Question

If I measure the recession velocities of 2 galaxies and galaxy A has a velocity twice that of galaxy B, how far away is B if A is 200 Mpc away? (Remember that V = H0 x D)

Answer 1

Answer: 100Mpc

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` is the approximate recession velocity of the galaxy

`H_(o)` is the Hubble constant

`D` is the distance

Now, we have two galaxies A and B ande we know the following:

Galaxy A has a velocity twice that of galaxy B

`V_(A)=2V_(B)` (2)

Galaxy A is 200 Mpc away

`D_(A)=200Mpc` (3)

According to Hubble Law:

`V_(A)=H_(o)D_(A)` (4)

`V_(A)=H_(o)200Mpc` (5)

Substituting (2) in (5):

`2V_(B)=H_(o)200Mpc` (6)

`V_(B)=(H_(o)200Mpc)/(2)` (7)

`V_(B)=H_(o)100Mpc` (8)

If
`V_(B)=H_(o)D_(B)` (9)

`D_(B)=(V_(B))/(H_(o))` (10)

Substituting (8) in (11):

`D_(B)=(H_(o)100Mpc)/(H_(o))` (11)

Finally:

`D_(B)=100Mpc`