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modphysnoob

  • one year ago

A spacecraft traveling out of the solar system at 0.92c sends back information at a rate of 400Hz. At what rate do we receive the information?

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  1. modphysnoob
    • one year ago
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    @Jemurray3

  2. Jello_Submarine
    • one year ago
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    According to my calculations - at the rate of 400 Hz, because after adding up speed of the radio waves(c) and speed of the spacecraft(0,92) regarding the theory of special relativity, the answer is c and therefore the rate stays the same because the speed of the radiowaves is same.

  3. Jello_Submarine
    • one year ago
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    0.92c should be behind the word "spacecraft"

  4. modphysnoob
    • one year ago
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    I was given following options A. 1960Hz B. 17Hz C. 9600Hz D. 82Hz.

  5. Jello_Submarine
    • one year ago
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    oh...let me rethink

  6. Jello_Submarine
    • one year ago
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    definitely not 1960 Hz nor 9600 Hz, because as the speed increases, the waves are further apart therefore frequency received has to be lower than frequency transmitted.

  7. Jello_Submarine
    • one year ago
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    hmm, whatever I do, it still adds up to c. \[\lambda = 750000m\] if it helps

  8. modphysnoob
    • one year ago
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    I appreciate any help

  9. Jemurray3
    • one year ago
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    This is a Doppler shift problem. Just to make sure you have the idea right, we'll derive the equation here. Question: At what frequency does the spaceship see the observer receiving information? The distance between crests in the ship frame is simply the wavelength lambda. However, in the time it takes for the light to travel one wavelength, the earth will have moved. The time between crests is therefore: \[ t = \frac{\lambda + vt}{c} \implies ct = \lambda + vt \implies t = \frac{\lambda}{c-v} \] Since we are talking about electromagnetic radiation, lambda = c / frequency, so \[ t = \frac{c}{c-v} \frac{1}{f} = \frac{1}{1-v/c} \frac{1}{f} \] Now what about the observer? The proper time for the reception of information is calculated in the observer's frame, because the receiver is stationary there. Therefore, the observer measures this time to be \[t' = t/\gamma = \frac{\sqrt{1-v^2/c^2}}{1 - v/c} \frac{1}{f} = \sqrt{\frac{1+v/c}{1-v/c} } \frac{1}{f} \] Therefore, the frequency measured by the observer will be \[ f_o = \sqrt{\frac{1-\beta}{1+\beta} } f_s\] where \[f_0, f_s, \beta \] are the observer frequency, the source frequency, and v/c respectively. Plugging numbers in, that gives me ~82 Hz, a redshift, as expected.

  10. modphysnoob
    • one year ago
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    thanks you man, I will read it and ask if anything is unclear

  11. modphysnoob
    • one year ago
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    where did the beta come from?

  12. Jemurray3
    • one year ago
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    beta = v/c is just a shorthand that we use

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