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Orbital Angular Momentum with eccentricity

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Considering a two-body problem, star-planet, prove that the angular momentum of the planet is given by:

    {\cal L}_{e} = {\mu}\sqrt{GMa(1-e^2)},

    [tex]$\mu$[/tex] is the reduced mass
    M is the star mass
    a and e are the semi-major axis and eccentricity of the planet

    3. The attempt at a solution
    I have no idea how to start the problem. Can someone give me a light? what consideration should I make?

    Thanks in advance
  2. jcsd
  3. Apr 10, 2010 #2
    here are some hints:
    do you know the expression for perigee distance in terms of mass, [itex] h [/itex] , [itex] k [/tex] and eccentricity?
    and as well an expression for perigee in terms of eccentricity and semi-major axis [itex] a [/itex] ,
    and the fact that [itex] h [/itex] is momentum per unit length , you can consider the two objects orbiting a common center and then the smaller mass replaced by a reduced mass [itex] \mu [/itex] , and from the above mention expressions find the angular momentum from the product of the momentum per unit mass and the reduced mass,
    Last edited: Apr 10, 2010
  4. Apr 10, 2010 #3


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    Mechdude's method doesn't make a whole lot of sense to me.

    As an alternative, take a look at the definition of the orbital angular momentum of the planet about the star's CoM, [itex]\mathbf{\mathcal{L}}=\textbf{r}\times\textbf{p}[/itex] (where [itex]\textbf{r}[/itex] is the planet's CoM's position relative to the star's CoM and [itex]\textbf{p}[/itex] is the planet's CoM's momentum relative to the star's CoM)

    Use the product rule to take the time derivative, and use the expression for the gravitational force between the planet's CoM and the star's CoM. You should find that the time derivative is zero, and hence the angular momentum is constant. From there, just pick any point ion the planet's orbit and calculate what that constant is (using the point of apogee or perigee is probably easiest).
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