Relativistic Doppler effect

  1. fluidistic

    fluidistic 3,313
    Gold Member

    1. The problem statement, all variables and given/known data
    A car is getting closer to a radar as a speed of 135 km/h. If the radar works at a [tex]2 \times 10 ^{9} Hz[/tex], what difference of frequency is observed for the radar?


    2. Relevant equations
    [tex]\mu=\frac{\mu _0 \sqrt {1- \frac{v^2}{c^2}}}{1-\frac{v}{c}}[/tex]


    3. The attempt at a solution
    I converted 135 km/h into m/s, which gave me [tex]\frac { 1350 m }{36 s}[/tex].
    Then I applied the formula and [tex]\mu - \mu _0[/tex] gave me [tex]1.024 \times 10^{13} Hz[/tex] which seems WAY too big to be realistic. Am I missing something? Is it a wrong formula?!
     
  2. jcsd
  3. kuruman

    kuruman 3,448
    Homework Helper
    Gold Member

    Chegg
    The formula is correct and I agree the number is way too large. Can you show how you got it? A word of caution: v/c is too small to plug in a calculator and expect something other than μ0, i.e. no frequency shift. I suggest that you try a Taylor expansion for small values of v/c.
     
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