EM Doppler effect: Two spaceships leave a planet in exactly opposite directions.

Question

EM Doppler effect: Two spaceships leave a planet in exactly opposite directions. Ship A has a speed of 2.0 times 10^7 m/s while ship B has a speed of 3.0 times 10 m/s (both speeds are measured with respect to the planet). Ship A sends a radio signal of a wavelength of 8.0 meters to ship B. a. What is the relative speed of one ship with respect to the other? b. What frequency does ship A send its radio signal at? c. What frequency does ship B receive the signal at? d. By what percentage is the frequency observed by (B) different than sent by (A)

Explanation / Answer

Ans:-

a.Relativistic formula

V(b/a) = (Va + Vb)/[1 – VaVb/c²],

where:

V(b/a) - velocity of particle B relative to particle A

V(b/a) =( 2*10^7 + 3*10^7)/[1-(2*10^7*3*10^7/(3*10^8)^2)]

V(b/a) = 5.034*10^7m/s

b. f = Va/ = 2*10^7/8 =2.5*10^6Hz

c. f= Vb/ = 3*10^7/8 = 3.75*10^6Hz                          

D. 3.75*10^6 -2.5*10^6 = 1.25*10^4%

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