift of the reflected wave received at the source of the wave is


delta f / f = 2v/c

Can anyone else chime in? I realize EK gets things wrong sometimes - but I just want to know whether the correct answer is 3.0x10^5 or 6.0x10^5 - and if it's the latter, I want to know why we don't divide the calculated velocity by 2 when in essence its the same situation as the stationary radar gun-moving vehicle scenario.

 

[sps27]

I took earth as the source and the interstellar gas as the detector.

W = observed wavelength
W(o) = original wavelength
V(o) = original velocity
V(d) = velocity of detector
V(s) = velocity of source

Case 1: for the wave travelling towards earth

W = W(o) {V(o) - V(s)) / V(o) + V(d)} --------- V(s) = 0 I.e., you as an observer on earth.
499 = W(o) { V(o) / V(o) + V(d) } ---------- 1

Case 2: for the wave travelling away from earth

W = W(o) {V(o) + V(s)) / V(o) - V(d)} --------- V(s) = 0 I.e., you as an observer on earth.
501 = W(o) { V(o) / V(o) - V(d) } ---------- 2

Divide 1 / 2
499/501 = V(o) - V(d) / V(o) + V(d)
Rearrange
499V(d) + 501V(d) = 501V(o) - 499V(o)
1000V(d) = 2 V(o)
1000V(d) = 2 * 3 * 10^8
V(d) = 6 * 10^5

This is how I worked out this problem using dopplers eqn. Obviously I did not follow any of EK's rules. This may still be wrong as it seems rather simplistic, so let me know.

 

[gettheleadout]

I don't even understand the scenario... What a terrible question.

Edit: Oh wow, this really is poor wording. "...the light reflecting off the gas coming toward the earth..." Is this referring to the light coming toward the earth after reflection, or the gas coming toward the earth reflecting the light? It's ambiguous. I realize they intend it to mean the latter, but they don't even specify a light source. Very poor quality.

 

[JingleChips]

I took earth as the source and the interstellar gas as the detector.

W = observed wavelength
W(o) = original wavelength
V(o) = original velocity
V(d) = velocity of detector
V(s) = velocity of source

Case 1: for the wave travelling towards earth

W = W(o) {V(o) - V(s)) / V(o) + V(d)} --------- V(s) = 0 I.e., you as an observer on earth.
499 = W(o) { V(o) / V(o) + V(d) } ---------- 1

Case 2: for the wave travelling away from earth

W = W(o) {V(o) + V(s)) / V(o) - V(d)} --------- V(s) = 0 I.e., you as an observer on earth.
501 = W(o) { V(o) / V(o) - V(d) } ---------- 2

Divide 1 / 2
499/501 = V(o) - V(d) / V(o) + V(d)
Rearrange
499V(d) + 501V(d) = 501V(o) - 499V(o)
1000V(d) = 2 V(o)
1000V(d) = 2 * 3 * 10^8
V(d) = 6 * 10^5

This is how I worked out this problem using dopplers eqn. Obviously I did not follow any of EK's rules. This may still be wrong as it seems rather simplistic, so let me know.

I think you did more work than you needed to. They said the gas is moving in a circular pattern. If you assume its moving at a constant speed, then the magnitude of change in wavelength when the gas moving away from earth should be the same as the magnitude of change in wavelength when the gas is moving towards the earth. The only way this can be true is if the earth is sending out a light with an initial wavelength of 500nm (with delta lamba = 1). You can quickly arrive at this qualitatively and just write one equation using 500nm as W(o) and picking either 499 or 501 as W, and just using the correct signs.

I don't even understand the scenario... What a terrible question.

Edit: Oh wow, this really is poor wording. "...the light reflecting off the gas coming toward the earth..." Is this referring to the light coming toward the earth after reflection, or the gas coming toward the earth reflecting the light? It's ambiguous. I realize they intend it to mean the latter, but they don't even specify a light source. Very poor quality.

Yeah its ambiguous. If they were talking about the source of the light being from the earth (as in, someone shining a laser from the earth to the gas, and measuring the wavelength of the laser as it returned to earth), then the answer would be 3x10^5 correct? This question is driving me crazy! I fail to see how it is different ( if it even is different) than a stationary police officer using a radar gun to determine the speed of a car. If you solve this like a normal doppler effect question without reflection, then you get 6x10^5. If you solve it like the police radar gun situation (reflection with doppler effect), you should get 3x10^5 due to a double doppler shift.

 

[sillyjoe]

I always love finding threads like these after I have been sitting for 35 minutes trying to figure out their explanation