Physicsandmathstutor Astronomers observe a quasar and measure its spectrum. A particular spectral line has a value of 279.8 nm when measured using a laboratory source. The equivalent line in the spectrum of this quasar is 382.5 nm. Calculate the speed of this quasar relative to the Earth, ignoring relativistic effects. Express your answer in meters/second.

v = -8.05 * 10⁷ ms⁻¹

Explanation:

Using the non-relativistic Doppler effect, which is the change in frequency of a wave as detected by an observer who's moving relative to the wave source. The shifted wavelength is:

λₓ = λ₀ (1 + v / c₀)

where  λ₀ is the wavelength from the source, v is the velocity of the source (the quasar) and, c₀ is the velocity of the wave which is equivalent to the speed of light, c.

Making v subject of the formula;

v = c₀ (λₓ / λ₀ - 1)

λₓ  = 279.8 * 10⁻⁹ m, λ₀ = 382.5 * 10⁻⁹ m, c = 3.0 * 10⁸

v = 3.0 * 10⁸ {(279.8 * 10⁻⁹ m / 382.5 * 10⁻⁹ m ) - 1}

v = -8.05 * 10⁷ ms⁻¹

The velocity of the quasar is negative because it is getting close to the Earth!

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