The Kepler problem

The Kepler problem corresponds to the study of the motion of two bodies which are influenced by their mutual gravitational attraction. In the center of mass and relative coordinates, the problem is reduced to the motion of one body of mass $m = m_1 m_2 / M$, which we shall refer as particle below, attracted gravitationally by another located at rest at the origin of mass $M=m_1+m_2$.

In cartesian coordinates, the equations of motion can be written as

\[\begin{aligned} \dot{x} & = v_x,\\ \dot{y} & = v_y,\\ \dot{v}_x & = - \frac{G M x}{(x^2 + y^2)^{3/2}},\\ \dot{v}_y & = - \frac{G M y}{(x^2 + y^2)^{3/2}}. \end{aligned}\]

For concreteness, we fix $\mu = G M = 1$. The coordinates $x$ and $y$ are the relative coordinates (to the center of mass) of the particle, and $v_x$ and $v_y$ its velocity. The function kepler_eqs! mutates the vectors corresponding to the LHS of the equations of motion.

function kepler_eqs!(dq, q, params, t)
    dq[1] = q[3]
    dq[2] = q[4]
    rr = ( q[1]^2 + q[2]^2 )^(3/2)
    dq[3] = - q[1] / rr
    dq[4] = - q[2] / rr
end;

kepler_eqs! (generic function with 1 method)

For suitable initial conditions (such that the total energy is negative) the solutions are ellipses with one focus at the origin, which can be parameterized in terms of its semi-major axis $a$ and its eccentricity $e$. We set the initial conditions for the particle at periapse, which we locate on the positive x-axis. Using the semimajor axis and the eccentricity, we write them as

\[\begin{aligned} x_0 & = a (1-e),\\ y_0 & = 0,\\ v_{x_0} & = 0,\\ v_{y_0} & = \frac{l_z}{x_0} = m \frac{\sqrt{\mu a (1-e^2)}}{x_0}, \end{aligned}\]

where $l_z$ is the angular momentum. We set the mass of the particle $m=1$, the semi-major axis $a=1$ and the eccentricity $e=0.8$. Kepler's third law defines the period of the motion as $T= 2\pi a^{3/2}$.

const mu = 1.0
const mass = 1.0
const aKep = 1.0
const eKep = 0.8;

0.8

The initial conditions are then set using ini_cond

function ini_cond(a, e)
    x0  = a*(one(e)-e)
    vy0 = mass * sqrt( mu * a * (1-e^2) ) / x0
    y0  = zero(vy0)
    vx0 = zero(vy0)
    return [x0, y0, vx0, vy0]
end
q0 = ini_cond(aKep, eKep)

4-element Vector{Float64}:
 0.19999999999999996
 0.0
 0.0
 3.0

We now perform the integration, using a 25 order expansion and absolute tolerance of $10^{-20}$.

using TaylorIntegration, Plots
t, q = taylorinteg(kepler_eqs!, q0, 0.0, 10000*2pi, 25, 1.0e-20, maxsteps=700_000);
t[end], q[end,:]

(62831.853071795864, [0.20000000000001486, 5.940560592488514e-9, -4.950456454488883e-8, 2.9999999999999])

We first plot the orbit. (For performance reasons only the first 10000 points are considered.)

x = view(q, :, 1)
y = view(q, :, 2)
vx = view(q, :, 3)
vy = view(q, :, 4)
plot(x[1:10_000], y[1:10_000], legend=false)
scatter!([0], [0], shape=:circle, ms=5)
xaxis!("x", (-2.0, 0.5))
yaxis!("y", (-1.0, 1.0))
title!("Fig. 1")

The following functions allow us to calculate the energy and angular momentum using cartesian coordinates.

function energy( x, y, vx, vy )
    kinetic = 0.5 * (vx*vx + vy*vy)
    r = sqrt( x*x + y*y)
    potential = - mu * mass / r
    return kinetic + potential
end
lz( x, y, vx, vy ) = mass * ( x*vy - y*vx ) ;

lz (generic function with 1 method)

We use the change in energy and angular momentum of the orbit with respect to the initial value of the corresponding quantity as a function of time. These quantities are expressed in units of the local epsilon of the initial energy or angular momentum, respectively. This serves to illustrate the accuracy of the calculation, shown in Figure 2 and 3.

e0 = energy(q0...)
δE = (energy.(x,y,vx,vy) .- e0) ./ eps(e0)
plot(t[1:3:end], δE[1:3:end])
xlabel!("t")
ylabel!("dE")
title!("Fig. 2")

lz0 = lz(q0...)
δlz = (lz.(x,y,vx,vy) .- lz0) ./ eps(lz0)
plot(t[1:3:end], δlz[1:3:end])
xlabel!("t")
ylabel!("dlz")
title!("Fig. 3")

These errors are reminiscent of random walks.

The maximum absolute errors of the energy and angular momentum are

maximum( abs.(energy.(x,y,vx,vy) .- e0) ), maximum( abs.(lz.(x,y,vx,vy) .- lz0) )

(9.947598300641403e-14, 2.808864252301646e-14)