Orbital Angular Momentum with eccentricity

In summary, the angular momentum of the planet is given by:$\mathbf{\mathcal{L}}={\mu}\sqrt{GMa(1-e^2)},$ where $\mu$ is the reduced mass, $\mathbf{r}$ is the planet's CoM's position relative to the star's CoM, and $\textbf{p}$ is the planet's CoM's momentum relative to the star's CoM. The angular momentum is constant, and can be found from any point on the planet's orbit.
  • #1
amarante
44
6

Homework Statement


Considering a two-body problem, star-planet, prove that the angular momentum of the planet is given by:

[tex]
\begin{equation}
{\cal L}_{e} = {\mu}\sqrt{GMa(1-e^2)},
\end{equation}
[/tex]

[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


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
 
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  • #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:
  • #3
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).
 

What is Orbital Angular Momentum with eccentricity?

Orbital Angular Momentum with eccentricity refers to the amount of angular momentum possessed by an object in an orbit around another object when the orbit is not perfectly circular. It takes into account the eccentricity, or shape, of the orbit.

How is Orbital Angular Momentum with eccentricity calculated?

Orbital Angular Momentum with eccentricity can be calculated using the equation L = mvr, where L is the angular momentum, m is the mass of the object, v is its velocity, and r is the distance from the center of rotation.

What is the significance of Orbital Angular Momentum with eccentricity in space exploration?

The amount of Orbital Angular Momentum with eccentricity can affect the stability and trajectory of an object in orbit. This is important in space exploration because it can impact the success of missions and the safety of spacecraft.

How does eccentricity of an orbit affect Orbital Angular Momentum?

The eccentricity of an orbit directly affects the amount of Orbital Angular Momentum. A higher eccentricity means the orbit is more elliptical, resulting in a higher angular momentum. A lower eccentricity means the orbit is closer to a perfect circle, resulting in a lower angular momentum.

Can Orbital Angular Momentum with eccentricity be changed?

Yes, Orbital Angular Momentum with eccentricity can be changed by altering the velocity or distance of the orbiting object. This can be done through the use of thrusters or gravitational assists from other objects in space.

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