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 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$^\dagger$ 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.

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.

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 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.

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 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$^\dagger$ 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.