[Birkhoff's Theorem][1] 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 when 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 and there are (small) effects like orbital precession that occur due to the oblateness of the Earth and the tidal effect 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 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.


  [1]: https://en.m.wikipedia.org/wiki/Birkhoff%27s_theorem_(relativity)#:~:text=In%20general%20relativity%2C%20Birkhoff's%20theorem,given%20by%20the%20Schwarzschild%20metric.