91Ó°ÊÓ

The recession velocity for a galaxy is given by,

\(v = {H_o}d\)

Here\({H_o}\)is the Hubble constant and\(d\)is the distance to the galaxy.

Calculating the relative velocity of a galaxy using Hubble's law as:

\(v = {H_0}d\)....... (i)

The value of\({H_0} = {\rm{ 20 km}}{{\rm{s}}^{{\rm{ - 1}}}}{\rm{ Ml}}{{\rm{y}}^{{\rm{ - 1}}}}\)is the Hubble constant. Here\(d\)is the distance to the galaxy.

Putting the values into the first equation and then we get:

\(\begin{array}{c}v{\rm{ }} = {\rm{ (20 km}}{{\rm{s}}^{{\rm{ - 1}}}}{\rm{ Ml}}{{\rm{y}}^{{\rm{ - 1}}}}{\rm{)(1 \times 1}}{{\rm{0}}^{\rm{4}}}{\rm{ Mly)}}\\ = {\rm{ 2 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

Therefore, the speed of the galaxy is \({\rm{2 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}{{\rm{s}}^{{\rm{ - 1}}}}\).

To evaluate the fraction of speed of light we divide and multiply the value by\({\rm{c}}\)as:

\(\begin{array}{c}v = \frac{v}{c}c\\ = \frac{{{\rm{(2 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}{{\rm{s}}^{{\rm{ - 1}}}}{\rm{)}}}}{{{\rm{(3 \times 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}{{\rm{s}}^{{\rm{ - 1}}}}{\rm{)}}}}{\rm{ \times c}}\\ = {\rm{0}}{\rm{.67c}}\end{array}\)

Therefore, the speed of the galaxy is \(0.67\) times that of the speed of light.