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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.

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cryssatemp

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:

[tex]V=H_{o}D[/tex] (1)

Where:

[tex]V=30,000 km/s[/tex] is the recession velocity of the galaxy

[tex]H_{o}=70 km/s/Mpc[/tex] is the Hubble constant  

[tex]D[/tex] is the distance

Isolating [tex]D[/tex] from (1):

[tex]D=\frac{V}{H_{o}}[/tex] (2)

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

Finally:

[tex]D=428.57 Mpc[/tex]

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