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Homework Help: Question using Hubble's Constant

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Converting Hubble's constant in SI units of s-1 into kms-1 Mpc-1 (kilometers per second per megaparsec)
    Number given is 2.31 x 10^-18 s-1.

    2. Relevant equations
    H = v/d (where H is Hubble's constant, v = velocity and d = distance)

    3. The attempt at a solution
    I can see how you get the answer but it took me a while to see the reasoning for what to divide/multiply by. I reasoned that if you use H = v/d then 2.31 x 10^-18 s-1 can be re-written as 2.31 x 10^-18 ms-1 m-1 (meters per second per metre). I then treated the numerator and denominator separately. So to get from m/s to km per second you divide by 1000. Then to get from per metre to per megaparsec you multiply by the number of metres in a megaparsec. So overall you get 2.31 x 10^-18 x 3.09 x 10^22 / 1000 which gives 71.4 kms-1 Mpc-1. Using the same logic I don't understand something on the additional notes in the mark scheme for this question. I have uploaded the question and markscheme. They have shown an example of if you forget to convert hubble's constant into s-1 otherwise it will be in years-1. They then work this answer through the same way and give less marks for it however they say that the units are still kms-1 Mpc-1 at the end but that seems wrong to me. If you take hubble's constant in years-1 instead of seconds-1 and then multiply by the same number and divide by the same number surely its impossible to end up with the same final units? or am I missing something?

    Attached Files:

  2. jcsd
  3. May 16, 2015 #2


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    Gold Member

    I think they just give half mark because the reasonning behind it is somewhat correct and the answer if off "only" because it confuses units - but you are right, that second answer is just wrong (unless 2×10^9=71:wink:).

    Note that calculating the Hubble constant from the age of the universe is rather dubious, and that constant is actually observed, not derived.
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