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Kepler, Newton, Gravity, error Please Help

  1. Nov 25, 2009 #1
    Looking at the gravitational equation

    F=G*(M*m)/r^2

    and centripetal force

    F = ((V^2)/r)*m

    If you set the two equal and solve for G you get:

    G = ((V^2)*r)/M

    Substituting (4*pi^2*r^2)/T^2 for V^2 you now have

    G = (4*pi^2*r^3)/(M*T^2)

    With solution for G, look at Kepler's law with Newton's update,

    (M+m)*P^2 = (4*pi^2*a^3)/G

    Substituting the G solved for into Kepler's equation and consider working with a perfect circle, r will equal a and T will equal P.

    Now, once you substitute in the solved G, simplify...

    You Get

    (M+m)*P^2 = P^2*m

    This is obviously not true!

    Please let me know where the error is in my observation.

    Thanks!
     
  2. jcsd
  3. Nov 27, 2009 #2

    jmb

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    No! The law is [itex]MP^2 = 4\pi^2a^3/G[/itex] (see here). In which case there is no contradiction (hardly surprising given the above form of Kepler's Law is derived from Newton's law of gravity).
     
  4. Nov 27, 2009 #3
    Well... I am confused, as the link you reference has 4π2a3 = P2G(M + m) under "Conversions for Unknowns", which is the same as what I referenced (M+m)*P^2 = (4*pi^2*a^3)/G.
     
  5. Nov 27, 2009 #4

    D H

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    A bit out of order. jmb's response first.

    Sorry, jmb, but NCStarGazer has the right expression.

    [tex]\tau^2 = \frac{4\pi^2}{G(M+m)} \,a^3[/tex]



    The source of NCStarGazer's error:
    You are assuming that the r in Newton's law of gravity and the r in the centripetal force equation are one and the same. They aren't. Both objects are orbiting their center of mass. Given that the distance between a pair of objects of mass m and M is r, the distance from the object with mass m to the center of mass is

    [tex]r_{cm} = \frac M{M+m}\,r[/tex]

    The centripetal force needed to sustain a circular orbit about the center of mass is thus

    [tex]F=\omega^2 r_{cm} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{M}{M+m}\,r[/tex]

    Combining this with the gravitational force yields

    [tex]\frac {GMm}{r^2} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{M}{M+m}\,r[/tex]

    Solving for the period,

    [tex]\tau^2 = \frac{4\pi^2}{G(M+m)}\,r^3[/tex]
     
  6. Nov 27, 2009 #5
    D H

    Thanks, I see how the center of mass can bring clarity. In working with your post I am still having trouble getting

    [tex]\frac{GMm}{r^2}= \left(\frac{2\pi}{\tau}\right)^2\frac{M}{M+m}\,r[/tex]

    To yield your answer, I placed it in Maple and it always returns a version that has Gm(M+m); not G(M+m) in denominator ( I cannot find how to reduce the Gm to be only G), I did it by hand and I came up with the same. Did I miss something or was there a typo or such?

    Appreciate the help!
     
  7. Nov 27, 2009 #6

    D H

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    A typo. ω2r has units of acceleration, not force. Restarting with the equation for the center of mass,

    [tex]r_{cm} = \frac M{M+m}\,r[/tex]

    The centripetal force needed to sustain a circular orbit about the center of mass is thus

    [tex]F=m\omega^2 r_{cm} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{Mm}{M+m}\,r[/tex]

    Combining this with the gravitational force yields

    [tex]\frac {GMm}{r^2} = \left(\frac{2\pi}{\tau}\right)^2\,\frac{Mm}{M+m}\,r[/tex]

    Solving for the period,

    [tex]\tau^2 = \frac{4\pi^2}{G(M+m)}\,r^3[/tex]
     
  8. Nov 28, 2009 #7

    jmb

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    Ughh, my bad. I read this when I was very tired and tried to answer the question too quicky (something I really shouldn't do!).

    I saw NCStarGar's use of [itex]r[/itex] in the centripetal and gravitational force equations and assumed he was making the assumption that [itex]M>>m[/itex] and ignoring centre of mass issues. Of course had I taken on the significance of what he/she meant by "Newton's update" I would have realised that wasn't the intent...

    Sorry NCStarGazer and thanks to D H for spotting this. Very embarrassed now!
     
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