The error is that you assume that the density distribution is "nearly spherically symmetric". It's far enough from spherical symmetry if you want to calculate first-order subleading effects such as the equatorial bulge. If your goal is to compute the deviations of the sea level away from the spherical symmetry (to the first order), it is inconsistent to neglect equally large, first-order corrections to the spherical symmetry on the other side - the source of gravity. In other words, the term $hg$ in your potential is wrong.
Just imagine that the Earth is an ellipsoid with an equatorial bulge, it's not spinning, and there's no water on the surface. What would be the potential on the surface or the potential at a fixed distance from the center of the ellipsoid? You have de facto assumed that in this case, it would be $-GM/R+h(\theta)g$ where $R$ is the fixed Earth's radius (of a spherical matter distribution) and $R+h(\theta)$ is the actual distance of the probe from the origin (center of Earth). However, by this Ansatz, you have only acknowledged the variable distance of the probe from a spherically symmetric source of gravity: you have still neglected the bulge's contribution to the non-sphericity of the gravitational field.
If you include the non-spherically-symmetric correction to the gravitational field of the Earth, $hg$ will approximately change to $hg-hg/2=hg/2$, and correspondingly, the required bulge $\Delta h$ will have to be doubled to compensate for the rotational potential. A heuristic explanation of the factor of $1/2$ is that the true potential above an ellipsoid depends on "something in between" the distance from the center of mass and the distance from the surface. In other words, a "constant potential surface" around an ellipsoidal source of matter is "exactly in between" the actual surface of the ellipsoid and the spherical $R={\rm const}$ surface.
I will try to add more accurate formulae for the gravitational field of the ellipsoid in an updated version of this answer.
Update: gravitational field of an ellipsoid
I have numerically verified that the gravitational field of the ellipsoid has exactly the halving effect I sketched above, using a Monte Carlo Mathematica code - to avoid double integrals which might be calculable analytically but I just found it annoying so far.
I took millions of random points inside a prolate ellipsoid with "radii" $(r_x,r_y,r_z)=(0.9,0.9,1.0)$; note that the difference between the two radii is $0.1$. The average value of $1/r$, the inverse distance between the random point of the ellipsoid and a chosen point above the ellipsoid, is $0.05=0.1/2$ smaller if the chosen point is above the equator than if it is above a pole, assuming that the distance from the origin is the same for both chosen points.
Code:
{xt, yt, zt} = {1.1, 0, 0};
runs = 200000;
totalRinverse = 0;
total = 0;
For[i = 1, i < runs, i++,
x = RandomReal[]*2 - 1;
y = RandomReal[]*2 - 1;
z = RandomReal[]*2 - 1;
inside = x^2/0.81 + y^2/0.81 + z^2 < 1;
total = If[inside, total + 1, total];
totalRinverse =
totalRinverse +
If[inside, 1/Sqrt[(x - xt)^2 + (y - yt)^2 + (z - zt)^2], 0];
]
res1 = N[total/runs / (4 Pi/3/8)]
res2 = N[totalRinverse/runs / (4 Pi/3/8)]
res2/res1
Description
Use the Mathematica code above: its goal is to calculate a single purely numerical constant because the proportionality of the non-sphericity of the gravitational field to the bulge; mass; Newton's constant is self-evident. The final number that is printed by the code is the average value of $1/r$. If {1.1, 0, 0} is chosen instead of {0, 0, 1.1} at the beginning, the program generates 0.89 instead of 0.94. That proves that the gravitational potential of the ellipsoid behaves as $-GM/R - hg/2$ at distance $R$ from the origin where $h$ is the local height of the surface relatively to the idealized spherical surface.
In the code above, I chose the ellipsoid with radii (0.9, 0.9, 1) which is a prolate spheroid (long, stick-like), unlike the Earth which is close to an oblate spheroid (flat, disk-like). So don't be confused by some signs - they work out OK.
Bonus from Isaac
Mariano C. has pointed out the following solution by a rather well-known author:
http://books.google.com/books?id=ySYULc7VEwsC&lpg=PP1&dq=principia%20mathematica&pg=PA424#v=onepage&q&f=false
For the block to move in a circle, there must be a centripetal force (as you have said)--something needs to be constantly pulling the block toward the center of the circle if it is to move in a circle. The thing that does the pulling in this case is the spring. (In other words, the source of the centripetal force is the spring. Without the spring, there is no centripetal force, assuming there is no friction between the block and the rod.)
If this is not clear, imagine what would happen if the spring was not attached to the block. When the rod begins to rotate, the block will soon slide off the end of the rod because we know from Newton's First Law that the block will try to move in a straight line unless we apply a force to change its direction. If the spring is attached to the block, however, when the block tries to go straight and slide off the end of the rod, the spring will stretch and apply a force toward the center of the circle. That will prevent the block from flying off the end of the rod.
Best Answer
Let's think of a simplified system: two point masses connected by a massless string. As the masses circle around the center of the string, the string provides the centripetal force required to make the masses turn in a circle. We can find the tension by using Newton's second law. Ignoring gravity, we recognize that the tension is the only force acting on each mass, and the acceleration is centripetal acceleration.
$$ F_\mathrm{net} = T = m a_c $$
The tension depends on the mass on each end $m$, the length of the string $\ell$, and the angular frequency $\omega$. The centripetal acceleration to turn is: $$a_c = \frac{v^2}{r} = \frac{(\omega r)^2}{r} = \omega^2 r = \omega^2 \frac{\ell}{2}. $$
So the tension in the string is: $$ T = \frac{m \omega^2 \ell}{2}. $$
A real string will eventually break if the tension gets too great. If it spins too fast (big $\omega$), then the string breaks.
From the point of view of an inertial frame the tension is required to change the direction of the masses velocity. The acceleration turns the masses in a circle.
Using a co-rotating frame, the the tension is balanced by the centrifugal pseudo-force.