Question

A star is moving away from an observer at 1% of the speed of light. At what wavelength would the observer find an emission line which would occur at a wavelength of 6000 Angstroms if the star were at rest?

Answer 1

λ = 5940 Angstroms

Step-by-step explanation:

This is an exercise of the relativistic Doppler effect

f’= f √((1- v / c) / (1 + v / c))

Where the speed in between the strr and the observer is positive if they move away

Let's use the relationship

c = λ f

f = c /λ

We replace

c /λ’ = c /λ √ ((1- v / c) / (1 + v / c))

λ = λ’ √ ((1- v / c) / (1 + v / c))

Let's calculate

v = 0.01 c

v = 0.01 3 10⁸

v= 3 10⁶ m / s

λ = 6000 √ [(1- 3 10⁶/3 10⁸) / (1+ 3 10⁶/3 10⁸)]

λ = 6000 √ [0.99 / 1.01]

λ = 5940 Angstroms