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Kepler's third law

  1. Aug 30, 2007 #1
    1. The problem statement, all variables and given/known data

    Using astronomical units as the unit of length , years as the time , and the mass of the Sun as the unit of mass , the value for k in kepler's third law is 1. In these units , what is the value of Newton's constant of gravitation G?


    2. Relevant equations
    P^2 =k*a^3
    P^2= 4*pi^2*a^3/(G(m(1)+m(2))
    Possibly k=4*pi^2/G*M(sun)
    P is the period

    3. The attempt at a solution


    since k = 1 and k =4*pi^2/G*M(sun), all I have to do is switch variables. k=4*pi^2/(G*M(sun)) => 4*pi^2/(M(sun))=G , but I found that method to be a problem because the G I calculated isn't equal the universal Gravitational constant we are all familar with.
     
  2. jcsd
  3. Aug 31, 2007 #2

    learningphysics

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    So taking the mass of the sun as 1 solar mass... your answer should be [tex]G = 4{\pi}^2 (years)^{-2}(AU)^3(solar masses)^{-1}[/tex]

    To check that this is the same G you need to convert units...
     
  4. Aug 31, 2007 #3

    D H

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    The universal gravitational constant has units. The value with which you are familiar, [tex]6.6730\times10^{-11}[/tex] is in units of [tex]m^3/kg/s^2[/tex]. The value differs when you change the system of units. For example, G is [tex]7.616\times10^{-5}[/tex] in the ever-popular furlongs-fortnights-stones system. You are asked to specify G in the AU-Msun-year system. The answer is not [tex]6.6730\times10^{-11}[/tex].
     
  5. Aug 31, 2007 #4
    Are my equations correct? I just need to convert my units to SI units?
     
  6. Aug 31, 2007 #5

    learningphysics

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    Yes, I think the equations are right. just convert the units.
     
  7. Aug 31, 2007 #6

    D H

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    Your instructor did not ask for the value of G in SI (MKS) units. He asked for it in AU-solar mass-year units. Learning physics gave you the answer in these units post #2.

    You only need to convert to SI if you want to verify that your answer is correct. If you do the conversion right, you will get [itex]6.672\times10^{-11}\text{m}^3/\text{sec}^2/\text{kg}[/tex], which compares very favorably with the published value of [itex]6.673\times10^{-11}\text{m}^3/\text{sec}^2/\text{kg}[/tex].
     
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