91Ó°ÊÓ

The speed of recession of a galaxy is proportional to its distance , and this relationship is known as Hubble law; represented as:

$\mathit{ν}\mathbf{=}{\mathit{H}}_{\mathbf{0}}\mathit{r}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(1\right)$

And,${\mathit{H}}_{\mathbf{0}}$ is known as Hubble constant.

The relationship between the wavelength of light${\mathit{λ}}_{\mathbf{0}}$calculated on the earth from the source at speed v and the wavelength${\mathit{λ}}_{\mathbf{s}}$calculated in the rest frame of the source when it was emitted is given by the expression:

${\mathit{λ}}_{\mathbf{0}}\mathbf{=}{\mathit{λ}}_{\mathbf{s}}\sqrt{\frac{\mathbf{c}\mathbf{+}\mathbf{v}}{\mathbf{c}\mathbf{-}\mathbf{v}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(2\right)$

Also,

$\mathit{v}\mathbf{=}\left[\frac{{\left({\mathrm{λ}}_0/{\mathrm{λ}}_s\right)}^2-1}{{\left({\mathrm{λ}}_0/{\mathrm{λ}}_s\right)}^2-1}\right]\mathit{c}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(3\right)$

a)

The redshifted ratio is$\frac{{\mathrm{λ}}_0}{{\mathrm{λ}}_s}$ . It can be calculated as:

$$\begin{gathered} \frac{{\mathrm{λ}}_0}{{\mathrm{λ}}_s}=\frac{658.5}{590} \\ =1.1161 \end{gathered}$$

From equation ( 3 )

$$\begin{gathered} v=\left[\frac{{\left({\mathrm{λ}}_0/{\mathrm{λ}}_s\right)}^2-1}{{\left({\mathrm{λ}}_0/{\mathrm{λ}}_s\right)}^2-1}\right]c \\ =\left[\frac{{\left(1.1161\right)}^2-1}{{\left(1.1161\right)}^2+1}\right]\left(3×{10}^8\right) \\ =0.328×{10}^{8}m/s \end{gathered}$$

Hence, the speed of the galaxy receding from the earth is$0.328×{10}^{8}m/s$ .

b)

From the equation ( 1 ) :

$v=H_{0}r$

Where, $H_0=2.18×{10}^{18}{s}^{-1}$

Now,

$$\begin{gathered} r=\frac{v}{H_0} \\ =\frac{0.328×{10}^8}{2.18×{10}^{-18}} \\ =1.51×{10}^{25}m \end{gathered}$$

Hence, the distance of the galaxy from the earth is$1.51×{10}^{25}m$ .