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

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)

$V_{B}=\frac{H_{o}200Mpc}{2}$ (7)

$V_{B}=H_{o}100Mpc$ (8)

If $V_{B}=H_{o}D_{B}$  (9)

$D_{B}=\frac{V_{B}}{H_{o}}$ (10)

Substituting (8) in (11):

$D_{B}=\frac{H_{o}100Mpc}{H_{o}}$ (11)

Finally:

$D_{B}=100Mpc$