Elliptic orbit from empty focus

[kaksmet]

hello, I am solving a problem about the angular velocity in an elliptic keppler orbit as seen from the empty focus of the ellipse.

Homework Statement

Show that the angular motion of a keppler orbit as seen from the empty focus is uniform to first order in the eccentricit.

The Attempt at a Solution

I have tried various ways to attack this problem but they all run into the same type of problem. What I am trying to get is a way to relate the constant angular momentum around the non-empty focus to the angular motion as seen from the empty focus. But all such attempts get into very complicated calculations, if not impossible.

Does anyone have a good way to relate these two angular motions to each other? any ideas of how to adress the problem to make it simpler? or any references where I could find something helpfull...?

I would appreciate it!

 

[Jhero]

Hello,

Thank you for bringing up this interesting problem. I have a suggestion for approaching this problem. First, let's define some variables to make it easier to discuss the problem. Let's call the angular motion as seen from the non-empty focus "ω1" and the angular motion as seen from the empty focus "ω2". Also, let's denote the eccentricity of the elliptic orbit as "e".

To relate ω1 and ω2, we can use the fact that the angular momentum is conserved in a Kepler orbit. The angular momentum is given by L = mvr, where m is the mass of the orbiting object, v is the velocity, and r is the distance from the focus. Since the orbit is elliptic, the distance r can be expressed as r = a(1-e^2)/(1+e cosθ), where a is the semi-major axis and θ is the true anomaly.

Now, let's consider a small change in the true anomaly, dθ. This will result in a small change in the distance from the focus, dr. Using the above equation for r, we can write this change as dr = a(1-e^2)/(1+e cosθ)^2 dθ. We can also express the change in the angular momentum as dL = mvdθ. Since the angular momentum is conserved, we can equate these two expressions and solve for v, which gives us v = L/(mr^2).

Substituting this value of v in the expression for ω1, we get ω1 = L/(mra^2(1-e^2)/(1+e cosθ)^2). Now, we can use the fact that the semi-major axis is related to the angular momentum by a = L^2/GMm^2, where G is the gravitational constant and M is the mass of the central body. Substituting this in the expression for ω1, we get ω1 = √(GM/a^3)(1+e cosθ)^2/(1-e^2).

Similarly, we can express ω2 in terms of a, e, and θ. Substituting these expressions for ω1 and ω2 in the equation ω2 = ω1 + dω, where dω is a small change in the angular velocity, we get dω = √(