Question

The light from galaxy NGC 221 consists of a recognizable spectrum of wavelengths. However, all are shifted towards the shorter-wavelength end of the spectrum. In particular, the calcium “line” ordinarily observed at $\mathbf{396}\mathbf{.}\mathbf{85}\mathit{n}\mathit{m}$is observed at $\mathbf{396}\mathbf{.}\mathbf{58}\mathit{n}\mathit{m}$. Is this galaxy moving toward or away from Earth? At what speed?

Short answer

The galaxy is approaching us and velocity at which galaxy is approaching us is $204\text{\hspace{0.17em}km}/\text{s}$.

Step-by-step solution

Write the given data from the question.

The wavelength of position of the calcium line,${\lambda }_{source}=396.85\text{\hspace{0.17em}nm}$

The wavelength of the shifted position of the calcium line,${\lambda }_{obs}=396.58\text{\hspace{0.17em}nm}$

Determine the equations to calculate the speed of the source.

The expression to calculate the observer frequency in terms of source frequency, speed and speed of light is given as follows.

$f_{obs}=f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}cos\theta }$ …(i)

Here,$\mathit{c}$ is the speed of light,$\mathit{\theta }$ is the angle between the source of light and observer and $\mathit{v}$is the velocity of the source.

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

Calculate the speed of the source and direction of moving of the galaxy.

According the Doppler effect, when the source is moving towards the observer, then the wavelength shifted toward the lower range of wavelength region.

In the given case the source wavelength$\left(396.85\text{\hspace{0.17em}nm}\right)$is greater than the observer wavelength$\left(396.58\text{\hspace{0.17em}nm}\right)$. thus light has been blue shifted. Therefore,galaxy is approaching us.

Hence the galaxy is approaching us.

The angle between the source and observer when the source moving towards the observe is equal to$180°$.

Calculate the observer frequency,

Substitute $180°$for$\theta$into equation (i).

$\begin{array}{l}{f}_{obs}=f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos \left(180\right)}\\ f_{obs}=f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}(-1)}\\ f_{obs}=f_{source}\frac{\sqrt{(1-\frac{v}{c})(1+\frac{v}{c})}}{1-\frac{v}{c}}\\ f_{obs}=f_{source}\frac{\sqrt{(1-\frac{v}{c})(1+\frac{v}{c})}}{\sqrt{{(1-\frac{v}{c})}^2}}\end{array}$

Solve further as,

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

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

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

Take a square of both the sides of the above equation.

$\begin{array}{c}{\lambda }_{source}^2(1-\frac{v}{c})={\lambda }_{obs}^2(1+\frac{v}{c})\\ {\lambda }_{source}^2-{\lambda }_{source}^2\frac{v}{c}={\lambda }_{obs}^2+{\lambda }_{obs}^2\frac{v}{c}\\ {\lambda }_{obs}^2\frac{v}{c}+{\lambda }_{source}^2\frac{v}{c}={\lambda }_{source}^2-{\lambda }_{obs}^2\\ \frac{v}{c}({\lambda }_{obs}^2+{\lambda }_{source}^2)={\lambda }_{source}^2-{\lambda }_{obs}^2\end{array}$

Solve further as,

$\frac{v}{c}=\frac{{\lambda }_{source}^2-{\lambda }_{obs}^2}{{\lambda }_{obs}^2+{\lambda }_{source}^2}$

Substitute$396.85\text{\hspace{0.17em}nm}$for ${\lambda }_{source}$and$396.58\text{\hspace{0.17em}nm}$for ${\lambda }_{obs}$into above equation.

$\begin{array}{l}\frac{v}{c}=\frac{{\left(396.85\text{\hspace{0.17em}nm}\right)}^2-{\left(396.58\text{\hspace{0.17em}nm}\right)}^2}{{\left(396.85\text{\hspace{0.17em}nm}\right)}^2+{\left(396.58\text{\hspace{0.17em}nm}\right)}^2}\\ \frac{v}{(3\times {10}^8\text{\hspace{0.17em}m/s})}=\frac{214.226}{314765.618}\\ v=\frac{214.226}{314765.618}\times 3\times {10}^8\text{m/s}\\ v=204\text{\hspace{0.17em}km}/\text{s}\end{array}$

Hence the velocity at which galaxy is approaching us is $204\text{\hspace{0.17em}km}/\text{s}$.