Kepler Problem – PHY 4241 Notes

Bernd A. Berg^{a}

a) Department of Physics, Florida State University, Tallahassee, FL 32306-4350 (Dated: November 4, 2010)

Kepler problem as discussed in class.

PACS numbers:

I. SOLUTION OF THE KEPLER PROBLEM FOR NUMERICAL IMPLEMENTATION

The purpose of this section is to cast the analytical solution of the Kepler problem into a form, which allows for easy implementation into a computer program (here done in Fortran). The final solutions are translated back to the initially given inertial system. There are many treatments of the Kepler problem, for instance [1, 2]. We follow to some extent by Landau and Lifschiz.

A. Central potential problem

Let t be the time and the masses be m_{1} and m_{2}. In a frame Σ defined by Cartesian orthonormal vectors ê_{1}, ê_{2}, ê_{3} positions and velocities of the masses m_{i}, (i = 1, 2) are given by:

r⃗_{i}(t) = (\begin{matrix}x_{i}^{1}(t) \\ x_{i}^{2}(t) \\ x_{i}^{3}(t) \end{matrix}) and v⃗_{i}(t) = (\begin{matrix}ẋ_{i}^{1}(t) \\ ẋ_{i}^{2}(t) \\ ẋ_{i}^{3}(t) \end{matrix})  (1)

where the dot denotes the time derivative. For positions and velocities the initial conditions at time t_{0} are

r⃗_{i}(t_{0}) and v⃗_{i}(t_{0}) , i = 1, 2  (2)

in the frame Σ. The computer program will calculate r⃗_{i}(t) and v⃗_{i}(t), i = 1, 2 for any desired time t.

We denote the total mass by

M = m_{1} + m_{2} ,  (3)

the Center of Mass (CM) position and velocity are defined by

r⃗_{cm}(t) = \frac{m_{1} r⃗_{1}(t) + m_{2} r⃗_{2}(t)}{M} ,  (4)
v⃗_{cm}(t) = \frac{m_{1} v⃗_{1}(t) + m_{2} v⃗_{2}(t)}{M} .  (5)

Due to momentum conservation one finds

v⃗_{cm}(t) = v⃗_{cm}(t_{0}) and  (6)
r⃗_{cm}(t) = r⃗_{cm}(t_{0}) + v⃗_{cm}(t_{0}) △t , △t = (t − t_{0}) . (7)

Defining the difference coordinates and velocities by

r⃗_{12}(t) = r⃗_{1}(t) − r⃗_{2}(t) ,  (8)
v⃗_{12}(t) = v⃗_{1}(t) − v⃗_{2}(t) ,  (9)

the positions of the particles are

r⃗_{1}(t) = r⃗_{cm}(t) + \frac{m_{2} r⃗_{12}(t)}{M} ,  (10)
r⃗_{2}(t) = r⃗_{cm}(t) − \frac{m_{1} r⃗_{12}(t)}{M} .  (11)

These equations show that the CM is located in between the particles on the straight line connecting them. Equations for the velocities are defined by taking the time derivatives.

So, the task has become to solve the equations of motion for the difference coordinates and to implement the solution numerically. In the following we used the notation

r_{12} = |r⃗_{12}(t)| .  (12)

For a central potential U(r_{12}) the energy

E = \frac{m_{1} v⃗_{1}(t)^{2}}{2} + \frac{m_{2} v⃗_{2}(t)^{2}}{2} + U(r_{12})  (13)

is conserved, i.e., does not depend on the time t.

As the difference coordinate r⃗_{12}(t) stays invariant under transformation to the CM frame Σ′ defined by

rcm^{′}(t) = 0 .  (14)

We can calculate its time dependence in the CM system and still use Eq. (10) and (11) to find the time dependence of the original coordinates. In the CM systems these equations simplify to

r1^{′}(t) = \frac{m_{2} r12^{′}}{M} and r2^{′}(t) = −\frac{m_{1} r12^{′}}{M} . (15)

In the following we drop the primes and continue to work in the CM frame (we shall use the primes for yet another frame soon).

Inserting the derivatives of equation (15) (without primes) into the energy conservation (13), we find

E_{cm} = \frac{μ v⃗_{12}(t)^{2}}{2} + U(r_{12}) where μ = \frac{m_{1} m_{2}}{M} . (16)

μ is called reduced mass. To simplify the notation, we drop the subscript 12:

r⃗(t) = r⃗_{12}(t) and v⃗(t) = v⃗_{12}(t) .  (17)