Suppose an astronomer discovers a distant quasar whose recessional velocity is 1/3 the speed of light. If Hubble's constant is 73.5 km/s per Mpc, how far away is the quasar?
A) 5 million Mpc
B) 145 Mpc
C) 73.5 Mpc
D) 1450 Mpc

Question

Quizplus · Accepted Answer

The distance to the quasar can be calculated using Hubble's Law, which states that the recessional velocity of a galaxy (v) is directly proportional to its distance from us (d), with the proportionality constant being Hubble's constant (H). The formula is v = H * d. Given that the recessional velocity is 1/3 the speed of light (c), we have v = c/3. The speed of light, c, is approximately 300,000 km/s. Therefore, v = 300,000 km/s / 3 = 100,000 km/s. With Hubble's constant H = 73.5 km/s per Mpc, we can rearrange the formula to solve for d: d = v / H = 100,000 km/s / 73.5 km/s per Mpc = approximately 1360 Mpc, which rounds to the closest given option, 1450 Mpc.

Suppose an Astronomer Discovers a Distant Quasar Whose Recessional Velocity