Circular Velocity & Escape Velocity

I have never heard of the term circular velocity - are you sure you don't mean rotational velocity (from force = m * v^2/r)?

Escape velocity generally means if a projectile is launched with usually an large and/or instaneous intial thrust (bullet or rocket) what velocity must it achieve to escape a planetary body's gravitional well (integrate GM1M2/x^2 dx from x = starting height to x = infinity, so its like G * Mass of planet * Mass of body / distance from point of suspension (gravitational centre) of planet.

For Earth that's 6.67 * 10^ -11 * 5.98 * 10 ^24 * Mass rocket / 6.4 * 10^6 or about 62 Mega Newton per kilogram of projectile you wish to launch.

If you are orbiting the Earth at velocity V and radius r, then escape velocity is V^2 / r > GMass Earth / r^2 or simply

V > square root of (G * Mass Earth / radius from Centre of Earth)

ok
This is exactly what I am seeking.

The point with the escape velocity of that of light (2.99792458*10^8 m/s) is said to be at the event horizon. In the case of the mass of the Sun (1.9891*10^30 Kg), the radius to the event horizon is 2953.518 m from the center of mass.

ok
Good stuff here.