Google "geoid",

Think about walking along the shore of the ocean. Everywhere you go is sea level, even if you are farther from the centre of the Earth. Ans even if you are rotating water.

Sea level tells you where all points are equal.

Now we need to determine the work done by which force and in which reference frame we wish to calculate. :)

If a person of mass ##m## walks from the pole to the equator at distance ##R## from the center of the Earth rotating with angular speed ##\omega##, then a good approximation to the work required is simply ##E_m##, the rotational energy of that person, that is,
$$W \approx E_m= \tfrac{1}{2}mR^2 \omega^2.$$
If ##m## becomes significant relative to the mass of the Earth, ##M##, then a more accurate expression using conservation of angular momentum (effectively slowing the Earth's rotation ever so slightly over the walk) and assuming spherical Earth would be
$$\begin{align*}W &= \frac{2}{2+5\frac{m}{M}} E_m \\ &\approx (1-\frac{5m}{2M}) E_m .\end{align*}$$
Note that ##E_m## is same value as before, still based on ##\omega## from when starting at the pole.

Did the wall do 200 J of work on you?

Did your body expend 200 J of energy?

Does this all add up for you?

Every molecule of water is subject to the same forces you are. If it did take took work to navigate from the pole to the equator, then sea water would pool at the poles.

The Mississippi's mouth is farther from the centre of the Earth than its headwaters (by about 3 miles), and yet the water still flows "downhill" to the Gulf of Mexico.

The global sea level

(Actually I'm not sure if Coriolis Foce is likewise exempt.)

One could argue that moving ##m## from the pole to equator also need to include linear momentum, but if we split the mass into four parts, two at each pole, and let these follow symmetric paths to keep the center of mass of these four parts to coincide with the center of the Earth at all times, then there should be no need to include conservation of linear momentum.

Did you see the advice I gave to the OP:

In the inertial frame:
- Coriolis Force doesn't exist.
- The lateral force from the Earth is doing positive or negative work to increase or decrease the kinetic energy due to the tangential motion.

But let's say the co-rotating frame is used to model the situation, then how does one then easily include conservation of linear/angular momentum and energy, which only really applies in inertial frames, to conclude there is no total energy difference between start and stop positions?

I mean, assume the small mass is a spherical uniform marble rolling (or even a point-like mass slipping) frictionless on the geoid surface, i.e. the surface contact forces are always orthogonal to both surfaces. After the marble has been given its initial (lets say very small impulsive) roll in any direct it will in the co-rotating frame have a non-zero speed in a direction perpendicular to the rotation axis and thus experience a Coriolis (and very small Euler) force giving the marbles initial south-bound roll an

To treat the rest frame of the planet as uniformly rotating, you obviously have to assume that the planet is much more massive than the mass you are moving around on it.

However, I would still like to see or do an inertial "geoid-free" model, but it seems tricky to establish a simple inertial model simple that "independently" captures the same effect as the geoid. Perhaps an ellipsoidal Earth using only the J₂ potential would do the trick.

So if you had a friction-less rail in an evacuated tunnel along a great circle containing both poles, and you give a cart (no engine) placed on it a push, then it would travel at constant speed along the north-south direction, and only the east-west component would vary in the inertial frame, as the planet is doing positive or negative work on it via the lateral rail forces. But the cart is completely passive and doesn't need to provide any energy.

In the earth centered inertial frame a mass at the equator has both a higher KE and a higher gravitational PE than the same mass at the pole. So why doesn’t ocean water flow to the poles?