Question

A particular galaxy is observed to have a recessional velocity (away from Earth) of 30,000 km/s. Assuming the Hubble constant to be 70 km/s/Mpc, Hubble's law gives the distance to this galaxy to be ____ Mpc.

Answer 1

Answer: 428.57 Mpc

Explanation:

Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum, and noted that all galaxies are "moving away from each other with a speed that increases with distance", and enunciated the now called Hubble–Lemaître Law.  

This is mathematically expressed as:

$V=H_{o}D$ (1)

Where:

$V=30,000 km/s$ is the recession velocity of the galaxy

$H_{o}=70 km/s/Mpc$ is the Hubble constant  

$D$ is the distance

Isolating $D$ from (1):

$D=\frac{V}{H_{o}}$ (2)

$D=\frac{30,000 km/s}{70 km/s/Mpc}$ (3)

Finally:

$D=428.57 Mpc$