# Position vector of an eccentric binary

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?

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.

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