The Sun orbits approximately 26,000 light years away from the galactic center. Are there any other known orbital parameters relative to the galactic plane for the Sun and other stars up to some distance away? For example, what are the orbital paramters ($a,e,i,\Omega,\omega,M,f$) of Alpha Centauri?

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2 Answers 2


The "orbits" of stars around the galaxy are not closed Keplerian orbits (e.g. they are not confined to a plane), so they are not characterised in that fashion. What you can do is get the present-day $U,V,W$ velocities ($U$ measured towards the Galactic centre, $V$ in the direction of Galactic rotation, $W$ out of the Galactic plane) and present-day positions and attempt to predict a trajectory based on a model for the Galactic gravitational potential

This is possible for many millions of stars now thanks to the data from the Gaia satellite. What is needed are proper motions and a parallax, along with a measurement fo the line-of-sight velocity.


For example, what are the orbital paramters (𝑎,𝑒,𝑖,Ω,𝜔,𝑀,𝑓) of Alpha Centauri?

These are Keplerian orbital elements which characterize Keplerian orbits which are orbits in a strict $\mathbf{r}/r^3$ force field.

Situations that approximate such a force field include planets around stars, moons around their planets, double stars1 that are not too close together etc.

Also included might be a star's trajectory just outside almost all of a relatively spherically distributed globular cluster or a spherically symmetric galaxy, cf. ProfRob's and HDE's answer to Why aren't there spherical galaxies? which state (spoiler alert):

There are!

That orbits occurring just outside of an extended but spherically symmetric distribution are still Keplerian can be understood through the lens of Newton's shell theorem which shows that as long as the distribution is spherically symmetric, for the purposes of calculating $\mathbf{r}/r^3$ forces like gravity and electrostatics one can replace the distribution with a single, central point source for which Keplerian orbits are defined.

Situations that do not approximate such a force field include mass distributions like the Milky Way spiral galaxy which is nothing like spherically symmetric whether you include all the dark matter or not, and trajectories that dive down into a spherically symmetric mass distribution.

Orbits in a uniform spherical mass distribution are still elliptical, but they are not Keplerian in that if you plot position vs time you see they are simply Hooke's law sinusoids with the center of the spherical mass at the center of the ellipse rather than at one focus. For more on that see:

As ProfRob's answer here already suggests, one can take Alpha Centauri's current state vector $(x, y, z, vx, vy, vz)$ and a tabulated (or parameterized) galactic potential model and use numerical integration to get trajectories.

You need to take the numerical gradient of the potential to get the force field, and depending on how the potential is defined (is it per unit mass, or is it a reduced potential) you may or may not need to divide the gradient by your star's mass to get acceleration.

For more about the Milky Way's gravitational potential and how it can be parameterized, see:

1mix and match normal stars, black holes, neutron stars, etc.


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