Question

By what factor would a star's characteristic wavelengths of light be shifted if it were moving away from Earth at $\mathbf{0}\mathbf{.}\mathbf{9}\mathit{c}$?

Short answer

The factor by which wavelength would be change is$4.358$ .

Step-by-step solution

Write the given data from the question.

The velocity of the source,$v=0.9c$

Determine the formulas to calculate the factor of characteristics wavelength.

The expression between the frequency, wavelength and speed of light is given as follows.

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{c}}{\mathbf{f}}$

Here,$\mathit{\lambda }$is the wavelength, $c$is the speed of light and $\mathit{f}$is the frequency.

The relationship between the observer and source frequency is given as follows.

${\mathit{f}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\sqrt{\frac{\mathbf{1}\mathbf{-}\frac{\mathbf{v}}{\mathbf{c}}}{\mathbf{1}\mathbf{+}\frac{\mathbf{v}}{\mathbf{c}}}}$

Calculate the factor of characteristics wavelength. 

Determine the wavelength equation for the observer.

${\lambda }_{obs}=\frac{c}{f_{obs}}$

Substitute $f_{source}\sqrt{\frac{1-v/c}{1+v/c}}$for$f_{obs}$ into above equation.

$\begin{array}{l}{\lambda }_{obs}=\frac{c}{f_{source}\sqrt{\frac{1-v/c}{1+v/c}}}\\ {\lambda }_{obs}=\frac{c}{f_{source}}\sqrt{\frac{1+v/c}{1-v/c}}\end{array}$

Substitute ${\lambda }_{source}$for$\frac{c}{f_{source}}$ into above equation.

$\begin{array}{c}{\lambda }_{obs}={\lambda }_{source}\sqrt{\frac{1+v/c}{1-v/c}}\\ \frac{{\lambda }_{obs}}{{\lambda }_{source}}=\sqrt{\frac{1+v/c}{1-v/c}}\end{array}$

Substitute $0.9c$for$v$ into above equation.

$\begin{array}{l}\frac{{\lambda }_{obs}}{{\lambda }_{source}}=\sqrt{\frac{1+0.9c/c}{1-0.9c/c}}\\ \frac{{\lambda }_{obs}}{{\lambda }_{source}}=\sqrt{\frac{1+0.9}{1-0.9}}\\ \frac{{\lambda }_{obs}}{{\lambda }_{source}}=\sqrt{\frac{1.9}{0.1}}\\ \frac{{\lambda }_{obs}}{{\lambda }_{source}}=4.358\end{array}$

Hence the factor by which wavelength would be change is$4.358$.