I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.
- 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?
- 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?
- A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.
