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Homework Help: Keplers law

  1. Nov 13, 2011 #1
    what is it exactly?

    and in my book they had an equation

    F = (4)(pi)^2(m)(r)/T^2

    then they said by using keplers law... they arrived to a new equation that relates the gravitational force exerted by the sun which is...
    F = m/r^2

    If Keplers law says T = r^3/2 how in the heck did they go from F = (4)(pi)^2(m)(r)/T^2
    to F = m/r^2 by substituting T for r?????
    Where did (4)(pi)^2(r) go ????????
  2. jcsd
  3. Nov 13, 2011 #2


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    Staff: Mentor

    It would appear that they started with the centripetal force:
    [tex] F = m\frac{v^2}{r}~~~~~\text{where:}~~~~v = \frac{2 \pi r}{T}[/tex]
    [tex] F = \frac{4 \pi^2 m r}{T^2}[/tex]
    Now, Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. For a circular orbit we identify the semi-major axis with the orbital radius. Note that the law as stated does not give us the constant of proportionality, so we write: [itex] r^3 \propto T^2[/itex]. So we can replace the T2 in the force equation with r3 but without a precise constant of proportionality it doesn't make sense to retain the others, so that:
    [tex] F \propto \frac{4 \pi^2 m r}{r^3} \propto \frac{m}{r^2}[/tex]
  4. Nov 13, 2011 #3


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    Staff: Mentor

    For different planets, Kepler's third Law says (R^3)/(T^2) = constant
    and equals unity when using units of years and astronomical units (A.U.).
    Last edited: Nov 13, 2011
  5. Nov 13, 2011 #4
    Ok so I see that I r will divide out and it will equal 4(pi)^2m/r^2... where did the 4 pi^2 go?
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