Let's say there's some satellite revolving around the Earth. Its orbital velocity would be $$v=\sqrt{\frac{GM}{r}}.$$
Evidently, this is independent of the distribution of the mass of the Earth, and that implies that even if we manage to compress the Earth to the Schwarzschild Radius, it will not affect the motion of the satellite. But I find it hard to believe that there's a black hole right there but the satellite is revolving as usual. Am I going wrong somewhere?
In fact, the satellite will continue to orbit as usual.
For each circular orbit with a certain radius ($r$), there is a certain speed ($v$) so that the object remains in the orbit, $$v = \sqrt {\frac{{GM}}{r}}, $$ where $M$ is the Earth mass and $G$ is the gravitational constant. As you mentioned, even if we manage to compress the Earth to the Schwarzschild Radius, it will not affect the motion of the satellite. This is true. In fact, according to the Newton's shell theorem, a spherically symmetric object gravitationally affects other objects as if all of its mass were concentrated at its center. Since the Earth is replaced by a black hole (with the same mass), at least, two new effects could be present due to the black hole: The extreme tidal forces (near the black hole) and Hawking radiation. But, in your question, none of them could have a sensible effect.
First, note that the Schwarzschild radius for the Earth is approximately one inch, so the tidal force becomes important here for small distances from the black hole. The tidal force along the radial coordinate ($r$) is proportional to $$ F_{\rm{tidal}} = \frac{GmMd}{r^3}, $$ where $m$ is the satellite's mass and $d$ is the satellite's height (along the radial coordinate). As long as $F_{\rm{tidal}} \ll F_{\rm{Earth-satellite}}$ (or equivalently $r \gg d$), the satellite is revolving as usual since the tidal force is the same as before, i.e., like the Earth-satellite system.
Furthermore, black holes emit black-body (Hawking) radiation as
$${T_{BH}} \approx 6.17 \times {10^{ - 8}}\left( {\frac{{{M_ \odot }}}{M}} \right)\,{\rm{K} },$$
where $M_\odot \approx 2×10^{30} \rm{kg}$ is the solar mass. Evidently, in this case (a black hole having the mass of the Earth), the Hawking radiation is negligible ($\sim 0.02$ $\rm{K}$); it is near the absolute zero!
When you are "far" from the black hole, the physics is the same whether you do the calculation using Newton's laws or General Relativity. This is a necessary condition since we know that Newton's laws are very good under "normal" situations. In this case "far" should be measured relative to the Schwarzschild radius of the black hole.
For a black hole with mass of the Earth, the Schwarzschild radius is very small, and so whatever satellite you're imaging is probably "far" for this purpose. It will therefore obey the orbital velocity relation that you posted in your question. It's living and operating in a region in which Newton's laws are very accurate - Or, stated another way, it's in a region where the predictions of Newton's laws and GR are very close to identical.
Now if you get "close enough" to the black hole, there are some differences. A major one is that black holes have an innermost circular orbit. As the name implies, there are no stable circular orbits inside that radial distance. So in that sense your intuition that there must be a difference is correct, but to realize it you'd either need a much more massive black hole or you'd need to be at an "orbital" distance that is unrealistically small.
But actually it's true, the shape or size (I guess that is what you mean with orientation?) doesn't change the eqution. You can think of the earth as a point having the entire mass. Important is that the $r$ in your equation is the radius of the satellite trajectory, not the radius of the earth.
The eauation arises from gravitational force = centrifugal force: $$\gamma \cdot \frac{M \cdot m}{r^2} = \frac{m \cdot v^2}{r}$$
$r$ is the the distance in which the satellite orbits the Earth.
Well actually, the formula you have used implicity assumes a spherically symmetric mass distribution, so it is not independent of how the mass is distributed.
Birkhoff's Theorem says that the spacetime outside any non-rotating, spherically symmetric mass distribution is unique. Therefore, if we approximate the Earth as spherically symmetric and non-rotating, then if it becomes a non-spinning Schwarzschild black hole, the spacetime that was originally above the Earth's surface is unchanged.
**That means the orbital dynamics of a satellite would be identical. **
However, the Earth is not spherically symmetric and it does spin, which means the formula $ v = \sqrt{GM/r}$ is an approximation for satellites and there are (small) effects like orbital precession that occur due to the oblateness of the Earth, the tidal effects of the Moon and the Sun on the Earth's mass distribution and small local accelerations due to the "lumpiness" of the Earth.
In order to collapse to a black hole it would be necessary for the Earth to lose angular momentum$^\dagger$ and it would become a spinning Kerr black hole. A Kerr black hole is a much simpler and axisymmetric object, so many of the small perturbations of the satellite's orbit would cease. i.e. the orbit would become much better behaved, than around an asymmetric, lumpy Earth; but the approximate orbital speed of the satellite for a fixed $r$ would not be changed.
$\dagger$ In principle, the change in angular momentum would change the size of the General Relativistic Lense-Thirring effect ("frame-dragging"). But this is tiny even compared with the orbital perturbations due to the asymmetry of the Earth.
