If Hubble's constant is taken to be 70 km/s/Mpc, and a quasar is found with a radial velocity of 0.95 times the speed of light, how far away is the quasar
A)4100 Mpc
B)0.014 Mpc
C)67 Mpc
D)74 Mpc
E)300,000 Mpc

Question

Quizplus · Accepted Answer

Hubble's Law states that the velocity at which an object is receding from us is proportional to its distance, and this proportionality constant is known as Hubble's constant (H0). The equation for Hubble's Law is:

v = H0 x d

Where:
v = radial velocity of the object
d = distance to the object

In this question, we are given the value of Hubble's constant as 70 km/s/Mpc and the radial velocity of the quasar as 0.95 times the speed of light, which is approximately 285,000 km/s. We can convert this to km/s/Mpc by dividing it by the speed of light, which is approximately 3 x 10^5 km/s. Therefore:

v = 0.95 x 3 x 10^5 / 1 Mpc = 285 km/s/Mpc

We can now use Hubble's Law to find the distance to the quasar:

d = v / H0 = 285 / 70 = 4.07 Mpc

Therefore, the answer is A) 4100 Mpc, which is the closest option to the calculated distance.

If Hubble's Constant Is Taken to Be 70 Km/s/Mpc, and a Quasar