A distant spaceship carrying a red headlight, emitting light with wavelength
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).
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.
To find the velocity of the spaceship, we can use the principle of the Doppler effect. The Doppler effect states that when a source of waves (in this case, light) is moving towards an observer, the wavelengths of the waves are compressed, causing an increase in frequency. Conversely, when the source is moving away from the observer, the wavelengths are stretched, causing a decrease in frequency.
Let's start by calculating the speed at which light is traveling:
The speed of light, denoted as c, is approximately 3 x 10^8 meters per second or 300,000 km/s.
Now, we can use the Doppler effect equation for a moving source of light:
λ_observed = λ_emitted * (c + v) / (c - v)
where:
λ_observed = observed wavelength on Earth (200 nm)
λ_emitted = emitted wavelength by the spaceship (600 nm)
v = velocity of the spaceship
Plugging in the values:
200 nm = 600 nm * (3 x 10^8 m/s + v) / (3 x 10^8 m/s - v)
Next, we can solve for v. Let's start by cross-multiplying:
200 nm * (3 x 10^8 m/s - v) = 600 nm * (3 x 10^8 m/s + v)
Now, let's convert the wavelengths from nanometers to meters by dividing by 10^9:
(200 nm * 10^-9) * (3 x 10^8 m/s - v) = (600 nm * 10^-9) * (3 x 10^8 m/s + v)
Simplifying further:
(200 * 10^-9) * (3 x 10^8 - v) = (600 * 10^-9) * (3 x 10^8 + v)
Now, let's multiply the terms:
(200 * 10^-9) * (3 x 10^8) - (200 * 10^-9) * v = (600 * 10^-9) * (3 x 10^8) + (600 * 10^-9) * v
Simplifying even further:
600 * 10^-1 - (200 * 10^-9) * v = 1800 * 10^-1 + (600 * 10^-9) * v
Now, let's collect the v terms on one side of the equation:
(200 * 10^-9) * v + (600 * 10^-9) * v = 1800 * 10^-1 - 600 * 10^-1
Combining like terms:
(800 * 10^-9) * v = 1200 * 10^-1
Now, let's divide both sides of the equation by (800 * 10^-9) to isolate v:
v = (1200 * 10^-1) / (800 * 10^-9)
Simplifying:
v = 1.5 x 10^9 m/s
Therefore, the velocity of the spaceship is 1.5 x 10^9 meters per second.
b) When the spaceship passes the Earth at the speed calculated in part (a), the velocity is perpendicular to the line of sight from Earth. In this case, there would be no Doppler effect, and the frequency of light observed remains the same as the emitted frequency.
The frequency of light is related to the wavelength by the equation:
f = c / λ
where:
f = frequency
c = speed of light (3 x 10^8 m/s)
λ = wavelength (600 nm in this case)
Plugging in the values:
f = (3 x 10^8 m/s) / (600 nm * 10^-9)
Simplifying:
f = (3 x 10^8 m/s) / (600 * 10^-9)
f = 5 x 10^14 Hz
Therefore, when the spaceship is passing Earth, the frequency of light observed by Earth observers is 5 x 10^14 Hz.