91Ó°ÊÓ

The percentage of wavelength greater than corresponding light from Earth sources is $\mathrm{x}=0.4\%=0.004$

The wavelength of the light shift toward increased wavelength if the source is moving away from the observer. If the wavelength shifts towards decreased wavelength then source moves toward observer.

(a)

The radial speed of galaxy relative to Earth is given as:

$\mathrm{v}=\left(\frac{{\left(1+\mathrm{x}\right)}^2-1}{{\left(1+\mathrm{x}\right)}^2+1}\right)\mathrm{c}$

Here, c is the speed of light and its value is $3×{10}^8\mathrm{m}/\mathrm{s}$.

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{v}=\left(\frac{{\left(1+0.004\right)}^2-1}{{\left(1+0.004\right)}^2+1}\right)\mathrm{c}\left(\frac{3×{10}^8\mathrm{m}/\mathrm{s}}{\mathrm{c}}\right) \\ \mathrm{v}=1.2×{10}^6\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the radial speed of galaxy relative to Earth is $1.2×{10}^6\mathrm{m}/\mathrm{s}$.

(b)

As the wavelengths of the light are observed longer than corresponding wavelength on Earth so the light is shifting toward larger wavelength. The galaxy is moving away from the Earth so galaxy is receding from Earth.

Therefore, the galaxy is receding from Earth.