When discussing the Doppler Effect in sound, we have relationships for the obser

Question

When discussing the Doppler Effect in sound, we have relationships for the observed frequency if the source was moving and if the observer was moving. f_0 = f_s(1/(1plusminus v_s/v_wave)) These same equations work for electromagnetic waves. They can also be simplified if the relative velocity between the source and the observer (v_re|) is much less than the speed of the waves. The speed of the electromagnetic waves is the speed of light, c = 3.00 times 10^8 m/s. The formulas can be simplified to: /, /,(1 plusminus v_rel/c), if v_rel c. where f_s is the frequency of the source and f_0 is the observed frequency. Suppose a space probe emits a radio signal at a frequency of 3.900000 MHz (Megahertz). It travels away from Earth with a speed of 2.200x10^4 m/s. What is the frequency of the signal when it is detected on Earth?

Explanation / Answer

from the equation given above

f = 3.9 ( 1 - 2.2(10000)/c)

f= 3.9 x0.9999266

f = 3.899714 Hz

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