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For a circular, equatorial ($z=0$) Newtonian binary, the position can be clearly written as,

$$ x_i = r(\cos \Omega t, \sin \Omega t, 0)$$

for orbital frequency $\Omega$.

My question is how would this change in the general case for e.g. an eccentric, inclined binary?

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For an elliptical orbit in the $z=0$ plane, the equations for the time-dependent position are

$$x(t)=a[\cos{E(t)}-e]\tag{1},$$

$$y(t)=b\sin{E(t)}\tag{2},$$

where the constants $a$, $b$, and $e$ are the semi-major axis, semi-minor axis, and eccentricity. The $x$-direction here is along the major axis of the ellipse.

The angular quantity $E(t)$ is called the "eccentric anomaly" and is related to the time by “Kepler’s equation”,

$$n(t-t_0)=E(t)-e\sin{E(t)}\tag{3},$$

where the constant $n$ is called the "mean motion". It’s basically an average angular velocity and is just $\Omega$ for a circular orbit.

Given a time $t$, you have to solve the transcendental equation (3) numerically, or using a series expansion, to get $E(t)$. You then put this into (1) and (2) to get the position at time $t$.

These equations are discussed in Wikipedia.

For an inclined orbit, you can simply apply a three-dimensional rotation matrix.

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  • $\begingroup$ Additional details can be found here. $\endgroup$ – G. Smith Nov 16 at 0:30

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