Respuesta :
Answer:
a) wavelengths seen decrease , b) λ’/λ = 0.9998
Explanation:
a) This is a relativistic doppler effect problem that is described by
f’= f₀ √ (1+ v / c) / √ (1- v / c)
The speed of light be related to wavelength and frequency
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)]
This expression gives us the wavelength depends on the speed of the premiere with respect to the Earth, as the estuary approaches the Earth the speed is positive
We can see that at the root it is less than 1 whereby the wavelengths seen decrease
b) we reduce the speed to m / s
v = 30 km / s = 3 10⁴ m / s
Change is
λ’/λ = √ [(1-v / c) / (1 + v / c)]
λ’/ λ = √ [(1- 3 10⁴/3 10⁸) / (1 + 3 10⁴/3 10⁸)]
λ’/ λ = √ (0.9999 / 1.0001)
λ’/λ = 0.9998
The wavelength of the given spectrum will be less than that of the star at rest when measured from the earth.
What is the Doppler effect?
An increase or decrease in wavelength as source and observer move relative to each other.
It can be given by,
[tex]f_{s}=f_{o} \sqrt {\dfrac{1+\frac vc}{1-\frac vc} }[/tex]
Where,
[tex]f_{o}[/tex] - observer frequency of sound
[tex]v[/tex] - speed of sound waves
[tex]f_{s}[/tex] - actual frequency of sound waves
c - speed of light
Since, [tex]f = \dfrac c\lambda[/tex]
Thus,
[tex]\dfrac c\lambda_s= \dfrac c\lambda_o\sqrt {\dfrac{1+\frac vc}{1-\frac vc} }\\\\\lambda_s=\lambda_o\sqrt {\dfrac{1-\frac vc}{1+\frac vc} }\\\\[/tex]
Form the equation, the root is less than 1.
Therefore, the wavelength of the given spectrum will be less than that of the star at rest.
Learn more about the Doppler effect:
https://brainly.com/question/5838520
