A distant spaceship carrying a red headlight, emitting light with wavelength

Question

600nm, is heading towards the earth. Observers on the earth detect the light from the
headlight as ultraviolet light with wavelength 200nm. What is the velocity of the
spaceship?
b) As the space ship gets closer to earth, it becomes clear that it is not heading straight at the earth.(Ignore this in part (a). It passes the earth at the speed you have calculated in part (a). What frequency
of light is seen by earth observers when the spaceship is just passing earth? (That is when the velocity
of the spaceship is perpendicular to the line of sight from earth).

Answer 1

L = V/Fs = 600*10^-9 m.
300*10^6/Fs = 600*10^-9
Fs = 300*10^6/600*10^-9 = 5.*10^14 Hz =
Freq. emitted by spaceship.

a. L = 300*10^6/Fr = 200*10^-9
Fr = 300*10^6/200*10^-9 = 1.5*10^15 Hz.
= Freq. observed by the receivers.

Fr = ((Vl+Vr)/(Vl-Vs))*Fs = 1.5*10^15
((3*10^8+0)/(3*10^8-Vs)*5*10^14 =
1.5*10^15 Hz.
15*10^22/(3*10^8-Vs) = 1.5*10^15
-1.5*10^15Vs + 45*10^22 = 15*10^22
-1.5*10^15Vs = -30*10^22
Vs = 20*10^7 = 2.0*10^8 m/s = Velocity of spaceship.

b. Fr = ((Vl-Vr)/(Vl+Vs))*Fs
Fr =((3*10^8-0)/(3*10^8+2*10^8))*5*10^14
Fr = 15*10^22/(5*10^8) = 3*10^14 Hz =
freq. seen by observers as spaceship
passes earth.