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Shortest possible period of revolution of two identical gravitating solid spheres

  1. Jan 16, 2013 #1
    1. The problem statement, all variables and given/known data
    From K&K's 'Intro to Mechanics'

    Find the shortest possible period of revolution of two identical gravitating solid spheres which are in circular orbit in free space about a point midway between them.

    2. Relevant equations



    3. The attempt at a solution

    So I figured the gravitational force exerted on each sphere by the other would be

    [tex]F=\frac{2mg}{r^2}[/tex]

    according to Newton's law of gravitation (m being each sphere's mass). This force would be providing the centripetal acceleration that's keeping them going in a circle so the angular velocity can't exceed a certain value and this is related to the period of revolution.

    [tex]F_c=\frac{2mG]{r^2}=m\frac{v^2}{r}[/tex]
    ∴[tex](\frac{2G}{r})^{1/2}=v[/tex]

    So plugging that into [tex]T=\frac{\omega}{2\pi}[/tex] gives me [tex]T=(\frac{G}{2\pi^2r^3})^{1/2}[/tex]


    Is this correct? If not, am I at least on the right track?

    Thanks in advance
     
  2. jcsd
  3. Jan 16, 2013 #2
    Your final equation suggests that as r tends to infinity then T tends to zero. Does that sound reasonable?
    Your method is essentially right but you have got errors in the details of your equations.

    Applying Newton's law of gravity gives a force of [tex]F=-\frac{Gm^2}{r_d^2}[/tex]
    Which should be equated to centripetal force (as you rightly did)
    [tex]F=-\frac{mv^2}{r_r^2}[/tex]
    Notice I have two different distances
    rd= distance between centre of spheres
    rr= distance between a spheres centre and the midpoint between the spheres
    clearly rd=2rr
    Your equation for T was wrong and it is more useful to use
    [tex]v=\frac{2\pi r_r}{T}[/tex]
     
  4. Jan 16, 2013 #3
    thanks apelling, redoing it i got:

    [tex]T=(\frac{4\pi^2r^3}{G})^{1/2}[/tex]

    but when I looked at the units, I seem to have a mass unit in there
     
  5. Jan 17, 2013 #4
    You should get

    [tex]T=(\frac{16\pi^2r^3}{Gm})^{1/2}[/tex]

    With r being the distance between the centre of a sphere and the midpoint about which the spheres rotate.


    Now I suspect you need to replace the mass m with density x volume of a sphere. This introduces another distance: the radius of the sphere. For minimum time period the orbital radius r should be a minimum. What is the smallest it can be?
     
  6. Jan 20, 2013 #5
    Oh, thanks...back to work then! Wow, you gotta be really careful when doing these problems in this book!
     
  7. Jan 22, 2013 #6
    alright i'm so close now...i realise the mistake i made before was adding the masses rather than squaring. my answer now is like yours except i have a 4 in place of your 16
     
  8. Jan 22, 2013 #7
    let r= distance between centre of sphere and midpoint between spheres.
    then distance between centres of spheres=2r

    We then get

    [tex]F=-\frac{Gm^2}{4r^2}=-\frac{mv^2}{r}[/tex]
    which simplifies to
    [tex]\frac{Gm}{4r}=v^2[/tex]
    Subbing in
    [tex]v^2=\frac{4\pi^2 r^2}{T^2}[/tex]

    leads to
    [tex]T=(\frac{16\pi^2r^3}{Gm})^{1/2}[/tex]
     
  9. Jan 24, 2013 #8
    Redoing it once again, I now have an 8. I've been taking r to be the distance between the two bodies and so have r/2 for the centripetal force side....but shouldn't the same answer pop out?

    [tex]\frac{Gm^2}{r^2}=\frac{2mv^2}{r}[/tex]
    [tex](\frac{Gm}{2r})^{1/2}=v[/tex]
    [tex]T=(\frac{8\pi^2r^3}{Gm})^{1/2}[/tex]
     
  10. Jan 24, 2013 #9
    Are you using

    [tex]v^2=\frac{\pi^2 r^2}{T^2}[/tex]
     
  11. Jan 24, 2013 #10
    No, I'm using

    But same answer if I use the squared version
     
  12. Jan 25, 2013 #11
    You should be using r/2 in this equation to be consistent with your definition of r.
     
  13. Jan 25, 2013 #12
    Hmmm...I've just redone it a few times now and I'm now getting a 2 (have tried another formula as well, T=2pi/w, which gives the same answer). I'll keep checking though...must've made another error somewhere.
     
  14. Jan 28, 2013 #13
    The 2 is correct. I got 16 because I used r=1/2 separation of masses. When cubed in the final equation this leads to a factor of 8 difference. 2 x 8 =16
     
  15. Jan 31, 2013 #14
    Ah yeah...thanks a lot for your help apelling!
     
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