Escape Velocity

When an object is thrown vertically upward, it rises to some height and then falls back, pulled down by the Earth's continuous gravitational attraction. The higher the initial launch speed, the greater the height reached before falling back. If the launch speed is increased further and further, a critical speed is eventually reached at which the object never falls back at all -- it escapes the Earth's gravitational influence entirely. This minimum launch speed, needed to escape from the surface all the way to infinity, is called the escape velocity, $v_e$.

Because the gravitational force due to the Earth becomes exactly zero only at INFINITE distance (it never becomes zero at any finite r), 'escaping' technically means the object must be able to reach $r\to\infty$, even if only with zero residual speed left over there. This gives a clean energy-conservation criterion for finding $v_e$: consider the object's total mechanical energy (kinetic + potential) at launch, on the Earth's surface, and compare it with its total energy once it has (just) reached infinity.

At the surface, launched with speed $v_e$: $KE=\dfrac{1}{2}mv_e^2,\qquad PE=-\dfrac{GMm}{R},\qquad \text{Total energy}=\dfrac{1}{2}mv_e^2-\dfrac{GMm}{R}.$ As the object travels outward, its kinetic energy is continuously converted into (increasing) potential energy by the Earth's pull, and for the MINIMUM (escape) launch speed, the object's kinetic energy is driven to exactly zero just as it reaches $r=\infty$; there, $PE=-GMm/\infty=0$ as well, so the total energy at infinity is exactly zero. Since total mechanical energy is conserved throughout the motion (gravity is a conservative force), the total energy at launch must ALSO equal zero: $\dfrac{1}{2}mv_e^2-\dfrac{GMm}{R}=0 \quad\Longrightarrow\quad v_e=\sqrt{\dfrac{2GM}{R}}.$ …