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Kepler's third law implies force proportional to mass

  1. Jul 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that Kepler's third law, [itex]\tau = a^{3/2}[/itex], implies that the force on a planet is proportional to its mass.

    2. Relevant equations
    3. The attempt at a solution

    I haven't really attempted anything. I'm not sure what the question is going for. What can we assume and use?
     
  2. jcsd
  3. Jul 25, 2016 #2
    Hint: You are given T^2 = k a^3 where T = period, k = constant, a = mean distance
    Also, you know F = m v^2 / a
    The trick is to eliminate v (how does this relate to T and a?)
     
  4. Jul 26, 2016 #3
    Well in this question a is the semi-major axis. The book tells me (without proof) that [itex]a[/itex] is also the median distance.

    I'm not sure why you said [itex] F = mv^2/a[/itex] when [itex] mv^2/r[/itex] is just the centripetal acceleration, and only equal to the force in the case of a circular orbit, right?
     
  5. Jul 26, 2016 #4
    in my point of view, the question start by asking you to show, therefore you should start with the derivation/proof for Kepler's third law.
    hint: Kepler third law is about the orbits of planets, so think of any equations that applicable to the orbit of a planet.
     
  6. Jul 26, 2016 #5
    Ok well the derivation of Kepler's third law requires knowing the expressions for the semi-major axis and semi-minor axis which also requires Kepler's first law. The derivation also makes use of Kepler's second law.
     
  7. Jul 26, 2016 #6
    i think it is safe to assume the mean distance of an elliptical orbit = [ (semi major + semi minor )/2 ] , is approximately equal to the radius of a circular orbit. hence, F=mv2/r or mv2/a is applicable.
     
  8. Jul 26, 2016 #7
    In that case the problem is a bit too easy I think, assuming I've done it correctly. If we are assuming the orbit is nearly circular so that [itex]\ddot{r}[/itex] is near zero then [tex] F = m \omega^2 r = \frac{m4 \pi^2 r}{\tau^2} = \frac{m 4 \pi^2 r}{k^2r^3} = \frac{m 4 \pi^2}{k^2r^2}[/tex]

    This is not really what they want I think.
     
  9. Jul 26, 2016 #8
    i don't think you can substitutes t2 = k2r3, because that's what they want to show, so you should have a series of steps which finally leads to t= a2/3.

    another hint: derivation of Kepler's third law start with F=mv2/r , or any equivalent formula for centripetal force, due to it's a circular orbit. and F=GMm/r2 due to force of attraction between 2 celestial object. work on the two equation.
     
  10. Jul 26, 2016 #9

    PeroK

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    Well, if Kepler's laws hold for any ellipical orbit, then they must hold for a circlular orbit; and, as a circle is symmetric, there must be constant speed; and, as the speed is constant, you can derive the force equation. QED
     
  11. Jul 26, 2016 #10
    I think the question is asking to do the opposite, to start from Kepler's third law and end up with the force equation.
     
  12. Jul 26, 2016 #11
    oh, then i must be misunderstanding the question
     
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