91影视

The wavelength of position of the calcium line,${位}_{source}=396.85\text{鈥塶尘}$

The wavelength of the shifted position of the calcium line,${位}_{obs}=396.58\text{鈥塶尘}$

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

$f_{obs}=f_{source}\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}cos胃}$ 鈥�(颈)

Here,$\mathit{c}$ is the speed of light,$\mathit{胃}$ 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.

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

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

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$\left(396.85\text{鈥塶尘}\right)$is greater than the observer wavelength$\left(396.58\text{鈥塶尘}\right)$. 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$180掳$.

Calculate the observer frequency,

Substitute $180掳$for$胃$into equation (i).

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

Solve further as,

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

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

$\begin{array}{c}\frac{c}{{位}_{obs}}=\frac{c}{{位}_{source}}\sqrt{\frac{(1+\frac{v}{c})}{(1鈭�\frac{v}{c})}}\\ \frac{1}{{位}_{obs}}=\frac{1}{{位}_{source}}\sqrt{\frac{(1+\frac{v}{c})}{(1鈭�\frac{v}{c})}}\\ {位}_{source}\sqrt{1鈭�\frac{v}{c}}={位}_{obs}\sqrt{1+\frac{v}{c}}\end{array}$

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

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

Solve further as,

$\frac{v}{c}=\frac{{位}_{source}^2鈭�{位}_{obs}^2}{{位}_{obs}^2+{位}_{source}^2}$

Substitute$396.85\text{鈥塶尘}$for ${位}_{source}$and$396.58\text{鈥塶尘}$for ${位}_{obs}$into above equation.

$\begin{array}{l}\frac{v}{c}=\frac{{\left(396.85\text{鈥塶尘}\right)}^2鈭�{\left(396.58\text{鈥塶尘}\right)}^2}{{\left(396.85\text{鈥塶尘}\right)}^2+{\left(396.58\text{鈥塶尘}\right)}^2}\\ \frac{v}{(3脳{10}^8\text{鈥尘/蝉})}=\frac{214.226}{314765.618}\\ v=\frac{214.226}{314765.618}脳3脳{10}^8\text{m/s}\\ v=204\text{鈥塳尘}/\text{s}\end{array}$

Hence the velocity at which galaxy is approaching us is $204\text{鈥塳尘}/\text{s}$.