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.
[Physics] Why does mechanical energy have to equal zero to find escape velocity
energy-conservationescape-velocitynewtonian-gravitynewtonian-mechanics
Best Answer
The easiest way to calculate escape velocity, is neglecting Earths rotation and assuming the object takes of in a radial direction. Then, indeed, you start from
$$E = K_1 + U_1 = K_2 + U_2$$
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.