This is meant to be handed out to the students.
Maybe you have heard an El train here in Chicago when it is going to pass up your stop. (This always seems to happen to me when the temperature outside is below zero!) The engineer blows the whistle. He hears just one, unchanging, pure tone. You, on the platform hear the note change. First, it is higher when the train approaches, then it quickly lowers as the train passes and heads away from you. What's happening is that the frequency of the wave is changing, like in the diagram above. Don't forget that higher frequency means higher notes and a lower frequency means a lower note.
There's a formula for this effect. It reads like this:
[Frequency Shift]/[Frequency] = [Speed of Source]/[Speed of Sound]
This may seem a little confusing, but the only thing we need to remember is that the radio dial measures the actual radio frequency received when the tone is the loudest. The satellite gives off radio waves, so the above equation must be modified for light waves. Let's also go to using symbols. Here they are:
The Doppler Effect now reads:
D/F = V/C
Here's the data for a pass. It shows the time and the actual observed frequency (the beacon is is 29,408 KHz = F in the above equation).
Wednesday, December 15, 1993 ------ Orbit #14347 ------ Local Doppler(KHz) 3:38:00 PM 29,408.618 3:39:00 PM 29,408.614 3:40:00 PM 29,408.606 3:41:00 PM 29,408.591 3:42:00 PM 29,408.563 3:43:00 PM 29,408.514 3:44:00 PM 29,408.424 3:45:00 PM 29,408.264 3:46:00 PM 29,408.24 3:47:00 PM 29,407.776 3:48:00 PM 29,407.601 3:49:00 PM 29,407.501 3:50:00 PM 29,407.446 3:51:00 PM 29,407.416 3:52:00 PM 29,407.398 3:53:00 PM 29,407.389 3:54:00 PM 29,407.384Plot these Doppler shifts versus time in the plot on the next page. (Hint! You probably should do some rounding on the numbers above.)
[Graph paper doens't work in hypertext.]
From the data, what is the maximum frequency shift?
Recalling that the F= 29,408 KHz, then what must D/F be?
Doppler tell us that this last number is just the difference in the satellite's speed divided by the speed of light.
We already figured out the satellite's speed last week. We figured it in Km/hour. Here we need it in Km/sec -- it is 7.4 Km/s. If we could see the satellite coming straight at us and straight away from us, this would mean that the total difference in speed of the satellite during the pass would be 2X 7.4 Km/s or 14.8 Km/s. In reality, the satellite moves at a slight angle to us when we first see it. This introduces a slight factor from trigonometry (which we won't get into). This number is 0.87. So, the V in the equation above is now 14.8 km/s x 0.87 = 12.9 km/s. This is the speed difference during the entire satellite pass. It is the V in the Doppler equation.
Now, using the space below, plug the last number as well as the ratio of frequencies you found above into the Doppler Equation. What value do you get for the speed of light?
Important Disclaimers and Caveats