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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.

Answer :

migueletorallam

Answer:

[tex]v=-80.5*10^{6}m/s[/tex]

Explanation:

For this particular problem, we are going to use the non-relativistic Doppler effect which, is the change in frequency of a wave concerning an observer who's moving relative to the wave source. The shifted wavelength is:

[tex]\lambda_{shifted}=\lambda_{0}\left(1+\dfrac{v}{c_{0}}\right)[/tex],

where [tex]\lambda_{0}[/tex] is the wavelenght as it comes from the source, [tex]v[/tex] is the velocity of the source (the quasar) and, [tex]c_{0}[/tex] is the velocity of the wave. In this case, the velocity of the wave is the speed of light, [tex]c_{0}=c=3*10^{8}\ m/s[/tex].

Solving for [tex]v[/tex],

[tex]v=c\left(\dfrac{\lambda_{shifted}}{\lambda_{0}}-1\right)[/tex],

now computing

[tex]v=3*10^{8}\left(\dfrac{279.8*10^{-9}}{382.5*10^{-9}}-1\right)[/tex],

[tex]v=-80.5*10^{6}m/s[/tex]

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

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