Tidal forces are experienced within a system moving in free fall around a bigger object because of the different strength of gravity over the system. The difference can be calculated between the far side and the near side.
Are there measurable tidal forces between the galactic centre and the solar system?
The magnitude of the gravitational force, $F$, between two masses, $M$ and $m$, separated by a distance, $r$, is given by: $$ F = \frac{ G \ M \ m }{ r^{2} } \tag{0} $$ where $G$ is the gravitational constant.
Suppose I move $m$ by a distance $\Delta r$, then the force changes to: $$ F_{\Delta} = \frac{ G \ M \ m }{ \left( r + \Delta r \right)^{2} } = F \ \left[ 1 + \frac{ \Delta r }{ r } \right]^{-2} . \tag{1} $$ Note that the difference is entirely encompassed in the brackets.
One lightyear is roughly 9.461 trillion km or roughly 63,241 astronomical units (AU). Suppose we want to compare the force from the center of the Milky Way to one edge of Earth's surface to the opposite edge (i.e., line from Milky Way center through middle of Earth). Let's ignore the non-uniform and non-spherical mass distribution of the galaxy and just treat it like a point source. Note that 1 AU is roughly 23,455 times the Earth's mean equatorial radius (or ~215 solar radii).
The Solar System is located roughly 27,000 lightyears from the galactic center, or ~1.7 billion AU ~ $4 \times 10^{13} \ R_{E}$
So if our $\Delta r = 2 R_{E}$ and $r \sim 4 \times 10^{13} \ R_{E}$, then $\tfrac{ \Delta r }{ r } \sim 5 \times 10^{-15}$, i.e., we're getting into the rounding error level of small.
Even if we changed $\Delta r$ to correspond to Earth's orbit about the Sun (i.e., ~2 AU), $\tfrac{ \Delta r }{ r }$ only increases up to $\sim 10^{-9}$.
Are there measurable tidal forces between the galactic centre and the solar system?
Are there tidal forces acting on massive objects due to the galactic center? Sure. Can we measure them? I am not aware of any instrument/detector accurate enough to do so (which may not mean much). This is further complicated by the fact that the galaxy is not a uniform, spherically symmetric, massive object. The geometry and distribution of mass outside the solar system's orbit about the galactic center is also not uniform or spherically symmetric. So all my estimates above would need to be altered accordingly. However, geometry is not going to increase the differences enough to make much of a difference.
