Newton's black hole

For a satellite leave the Earth and enter into orbit the satellite lauuncher must achieve some speed that is called  "escape velocity". if something travels at a lower speed is impossible to leave the Earth. Which is this velocity?
By simple Newton's physics

1/2 m.v^2 = G.m.M/r
Simplifying
1/2 v^2 = GM/r
or
v^2 = 2GM/r
When the gravity originating from a mass M becomes bigger we shall need bigger velocity to escape, the maximun speed that we know that something can travel is light c, so in one moment when the gravity is so strong nothing will be able to leave the "Earth", now in equations:
 c^2 = 2GM/r
We have now a black hole, if we invert the eqaution for the radius, something below this radius will need to travel faster than light, inside this critical radius the speed of light and above it less than light, so it is possible to leave.
The value of this radius, nothing more than simple Newton applied is:

r0 = 2GM/c^2

And this value is exactly the same that General Relativity gives for the horizon of events, but without any tensor or absolute differential calculus.

Stupidity also evolves!