The initial premise of your question is not, in general, correct.
Consider a 1000 kg car driving at a constant speed of 20 m/s around a flat circular racetrack with a radius of 500 meters. Note the constant speed: the car will take the same time to drive around the track, 157.1 seconds, hour after hour. The only change in the car's velocity is in the direction of the velocity, not its size. The only acceleration is the centripetal acceleration, and the only horizontal force needed is the centripetal force $$a_{cp}=\frac {v^2}{r}=0.8 \frac{m}{sec^2}$$ $$F_{cp}=m \times a_{cp}=800 \ newtons$$
The force exerted by the car's engine serves only to balance the various drag forces.
This force is supplied by the friction between the tires and the pavement; pour out some oil on the track to see what happens when the required centripetal force is not present!
The only acceleration is directed exactly towards the center of the circular track; there is no tangential acceleration. The car, at some moment, is travelling North at 20 m/s, and one half-circuit later, it is travelling South at the same speed. Clearly, it has accelerated.
Assume now that the driver presses on the brake pedal in a manner that she knows will bring the car to a stop in 40 seconds. There is now a tangential acceleration, at $-0.5 \frac{m}{sec^2}$, in addition to the centripetal acceleration above, and a tangential force of 500 newtons directed towards the back of the car. The total force needed from the tires is now the resultant of these two forces: 943.4 newtons directed around 32 degrees aft of inward. Touching the brakes could throw you into a skid! Of course, as the braking changes the car's speed, the centripetal force will decrease...
You don't even need a banked road. The analogy of your orbit example would simply be a car moving at a constant speed in a circle (for simplicity we, for example, restrict to the case without skidding). In that case, there is no tangential force, and the force keeping the car in a circle is the force of static friction. We can see that this is possible in real life as follows. Let $m$ be the mass of the car, then a free body diagram shows that the normal force on the car has magnitude
\begin{align}
N = mg.
\end{align}
It follows that the force of static friction on the tires satisfies
\begin{align}
F_\mathrm s\leq \mu_\mathrm s N = \mu_\mathrm s mg
\end{align}
In other words, the force of static friction can have any value between $0$ and the product of the coefficient of static friction and the weight of the car.
Now, suppose that the car wants to drive in a circle of radius $r$. In this case, it will experience an acceleration with magnitude $v^2/r$ in the radial direction, and it will require a corresponding force
\begin{align}
F = \frac{mv^2}{r}
\end{align}
to do so. Since the force of static friction will be supplying this force, we require $F_\mathrm s = F$ which tell us that the car can move at a constant speed $v$ in a circle of radius $r$ without skidding provided
\begin{align}
\frac{v^2}{r}\leq \mu_s g.
\end{align}
It's easy to make this inequality satisfied; you just need to make sure that the circle has large radius, or the car has small speed, or some appropriate combination of the two.
Best Answer
The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$.
$\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital speed $v$ that depends on distance $r$ through the relation $a_c=\frac{v^2}r$. This applies to any uniform circular motion.
The force of gravity at a given distance gives you the acceleration, and with a suitable orbital speed the motion will be uniform. However, as you've noticed, the angular speed for uniform circular motion due to a force $\propto \frac1{r^2}$ (like gravity) will have decreasing angular speed with increasing $r$.
If you're only concerned with the magnitude of $a_c$, this can be shown as follows:
$$a_c = \omega^2r = \frac{v^2}r$$
First, you can observe the following:
$$v^2 = \omega^2r^2$$
$$v = \omega r$$
But you also need to be mindful that $a_c$ itself depends on $r^2$ via $a_c = \frac {\mathbf F_g} m = \frac{GM_e}{r^2}$.
So orbital speed depends on $r$, and angular speed depends on $r$, for uniform orbital motion due to gravity, but in different ways:
$$\sqrt \frac{GM_e}{r} = v$$
$$\sqrt \frac{GM_e}{r^3} = \omega$$
So in a way you're right, if the angular speed is kept constant, you need a larger $a_c$ with increasing $r$ to create uniform orbital (circular) motion. However this is only because the orbital speed increases with distance for constant angular speed $\omega$, which in turn requires a larger $a_c$ in order for the motion to still be uniform and circular. But under gravitation, $\omega$ is not constant because the force doesn't grow linearly with distance (it shrinks in fact), and $a_c$ is lower with increasing $r$ as a result.
Regarding your two probes, let's call them E (on earth) and S (in space)... Consider that E is not actually in uniform circular motion at all - it is on the surface of the earth, which is itself spinning. In fact, it is the other way around, probe E is moving too slow to achieve uniform circular motion around the earth, and would fall down.
Think of it this way - if the acceleration felt is larger than what is required for uniform circular motion at a given distance and speed (as is the case near the earth's surface for every-day situations), the object would approach the center, in other words fall to the ground. Consider how far and hard you have to throw something at ground level into the horizon for it to never hit the ground; geosynchronous / geostationary orbits are exactly this, but at a height where their required speed for uniform motion is small enough to match the speed at which the earth rotates.
It may help to try and work out the orbital speed (not in terms of $\omega$) for S, and then compare it to the orbital speed needed for uniform motion under gravity at smaller and smaller distances to the earth - you should find that as you come closer, you need to be moving considerably faster to maintain uniform circular motion ($v^2 \propto \frac1r$). And with any increase in speed, for the motion to remain uniform and circular at the same radius, the acceleration must increase ($v^2 \propto a_c$).