Question

The average radial velocity of galaxies in the Hercules cluster is 10,800 km/s. (a) using H0 = 73 km/s/Mpc, find the distance to the Hercules cluster in Mpc. (b) How would your answer to (a) differ if the Hubble constant had a smaller value?

Answer 1

Answer:

a) 147.95 Mpc

Explanation:

Using Hubble's formula

$v = H_od$

where;

v = radical velocity

$H_o$ = Hubble's constant

d = distance

Given that:

The average radial velocity of galaxies in the Hercules cluster v =  10,800 km/s

Also using $H_o$ = 73 km/s/ Mpc ; we  make distance d the subject of the formula:

Then distance d can be written as:

$d = \frac{v}{H_o}$

$d = \frac{10,800 \ km/s}{73 \ km/s/Mpc}$

d = 147.95 Mpc

b)

Now, if the Hubble constant had a smaller value, then for a  given velocity the distance to the galaxy will increase because distance d is inversely proportional to $H_o$

i.e  

d  ∝ $\frac{1}{H_0}$