Special Relativity - Doppler Effect

The Lyman alpha spectral line of Hydrogen (λ = 122 nanometers) differs by 1.8 × 10⁻¹² meter in spectra taken at opposite ends of the Sun’s equator. What is the speed of a particle on the equator due to the Sun’s rotation, in kilometers per second?

A. 0.22
B. 2.2
C. 22
D. 220
E. 2200

(GR9677 #60)

Solution:

The problem deals with v ≪ c since the  answer is in km/s.

For v ≪ c, Redshift parameter, z = Δλ/λ ≈ v/c

v =  cΔλ/λ

Given:
Δλ = 1.8 × 10⁻¹² m 
λ = 122 nm = 1.22 × 10⁻⁷ m
c = 3 × 10⁸ m/s

v = (3 × 10⁸)(1.8 × 10⁻¹²)/(1.22 × 10⁻⁷) 

= (5.4/1.22) × 10³ m/s  
= (5.4/1.22) km/s ≈ 2.2 km/s

Answer: B

Sound and Wave - Doppler Effect

A source of 1-kilohertz sound is moving straight toward you at a speed 0.9 times the speed of sound. The frequency you receive is 

A. 0.1 kHz
B. 0.5 kHz
C. 1.1 kHz
D. 1.9 kHz
E. 10 kHz

(GR8677 #12)

Solution:

Moving toward observer
→ f increases
→ f  > 1 kHz
→ A and B are FALSE

Speed 0.9 times the speed of sound
→ f increases greatly
→ E. TRUE

Answer: E

Calculation:





+ sign = receding
− sign = approaching

Special Relativity - Doppler Effect

The ultraviolet Lyman alpha line of hydrogen with wavelength 121.5 nanometers is emitted by an astronomical object. An observer on earth measures the wavelength of the light received from the object to be 607.5 nanometers. The observer can conclude that the object is moving with radial velocity of

A. m/s toward Earth
B. m/s toward Earth
C. m/s away from Earth
D. m/s away from Earth
E. m/s away from Earth

(GR0177 #71)

Solution:

Given:
λ₀ = 121.5 nm
λ = 607.5 nm

λ  λ₀ → the object is moving away (receding)

(A) and (B) are FALSE.

(E) is FALSE since v is larger than c.

Doppler Effect for light:


with

and
+ sign = approaching
− sign = receding

Since the object and the source are receding:

λ = λ₀(1 + β)^1/2/(1 − β)^1/2
λ/λ₀ = (1 + β)^1/2/(1 − β)^1/2
607.5/121.5 = 5 = (1 + β)^1/2/(1 − β)^1/2
25(1 − β) = (1 + β)
24 = 26β
β = 24/26 = 12/13 = v_source/c
v_source = (12/13)c = (12/13)(3 × 10⁸) = (36/13) × 10⁸ = 2.76 × 10⁸ m/s

Answer: D

Special Relativity - Doppler Effect

A galaxy in the constellation Ursa Major is receding from the earth at 15.000 km/s. If one of the character wavelength of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by the astronomers on the earth?

Solution:

Given:
v = 15.000 km/s = 1.5 × 10⁷ m/s
c = 3 × 10⁸ m/s
λ₀ = 550 nm
λ = ?

β =  v / c =  − 0.05 ( − sign for receding)

Doppler Effect for light:



λ = 550 nm × √(1 + 0.05) / √(1 − 0.05) = 578 nm