Is relativistic event horizon half of Newtonian event horizon? Is Relativistic event horizon half of Newtonian event horizon?
relativistic escape velocity formula (from $m\phi=E-E_0$) is $v_e=\sqrt{2\phi-(\frac{\phi}{c})^2}$ and the Newtonian version of the formula (from $m\phi=\frac{1}{2}mv_e^2$) is just $v_e=\sqrt{2\phi}$, in both cases $v_e$ is the escape velocity, and $\phi$ is the gravitational potential. (for black hole $v_e=c$)
$R=\frac{GM}{c^2}$ Relativistic event horizon
where R is the radius of the black hole, or the distance from its center to its Relativistic event horizon. This formula gives exactly half the value as that of the standard Newtonian formula $R=\frac{1}{2}R_s$
$R_s=\frac{2GM}{c^2}$ Newtonian event horizon
or
Is the Schwarzschild radius twice the distance from the center of a black hole to its real event horizon?
or
are black holes only half as large as previously believed (according to relativistic effects)?

(source: cloudfront.net)
 A: No. You must have made an error in determining the GR horizon radius. 
The radius of the event horizon determined from Newtonian theory (simply determining the distance from a point mass at which the escape velocity equals the speed of light) happens to be the same as the radius rigorously derived from the General Relativistic equations. 
This, by the way, is a 'coincidence'. Would you calculate the radius of the photon sphere from Newtonian theory (determining the distance from a point mass at which the centripetal force equates the force of gravity for a test particle moving at the speed of light), you would find a distance that is three times too small compared to the correct General Relativistic expression.
A: Your question is meaningless. In Newtonian physics, there is no limit to the possible velocities. Light has velocity $c$ but higher velocities can exist. Therefore, you can send something from any distance from any mass fast enough to have escape velocity.
Black holes are a purely GR concept.
And it is easy to prove.
Consider a Newtonian compact object, of mass M and radius not necessarily zero, but small enough to be what you'd call a "Newtonian black hole", say radius less than $\frac {MG}{10c^2}$
Hovering very far from this object, I send in its direction a projectile with a reasonable velocity, much larger than escape velocity from its attraction at this distance. I aim extremely carefully, to have a hyperbolic trajectory that passes, say at a distance $\frac {MG}{5c^2}$, so it does not crash into the object. Well within what you would like to call its "horizon". Of course my projectile has acquired in the process a speed much higher than $c$, but we are in Newtonian physics, aren't we ? So it is OK.
Since the trajectory is hyperbolic, it will get out of the gravitational potential with more than the escape velocity.
There can be no horizon in Newtonian physics
