But the moon subtends 0.5 degrees at the earth!
Please explain.
The explanation is not very interesting. I know the size of
the moon; I just can't reliably do arithmetic. Sorry.
In temperate latitudes, the bulge in the earth causes the
vertical direction to be tilted about 0.1 degree relative
to what you would expect for a non-rotating earth, i.e.
relative to the straight line through the center of the
earth.
That's 20% of the diameter of the full moon.
The horizontal direction is affected in the same way, of
course. That's one inch per 50 feet, which is noticeable
if (say) you are building a swimming pool. You really
want the structure to be aligned with the water, and the
water is going to align itself horizontally relative to
the _total_ local acceleration of the reference frame,
_including_ the centrifugal contribution.
The /magnitude/ of the gravitational acceleration is affected
also. Let's be clear:
a) For ordinary terrestrial applications, the change in
direction of g is usually more noticeable than the
change in magnitude of g.
b) In the frame comoving with the orbiting space station,
the change in magnitude is very directly noticeable.
That is, the astronauts are weightless in this frame.
This is a 100% change in magnitude relative to an
earth-centered frame, in which the astronauts are of
course not weightless.