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

Step-by-step 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=(V)/(H_(o))` (2)

`D=(30,000 km/s)/(70 km/s/Mpc)` (3)

Finally:

`D=428.57 Mpc`