I have several confusions regarding [escape velocity](http://en.wikipedia.org/wiki/Escape_velocity). I am sure I am missing something(s) obvious or maybe I am taught wrong.  

 1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11 \text{km s}^{-1} $ but I am talking about exact escape speed.   That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$.
Wikipedia says and I quote 
>In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.  

 How is that possible?

 2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower _speed will never reaches zero_. Am I right?
 3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.