1. Limited time only! Sign up for a free 30min personal 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!

Kepler's third law: units

  1. Jun 10, 2014 #1
    For two two bodies of mass M1 and M2 in circular orbits of radius a1, a2 about their common centre of mass, the Newtonian modification of Kepler's third law is
    a3/P2=G(M1+M2)/4π2.
    Where a=a1+a2.

    The problem is that I have been told that when using the units of years, solar masses and astronomical units, this reduces to
    a3/P2=M1+M2.
    I'm not sure how to show this is true, and find it quite strange that such a unit change could manage to perfectly cancel out G/4π2. The internet and textbooks don't seem to be very helpful about this so I was hoping somebody could point me in the right way, thanks!
     
  2. jcsd
  3. Jun 10, 2014 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Well, these types of problems don't solve themselves. Have you tried plugging in the different units to see if you obtain the supposed simplification?
     
  4. Jun 10, 2014 #3
    Try applying your formula for the earth's orbit and you will understand that miraculous cancelation
     
  5. Jun 10, 2014 #4
    Oh wow, so simple (and slightly obvious). Thanks!

    As it happens I'm having more astrophysics unit problems which I was hoping somebody could help with, which appear in the following problem:
    Two stars in a binary system have a separation of s=3'' and a trigonometric parallax of p=0.1''. They have a orbital period of 30 years and the secondary star is five times further from the centre of mass than the primary star. Find the star masses for an inclination of zero degrees (face on orbit).

    m1r1=m2r2 gives m1=5m2.
    The distance to the system in parsecs is one over the parallax i.e 1/p=10. The solution then states that the separation in AU a=s/p=30, which I don't understand.
    Kepler's third law then gives the solution.

    1/p is in parsecs and then s is in arcseconds, so a has units arcseconds per parsec. I can't see how that can be AU...
     
  6. Jun 10, 2014 #5

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    You need to refresh yourself on the definition of the parsec unit:

    http://en.wikipedia.org/wiki/Stellar_parallax

    http://en.wikipedia.org/wiki/Parsec
     
  7. Jun 10, 2014 #6
    Defined as the parallax of one arcsecond? There doesn't seem to be anything there that relates it to astronomical units as there seems to be above...

    Edit: Parallax of one arcsecond when the baseline is 1AU! That might help, I'll get back to you if I still have any issues, thanks!
     
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: Kepler's third law: units
  1. Kepler's Third Law (Replies: 11)

Loading...