# Why does mechanical energy have to equal zero to find escape velocity?

A object orbiting the earth has total mechanical energy equal to \begin{align*} E^{mech} = \frac{1}{2} m v^2 - \frac{GMm}{r} \end{align*} with $M$ the mass of the earth and $r$ the distance. My course notes say we have to equal $E^{mech} = 0$ find the escape velocity, which then gives \begin{align*} v = \sqrt{\frac{2GM}{r}} \end{align*} But I don't understand why we should do this. In general we have $E = K_1 + U_1 = K_2 + U_2$. Now I see that if $U(r)$ with $r \rightarrow \infty$, then $U_2$ becomes zero. But why should $K_2$ ever be set the zero? That means the object would come to rest somewhere, which we cannot know.

• The escape velocity is defined to be the minimum velocity required to escape the gravitational well. What do you think the physical significance of "minimum" is? – lemon Jul 6 '15 at 12:12
• $K_2$ does not have to be set to zero. It represents the kinetic energy the particle would have once at infinity: if you want it to be $\neq 0$ then set it to be $\neq 0$. The usual terminology is, though, that the minimum energy is what's needed to be set at infinity with zero kinetic energy remaining. – gented Jul 6 '15 at 12:55

$$E = K_1 + U_1 = K_2 + U_2$$
where $K_1=\frac{mv^2}{2}$ and $U_1=- \frac{GMm}{r}$.
Since the range of gravitional forces is infinite, you say (theoretically, not practically) that an object has escaped Earths gravition when it is infinity far away, so $U_2 = 0$. Now, if the object would have velocity = 0 before it is infinity far away, then (neglecting the rest of the universe), it would fall back to Earth and hence didn't escape. So it should still have a velocity when it is infinity far away. This velocity may be as small as you want, so the border point between falling back to earth and escaping is velocity =0. So take $v_2 =0$ and you find the minimal value such that the objects velocity doesn't become zero before reaching infinity.
When a rocket is fired from Earth with a sudden impulse, its total energy is given by: $$E_k \text{ (kinetic energy)} + E_p \text{ (potential energy)}= \frac{1}{2}mv^2 - \frac{GMm}{r} = constant$$ The potential energy here is taken to be negative because the reference point chosen for potential energy to be zero is when the rocket is unbound in Earth's orbit. Hence, after the rocket is fired (with no propulsion after the initial impulse) it is bound if its: $$E_{total} < 0$$ and unbound if its: $$E_{total} \geq 0$$ In your case $E_{mech}$ is $E_{total}$. Setting $E_{total} = 0$ you can calculate what the escape velocity must be.