Earth's angular velocity when axis is aligned with apparent gravity

[Bastian1978]

TL;DR Summary

A question about finding the components of Earth's angular velocity when the vertical axis is aligned with apparent gravity (with regards to John Taylor's book "Classical Mechanics")

I've been reading the book "Classical Mechanics" by John Taylor, and in the chapter about noninertial reference frames, it states that the direction of 𝑔 (the apparent gravity, which includes the centrifugal force) is not necessarily aligned with the direction of 𝑔0 (the true gravity direction, directed towards the centre of the Earth).

It goes on to establish a coordinate system for a point on the surface of the Earth at colatitude 𝜃, where the vertical axis is aligned with 𝑔, and the horizontal axes are aligned with north and east. It then states that if Ω (the angular velocity of the Earth) is vertical in the standard coordinate system, we can calculate the components of Ω in the coordinate system described above as:

(0, Ωsin(𝜃), Ωcos(𝜃))

See attached diagram.

I'm confused by this because, as far as I understand, this calculation would only make sense if the coordinate system's vertical direction was aligned with 𝑔0 (so that a vector from the point on the Earth's surface along this direction would be directed towards the centre of the Earth), rather than along 𝑔 (which would mean a vector from the point on the Earth's surface along this direction would not be directed towards the centre of the Earth, due to the included centrifugal force).

Is this just an unstated approximation being made (because the directions of 𝑔 and 𝑔0 are similar), or have I just misunderstood what the book means?

Any help would be really appreciated.

 

[Tazerfish]

I think you're basically right.
The coordinate system is pointing along g0.

Maybe it's unfortunate that they tell you about lumping centrifugal force in with g in the chapter before.
That sounds like s strange practice. More likely than not, they'd just include another centrifugal force term in the acceleration.

Fun fact: Since water follows an equipotential surface perpendicular to g, the ocean isn't level with this coordinate system anymore.