1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Units for Kepler's 3rd Law

  1. Oct 10, 2015 #1
    1. The problem statement, all variables and given/known data
    Kepler's 3rd law can be written as ##p^2=a^3##
    If p, the period, is given in years, what are the units for a?

    2. Relevant equations
    n/a

    3. The attempt at a solution
    The answer is AU. Is there a proof for this or is this merely a definition? Thank you.
     
  2. jcsd
  3. Oct 10, 2015 #2
    Let's consider the Earth.

    If we choose the "year" for period, then the value of ## a ## also have to be ## 1 ## with some unit and the only thing is ## 1 \text{AU}. ## You may think that this is the definition for ## \text{AU}. ##

    BTW, be careful about dimensions of each side. ## (\text{Year})^2 \neq (\text{AU})^3, ## and so usually we write the law as ## p^2 \propto a^3. ##
     
  4. Oct 10, 2015 #3

    Krylov

    User Avatar
    Science Advisor
    Education Advisor

    Please pardon my ignorance, but I would feel that, strictly speaking, we cannot know the units of ##a## as long as we do not know the units of the proportionality constant ##C## in the relation ## \frac{a^3}{p^2} = C##? And vice versa, of course.
     
  5. Oct 10, 2015 #4
    Otherwise, you may fix first the units of ## a## and ## p##, then you can get the constant ## C ## with some unit.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Units for Kepler's 3rd Law
  1. Kepler's 3rd Law (Replies: 6)

  2. Keplers 3rd Law (Replies: 5)

  3. Keplers 3rd Law (Replies: 11)

Loading...