How long will it take an electromagnetic wave to reach a satellite

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Homework Help Overview

The discussion revolves around a problem involving the calculation of the time it takes for an electromagnetic wave to travel from a satellite to a receiver, as well as the frequency difference observed due to the satellite's motion. The subject area includes concepts from electromagnetism and the theory of relativity, particularly in the context of GPS technology.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the speed of light and the time calculation, questioning the relevance of the satellite's speed. There is also a focus on the accuracy of frequency calculations and the appropriateness of the formulas used for Doppler shift.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made in the original attempts and suggesting the need for more accurate formulas. There is a recognition of differing levels of accuracy required for the calculations, and some guidance has been offered regarding the use of relativistic versus non-relativistic formulas.

Contextual Notes

There are indications that the original poster may not have been aware of the complexities involved in the Doppler shift calculations, and the discussion reflects varying interpretations of the problem's requirements for accuracy.

aChordate
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Homework Statement



GPS determines your distance from a satellite by measuring how long it takes an electromagnetic wave to travel from the satellite to you. Imagine that a satellite is located at a distance of 127.2km and is moving away from you at a speed of 12.0km/s.
a)how long will it take an electromagnetic wave emitted by the satellite to reach you?
b)if the frequency of the wave emitted by the satellite is precisely 10.0 GHz, what is the frequency difference between this frequency and the frequency that you observe?

Homework Equations



fo=fs(1+/-vrel/c) if vrel<<c


The Attempt at a Solution



d=127.2km=127.2*103m
v=12km/s=12*103m/s
fs= 10GHz= 10*10^9Hz
c=3.00*108m/s

Part A: 10.6 seconds
Part B:

f0=10*109 Hz (1-(12*103m/s/3.00*108m/s))=10*109 Hz
 
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aChordate said:

Homework Statement



GPS determines your distance from a satellite by measuring how long it takes an electromagnetic wave to travel from the satellite to you. Imagine that a satellite is located at a distance of 127.2km and is moving away from you at a speed of 12.0km/s.
a)how long will it take an electromagnetic wave emitted by the satellite to reach you?
b)if the frequency of the wave emitted by the satellite is precisely 10.0 GHz, what is the frequency difference between this frequency and the frequency that you observe?

Homework Equations



fo=fs(1+/-vrel/c) if vrel<<c


The Attempt at a Solution



d=127.2km=127.2*103m
v=12km/s=12*103m/s
fs= 10GHz= 10*10^9Hz
c=3.00*108m/s

Part A: 10.6 seconds
Part B:

f0=10*109 Hz (1-(12*103m/s/3.00*108m/s))=10*109 Hz

For Part A, don't you think the time will have rather more to do with the speed of light than the speed of the satellite? For Part B, don't you think keeping more accuracy in the answer might be useful?
 
In addition to the problems pointed out in post #2 you have the wrong formula for the Doppler shift.

It's based on the theory of relativity so don't try to derive it. Look it up.
 
rude man said:
In addition to the problems pointed out in post #2 you have the wrong formula for the Doppler shift.

It's based on the theory of relativity so don't try to derive it. Look it up.

Depends on how much accuracy you need. It certainly doesn't matter if you round to the nearest GHz, like aChordate. But if I use the nonrelativistic formula and compare the results with the relativistic the frequency shifts are the same to 4 decimal places.
 
Dick said:
Depends on how much accuracy you need. It certainly doesn't matter if you round to the nearest GHz, like aChordate. But if I use the nonrelativistic formula and compare the results with the relativistic the frequency shifts are the same to 4 decimal places.

I doubt that the OP was aware of that. In any case, he/she might as well use the correct formula in case a future problem involves speeds closer to c.

EDIT: I have to admit that, if the emphasis was on satellites, the OP might never encounter the relativisic formula, so on second thoughts I think you were right to point out what you did.
 
Last edited:

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