Anti-Realism Euler Force Examples:
Let's go back to the analogy of the rotating asteroid. For the sake of simplicity we'll just assume that it had a rectangular cuboid shape where it was rotating clockwise about a point. The length and breadth of the asteroid were large but the height from the top surface to the bottom surface was thin. We'll also ignore the orbit of the asteroid around the Sun. An astronaut not too far away from the top left edge of the asteroid who throws a rock sideways further to the left near the cliff beneath them will see the rock appear to fall diagonally downwards relative to their sightline. This is due to the rising centripetal acceleration of the radius where the speed of the asteroid's superficial plane is varied with the outer circumferences moving faster than the inner circumferences. Centripetal force is perpendicular to circular motion and so the rock won't preserve all of the rotational momentum of the asteroid. The slower the overall asteroid rotates, the milder the downward angle will be where it will have a stronger sideways vector from the force of your throw. The faster the outer edges of the asteroid rotates, the steeper the rock will fall to the ground because the centripetal speed of the asteroid's floor will increasingly outweigh the sideways vector from your throw. If the asteroid rotated at an extreme speed then the sideways vector will be negligible in comparison where it'd seem to drop vertically downwards.
Let's asteroid-hop to one with a small oval shape that's also moving clockwise. This time you were standing on the underside of the asteroid. You were placed towards the left again except now you're only half-way to the steep outer edge. So when you throw a rock further to the left in a horizontal direction we'll need something else to happen since both you and the floor would be moving away from the rock with the rock possibly appearing to go higher and higher into outer space. However there's a steep incline on the plane with an average downward chord of -30 degrees returning rightwards to the centre point owing to the oval shape. Thus when you throw the ball to the left, the ground behind you to the right will be higher than your position if your vision reorients itself to see the light reflected off the bottom of the asteroid as being upwards*. For the sake of argument, let's further assume that the asteroid is in an orbit around the Sun such that entire asteroid is still moving leftwards while it also rotates clockwise about a point. The trajectory of the thrown rock is now considered in absolute space with the Solar System instead of it being relative to the surface of the asteroid alone. Owing to the leftwards orbit around the Sun, the higher ground behind you which is also moving leftwards will move even faster when both velocities are combined. However the same ground in front of you (radial length of the asteroid in a 7 o'clock angle) that already appears to be sloping (due to the oval shape) diagonally downhill will slowly begin to move in unison vertically downwards (reversed vision where it'd be upwards for an outside observer) as it crosses (with the rest of the radius) over the x-axis (9 o'clock position) where the clockwise motion of the asteroid will then be going in the opposite direction (upwards and rightwards from 10 to 11 o'clock) compared to its leftwards orbit around the Sun. This time the two velocities oppose each other resulting in a smaller net velocity. Therefore the ground behind you will actually be moving mostly horizontally forwards more than the height direction in absolute space and whack into the thrown rock. The faster the oval asteroid rotates around the Sun, the greater the degree to which the rock will fall straight backwards which is rightwards relative to the asteroid's midpoint even if your initial throw was leftwards. The slower the asteroid orbits the Sun, the longer it will take for the rock to hit the floor. In this scenario the same leftward clockwise motion (7 o'clock position) might seem to have slightly increased relative to the speed of the leftward orbit in the Solar System. Moreover the ground in front of you is objectively moving faster only in terms of the clockwise rotation due to the increased length of the radius. The rock might remain travelling in the leftwards trajectory after you throw it and land diagonally downwards if the orbit of the asteroid around the Sun has a slight upwards vector in addition to its predominantly leftwards orbit. In other words the asteroid could pivot downwards and upwards around the thrown rock with the rock appearing to move in a semi-circular path to the ground in front of you. If there's no upward trajectory on the asteroid's orbit around the Sun, then the rock throw which was originally attempted to be leftwards at slow speed will visibly keep going diagonally backwards and upwards until it eventually crashes really far back rightwards towards the midpoint or even near the right edge on the opposite end. (Anyone who's confused could instead imagine the diametrical opposite with you on the top surface of an asteroid balanced in a 1 o'clock position moving downwards to 4 o'clock as rotates about the midpoint where the asteroid also assumes a rightwards and downwards orbit around the Sun. If you're also puzzled by the inverted vision example then you could think of a ball machine throwing the ball where you're standing the right way up on a nearby spaceship.)
* "Under normal circumstances, an inverted image is formed on the retina of the eye. With the help of upside down goggles, the image on the retina of the observer's eyes is turned back (straightened) and thus the space around the observer looks upside down."
https://en.wikipedia.org/wiki/Upside_down_goggles
I think I'm owed an award of some kind:
Father Ted Acceptance speech