Kepler Orbits

Math

First we have the equation for the distance of the planet

$r = \frac{p}{1 + ε \cos θ}$

(1.1)

where r is the distance between sun and planet where the sun is implicitly assumed to stay at the focus of planet's ellipse. The sun cannot remain at the focus of the ellipse in inertial space unless the mass ratio sun/planet is infinite which is never the case. So r is the distance from the ellipse focus to the planet while another variable r_S is the distance between that same point in space and the position of the sun.

The definition of p is:

$p = a (1 - ε^{2})$

(1.2)

where $ε = \sqrt{1 - b^{2} / a^{2}}$ . Using the reference Kepler Math we find the equation for the rate of change of the azimuthal angle is:

$\dot{θ} = \frac{2 π}{P} \frac{b}{a (1 - ε^{2})} {(1 + ε \cos θ)}^{2}$

(1.3)

Here r is the distance from the sun to the planet, a and b are the ellipse parameters, and P is the period of the planet's and sun's rotation.

The foci of an ellipse with major axis along y=0 are at

$c = x_{c e n t e r} \pm \sqrt{a^{2} - b^{2}}$

(1.4)

where is the symmetric center of the ellipse.

To plot the location of the planet on the ellipse, we just need to iterate using the expression and then use to convert r to x and y of the ellipse.

$\begin{array}{l} x (θ) = x_{c e n t e r} + r \cos θ \pm c \\ y (θ) = y_{c e n t e r} + r \sin θ \end{array}$

(1.5)

So we just iterate and plot the results. Obviously, r as used in the above expressions is not the distance between the sun an the planet unless the sun/planet mass ratio is infinite so that the sun does not have to move in inertial space.

Derivation of Kepler Orbits

It's very important to understand that the shape and size of the orbits depend strictly on the initial total angular momentum and energies of the two bodies.

To get started we will assume a circular orbit. Then the planet distance from the barycenter (which is the center of the orbit) will be given by

(1.7)

where is the rotation rate and T is the period of rotation. Then

(1.8)

The rotation rate of the sun has to be the same as that of the planet but 180 degrees out of synchronization with the planet, and, to keep the center of mass fixed, the radius of the sun's orbit has to be

(1.9)

Motion Around the Center of Mass When Masses are Comparable

The acceleration equations are:

$\begin{array}{l} m_{1} {\ddot{r}}_{1} = - \frac{G m_{1} m_{2}}{| r_{1} - r_{2} |^{3}} (r_{1} - r_{2}) \\ m_{2} {\ddot{r}}_{2} = \frac{G m_{1} m_{2}}{| r_{1} - r_{2} |^{3}} (r_{1} - r_{2}) \end{array}$

(1.10)

${\ddot{r}}_{1} - {\ddot{r}}_{2} = - \frac{G (m_{1} + m_{2})}{| r_{1} - r_{2} |^{3}} (r_{1} - r_{2})$

(1.11)

$\ddot{r} = - \frac{G (m_{1} + m_{2})}{r^{3}} r$

(1.12)

This is the Kepler equation and it has the law for the motion of the masses about their center of mass:

$P^{2} = \frac{4 π^{2}}{G (m_{1} + m_{2})} a^{3}$

(1.13)

Since we want to know the motion of both masses individually we need to use the fact that the center of mass remains stationary:

$m_{1} r_{1} + m_{2} r_{2} = 0$

and then

$r_{1} - r_{2} = \frac{m_{1} + m_{2}}{m_{2}} r_{1} = - \frac{m_{1} + m_{2}}{m_{1}} r_{2}$

(1.14)

Then using the expression (1.12) for r we have the modified equations for the accelerations:

$\begin{array}{l} m_{1} {\ddot{r}}_{1} = - G m_{1} m_{2} (\frac{m_{2}^{3}}{{(m_{1} + m_{2})}^{3} r_{1}^{3}}) \frac{m_{1} + m_{2}}{m_{2}} r_{1} \\ m_{2} {\ddot{r}}_{2} = G m_{1} m_{2} (\frac{m_{1}^{3}}{{(m_{1} + m_{2})}^{3} r_{1}^{3}}) \frac{m_{1} + m_{2}}{m_{1}} r_{2} \end{array}$

(1.15)

These are two Kepler equations and the equations for the periods are:

$\begin{array}{l} P_{}^{2} = \frac{4 π^{2}}{(\frac{G m_{2}^{3}}{{(m_{1} + m_{2})}^{2}})} a_{1}^{3} \\ P_{}^{2} = \frac{4 π^{2}}{(\frac{G m_{1}^{3}}{{(m_{1} + m_{2})}^{2}})} a_{2}^{3} \end{array}$

(1.16)

Since these periods are the same we must have

$a_{1} = \frac{m_{2}}{m_{1}} a_{2}$

(1.17)

as expected to keep the center of mass stationary. These expressions for the periods are taken from the following link:

https://physics.stackexchange.com/questions/382847/keplers-3rd-law-applied-to-binary-systems-how-can-the-two-orbits-have-differen