4
$\begingroup$

Many people concern themselves with understanding the Galactic potential. To me, the most obvious way is to measure the acceleration of a star, as this is directly related to the potential. So, one could measure a LOS velocity (or proper motion) at different years and measure its change. I understand that orbital timescales of stars are huge however, so can this be done? If not, how far (order of magnitude) away are we from having the required uncertainties for achieving this?

$\endgroup$
5
  • 1
    $\begingroup$ What is that "galactic potential"? $\endgroup$
    – peterh
    Nov 24, 2019 at 23:45
  • 2
    $\begingroup$ Galactic potential = the gravitational potential of the Milky Way $\endgroup$
    – user31040
    Nov 25, 2019 at 0:09
  • $\begingroup$ Not even with Gaia-level precision will we be able to measure accelerations to a meaningful level. However proper motions, translated into 3D velocities, paired with a bit of theory, i.e. the Virial theorem and such, will be able to achieve that to a degree. $\endgroup$ Nov 25, 2019 at 1:57
  • 2
    $\begingroup$ Those accelerations are pretty small, which makes them hard to measure accurately. Eg, the Sun's galactic orbital speed is roughly 200 km/s and its distance from the galactic centre is around 8 kpc. Using $a=v^2/r$ that gives an acceleration of roughly $\mathrm{1.6 \times 10^{-10} m/s^2}$ $\endgroup$
    – PM 2Ring
    Nov 25, 2019 at 3:44
  • 1
    $\begingroup$ Why the close vote? It's a legitimate question even if we can't answer it $\endgroup$
    – user21
    Nov 26, 2019 at 15:35

1 Answer 1

1
$\begingroup$

A commenter points out that accelerations in the galactic gravitational potential are extremely small:

The Sun's galactic orbital speed is roughly 200 km/s and its distance from the galactic centre is around 8 kpc. Using $a=v^2/r$ that gives an acceleration of roughly $1.6×10^{−10}\rm\,m/s^2$.

A century is only $\pi×10^9$ seconds, so in a hundred years the $\Delta\vec v$ for the Sun and its neighbors is only half a meter per second --- and that's mostly a difference of direction, not magnitude. We can't measure three-dimensional stellar motions to a precision of a half-meter per second, and the career of a professional astronomer is usually shorter than a century.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .