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From Kepler's law to gravitation force

  1. Jun 18, 2009 #1
    Hi all! It is often said that Newton deduced from Kepler's laws the theory of gravitation. Particularly from (T^2)/(a^3)=const for differnet planets he deduced that gravitational force must look like '1/(r^2)'. I can also do that by writing F=ma and integrating it in spherical coordinates, but back when Newton lived there was nothing like integrals and differentials as I know.

    Any ideas or info about how Newton came to that conclusion?
     
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  3. Jun 18, 2009 #2

    Nabeshin

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    For a circular orbit it's fairly trivial to show that the gravitational force produces a T^2/r^3 relationship. I'm not sure about working it the other way, though. In general, the procedure is as follows:
    [tex]F=\frac{Gm_1 m_2}{r^2}=m_2 a[/tex]
    [tex]\frac{Gm_1}{r^2}=\frac{v^2}{r}[/tex]
    [tex]\frac{Gm_1}{r^2}=\frac{(\frac{2 \pi r}{T})^2}{r}=\frac{4 \pi ^2 r}{T^2}[/tex]
    [tex]\frac{Gm_1}{4\pi^2}=\frac{r^3}{T^2}[/tex]

    For ellipses the derivation is similar, but a bit more complicated. I imagine you could work backward from the above finding and work out that the force should be a 1/r^2 relationship with a bit of imagination.
     
  4. Jun 18, 2009 #3

    diazona

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    Ah, but Newton invented them for the purpose of doing physics. (For all I know, it may have been for this exact problem :wink:)
     
  5. Jun 18, 2009 #4

    Vanadium 50

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    You might want to look up who invented them. :wink:
     
  6. Jun 18, 2009 #5
    LOL! thanks, just checked and was surprised that actually he was one of the founders of infinitesimal calculus, I was sure it was developed not until 19th century, thanks for pointing that! And thanks to 'Nabeshin' too for detailed formulas
     
  7. Jun 18, 2009 #6
    Nabeshin, your argument works the other way too.

    For a circular orbit, a = v^2/r (that's universal, nothing to do with gravitation). Plug in

    [itex]v=\omega r = \frac{2\pi r}{T}[/itex]

    and obtain

    [itex]a = \frac{4 \pi^2 r}{T^2}[/itex]

    On the other hand, from Kepler's law, T^2 ~ r^3, so

    [itex]F = ma \propto \frac{4 \pi^2 }{r^2} [/itex]

    Generalizing from this is just a matter of induction (and courage!).

    -----
    Assaf
    http://www.physicallyincorrect.com" [Broken]
     
    Last edited by a moderator: May 4, 2017
  8. Jun 18, 2009 #7
    Newton did not use calculus to derive Kepler's laws from his own. He relied on what we today would consider very arcane properties of conic sections in his proof. Feynman examined Newton's proof and found it took for granted these properties that were not at all obvious. So, Feynman created his own "elementary" proof, which is a bit more accessible, but still does not use calculus. This is the famous "lost lecture" on gravitation. http://en.wikipedia.org/wiki/Feynman’s_Lost_Lecture If you can find the audio for this lecture, it's worth a listen.
     
  9. Jun 19, 2009 #8
    wow, so does that mean he derived those laws based on some philosophical thinking and geometry, or did I untderstand wrongly the phrase 'did not use calculus'?
     
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