If a quasar has a recessional velocity of $2.7 × 10^5$ km/s and is determined to be 3600 Mpc, what is the Hubble constant based on this quasar alone Hint: Note that the units of your result should be (km/s)/Mpc to see what arithmetic process you need to perform.
A)50 km/s/Mpc
B)$9.7 × 10^8$ (km/s)Mpc
C)75 km/s/Mpc
D)0.013 km/s/Mpc
E)273,600 km/s/Mpc

Question

Quizplus · Accepted Answer

Using Hubble's law, we have:

v = H0d

where v is the recessional velocity of the quasar and d is its distance in Mpc. Rearranging this equation gives:

H0 = v/d

Substituting the given values, we get:

H0 = (2.7 × 10^5 km/s) / (3600 Mpc)

H0 ≈ 75 km/s/Mpc

Therefore, the best choice is C.

If a Quasar Has a Recessional Velocity Of 2.7×1052.7 × 10^52.7×105

If a Quasar Has a Recessional Velocity Of $2.7 × 10^5$