newtonian mechanics - Why do we get 2 different answers for escape velocity, when we apply different laws to calculate it?

To measure the escape velocity, if I use the equation -

$$\frac{-GMm}{r^2}=\frac{v.dv}{dr} $$

and I put my final distance to be $\infty$ , then I get the answer,

$$u = \sqrt\frac{2GM}{R} \;. \tag{1}$$ Which is quite obvious!

But, if I use the equations -

$$U_{\infty} - U_i = \frac{GMm}{R}$$

and

$$K_{\infty}+U_{\infty} = K_R+U_R$$

Where, $K_R$ and $U_R$ are kinetic energies and potential energies at R respectively , where R is the radius of the earth.

Now in these two equations, if I put $U_{\infty} = 0$, then I get,

$$K_{\infty} = K_R - \frac{GMm}{R}\tag{3} $$

And since, $K_{\infty}$ is positive and not zero, we end up getting that

$$u > \sqrt\frac{2GM}{R} \tag{2} $$

That is to say that, if I want to take the object at infinity, the speed must be (1), but if I want to make it's final potential energy 0 , then it's speed must be greater than (1). Also, (1) cannot be used in (3) because that would mean the final $KE$ is 0, which obviously is not!

What is generating this hoax? Please tell me where am I wrong?

Answer

The escape velocity is the minimum velocity possible such that it escapes a gravitational well. We have from conservation of energy: $$ E = K+U = \frac{1}{2}mv^2 - \frac{GMm}{r} $$

Lets consider initial and final energy as $E_1$ and $E_2$. To escape, you need to make you sure your speed fully overrules the potential energy. The maximum possible potential energy there is, is the case $U = 0$, that happens a $r\to\infty$. At all other cases, we have $U < 0$, and thus $U = 0$ is maximum. If your kinetic energy is greater than this maximum, you escaped. So, lets make sure of this. Assuming your initial energy to be: $$ E_1 = K_1 + U_1 = K_2 + U_2 = E_2 $$

At maximum, we have $U_2 = 0$, then: $$ K_1 + U_1 = K_2 \quad\implies\quad K_1 = K_2 - U_1 $$

So, if you have kinetic energy $K_1$, you escape. As you pointed out, the things at the end happens as $r\to\infty$ and therefore $E_2 = E_\infty = K_\infty + U_\infty$. Thus: $K_2 = K_\infty$. That is, that will be your kinetic energy at infinity. $$ K_1 = K_\infty - U_1\quad\implies\quad \frac{1}{2}mv^2 = \frac{1}{2} mv_\infty^2 + \frac{GMm}{r_1} $$

The velocity you need to escape such that, at infinity, you have velocity $v_\infty$ is: $$ v = \sqrt{v_\infty^2 + \frac{2GM}{r_1}} $$

At $v_\infty = 0$, you recover the minimum velocity you need to have in order to escape. Since it is minimum, your velocity at infinity will be zero. Once you arrive at infinity, regardless if you have velocity or not, you escape, and then your potential energy is zero.