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17 June, 12:58

The hydrogen atoms in a star are also moving at high velocity because of the random motions caused by their high temperature. As a result, each atom is Doppler shifted a little bit differently, leading to a finite width of each spectral line, such as the 656.46-nm line we were just discussing. For a star like our sun, this leads to a finite width of the spectral lines of roughly Δλ=0.04nm. If our instruments can only resolve to this accuracy, what is the lowest speed V, greater than 0, that we can measure a star to be moving?

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  1. 17 June, 13:17
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    the lowest speed that can be measured is ± 1,828 10⁴ m/s

    Explanation:

    The expression for the relativistic doppler effect is

    f₀ = fs √ (1-v / c) / (1 + v / c)

    Where f₀ and fs are the observed and emitted frequency respectively, v is the velocity of the wave or particle

    Let's use the relationship of the speed of the wave with its frequency and wavelength

    c = λ f

    f = c / λ

    Let's replace

    c / λ₀ = c / λs √ (1-v / c) / (1 + v / c)

    λs/λ₀ = √ (1-v / c) / (1 + v / c)

    We cleared the speed

    (λs / λ₀) ² = (1-v / c) / (1 + v / c)

    (λs/λ₀) ²2 = (c-v) / c + v)

    (c + v) (λs / λ₀) ² = c-v

    v [1 + (λs / λ₀) ²] = c [1 - (λs / λ₀) ²]

    v = c [1 - (λs / λ₀) ²] / [1 + (λs / λ₀) ²]

    v = c (λ₀² - λs²) / (λ₀² + λs²)

    Let's calculate the speeds for the two possible cases

    When the wavelength increases

    λs = 656.46 + 0.04 = 656.50 nm

    λ₀ = 656.46 nm

    (λ₀² - λs²) = 656.5² - 656.46² = 52.518

    (λ₀² + λs²) = 656.5² + 656.46² = 861931.98

    v = 3 10⁸ 52.518 / 861931.98

    v = 1.828 10⁴ m / s

    Case 2 the wavelength decreases

    λ₀ = 656.46 - 0.04 = 656.42 nm

    (λ₀² - λs²) = 656.42² - 656.46² = - 52.5151

    (λ₀² + λs²) = 656.42² + 656.46² = 861826.948

    v = 3 10⁸ ( - 52.5151) / 861826.948

    v = - 1,828 10⁴ m / s

    The negative sign indicates that the Source moves away from the observer
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