Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Derivation of Kepler's Third Law

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data

    I am studying gravitation and I have been trying to derive Kepler's Third Law using Kepler's Second Law.

    2. Relevant equations

    The second law :

    [tex]\frac{dA}{dt}[/tex] = [tex]\frac{L}{2m}[/tex]

    3. The attempt at a solution

    To start with, I thought if we take [tex]\int[/tex] [tex]\frac{L}{2m}dt[/tex] from 0 to T; we would find Area of the ellipse.

    We also know that [tex]A = \Pi * a * b[/tex]. So, to find T, I could use these two equations.

    Moreover, to eliminate b, eccentricity could be used.
    [tex]e^{2} = 1 - \frac{b^{2}}{a^{2}}[/tex]

    substituting b would make

    [tex]A = \Pi * a^{2} * \sqrt{1 - e^{2}}[/tex]

    [tex]\frac{L.T}{2m} = \Pi * a^{2} * \sqrt{1 - e^{2}} [/tex]

    and we know that

    [tex] L = m.Vtan.R[/tex]

    where Vtan is tangential component of V. (to the radius R)

    And we also know that since T is period, [tex]R = ea + a[/tex]

    Thus we can write

    [tex] \frac{m.Vtan.(ea + a)}{2m} .T= \Pi * a^{2} * \sqrt{1 - e^{2}} (1)[/tex]

    Now, we need to express Vtan. If we write centripetal force, we can conclude that

    [tex]Fnet = m.\frac{Vtan^{2}}{R} = \frac{G.m.Msun}{R^{2}}[/tex]


    [tex]R = ea + a[/tex]

    and Msun is the mass of the sun.

    [tex] Vtan = \sqrt{\frac{G Msun}{ea + a} [/tex]

    and finally, we can rearrange equation (1) as

    [tex] T = \frac{2.\Pi.a^{3/2}\sqrt{1-e^{2}}}{\sqrt{GMsun}.\sqrt{1+e}}[/tex]

    simplfying a little bit would yield to the result

    [tex] T = \frac{2.\Pi.a^{3/2}\sqrt{1-e}}{\sqrt{GMsun}}[/tex]

    I seems to be close, but wrong!

    Kepler's third law :

    [tex] T = \frac{2.\Pi.a^{3/2}}{\sqrt{GMsun}}[/tex]

    So the term with squareroot and 1-e shouldn't have been there... I have been trying to find what I'm doing wrong, and I thought my assumption that

    [tex] Vtan = \sqrt{\frac{G Msun}{ea + a} [/tex]

    seems to be wrong. Although it sounds logical to me... I am stuck here and need your help.

    Thanks in advance.
    Last edited: Jan 2, 2009
  2. jcsd
  3. Jan 2, 2009 #2


    User Avatar
    Homework Helper

    Perhaps this link will help?
    http://astro.berkeley.edu/~converse/Lagrange/Kepler%27sThirdLaw.htm [Broken]
    Last edited by a moderator: May 3, 2017
  4. Jan 2, 2009 #3
    Hello, thanks for your help.

    I have checked it out. But I would like to correct my attempt.

    Expressing Vtan in terms of centripetal force didn't work... I wonder why it is incorrect. Please help me expressing Vtan somehow that would make my attempt correct.
  5. Jan 2, 2009 #4


    User Avatar
    Homework Helper

    Well for one thing your L = mvr is for circular orbits isn't it?
  6. Jan 2, 2009 #5
    L = m V r is incorrect, what I assume is;

    L = m Vtan r

    where Vtan is the component of the velocity vector V which is momentarily perpendicular to the radius; since each planet can actually make circular motion, momentarily.

    Moreover, L = mVtanr is used to derive kepler's second law, so I don't think that one is problematic.
  7. Jan 3, 2009 #6
    Anyone, please ?
  8. Jan 4, 2009 #7
    I am desperately waiting for some answers...
  9. Jan 29, 2010 #8


    User Avatar


    Wow, I was solving the same problem and I ended up here in this forum, so i joined just to be able to answer your question. It was the same problem I had and it took me some research to figure it out.

    Indeed, your expression for Vtan is not quite right, as the one you used for the centripetal force, and that happens for a very subtle reason.

    First of all, as you're dealing with the general case of an elliptical orbit, you have to remember that the distance R from the planet to the Sun IS NOT a constant, and therefore you can't really say R = ea + a because that is not true through the entire orbit. (unless you have the special case of a circle, and you can easily check that your derivation indeed holds for kepler's for e = 0).

    Now, the actual Vtan for the generalized elliptical case is a bit trickier. As the distance planet-Sun (which I will call r) changes, there's an extra r'' term for the centripetal acceleration besides V²/r. [in fact, since for an ellipse V is not always perpendicular to R, the correct formula would be V²/(radius of curvature) instead of simply r.].

    (r" stands for the second time derivative of r). I will also write V²/r as r*[tex]\omega^2[/tex], where [tex]\omega[/tex] stands for the angular velocity (which is not a constant).

    [tex]a = - \frac{GM}{r^2}[/tex]

    [tex]\frac{d^2r}{dt^2} - r*\omega^2[/tex] [tex] = - \frac{GM}{r^2} [/tex]

    This is a Non-linear second order ordinary differential equation. Its solution, in fact, leads to Kepler's first law (if you solve it in terms of r as a function of [tex]\theta[/tex] you will find the equation of an ellipse).

    I will not solve it here, but you can find its solution on wikipedia if I'm not mistaken, just search Kepler's Laws. From it's solution, you'd find that

    [tex] L^2 = m^2*a*(1 - \epsilon^2)*GM[/tex] which will lead to the correct result.

    I hope I was able to help :) English is not really my first language, so i'm not sure i was clear. Please let me know if there's anything you'd like me to rephrase.
  10. Jan 29, 2010 #9
    GMm/r^2 = mrw^2

    GM/r^3 = 4pi^2 / T^2

    T^2 / r^3 = 4pi^2/GM
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook