How come escape velocity isn't imaginary?

[Jules Winnfield]

Going through several definitions, it appears that escape velocity is equal to the potential energy. That is:$$\frac{1}{2}m v^2=-\frac{G M m}{r}$$but if I solve for velocity, $v$, I get:$$v=\sqrt{-2\frac{G M}{r}}$$So how do I get an escape velocity that isn't imaginary?

 

[Ibix]

It's the velocity such that the kinetic energy at launch is equal to the potential energy difference between infinity and the point of launch. So$$\frac 12mv^2=0-\left(-\frac{GMm}r\right)$$

 

[DrStupid]

Going through several definitions, it appears that escape velocity is equal to the potential energy.

No, at escape velocity the sum of kinetic and potential energy is zero.

 

[Jules Winnfield]

It's the velocity such that the kinetic energy at launch is equal to the potential energy difference between infinity and the point of launch. So$$\frac 12mv^2=0-\left(-\frac{GMm}r\right)$$

That makes sense. Thank you for the clarification.

 

[Dale]

No, at escape velocity the sum of kinetic and potential energy is zero.

And in this sum PE is always negative so KE is always positive.