Q.An object is thrown from Earth in such a way that it reaches a point at infinity with non-zero kinetic energy $K.E._{(r\to\infty)} = \dfrac{1}{2}Mv_\infty^2$. With what velocity should the object be thrown from Earth?

Definition. Escape speed is the minimum speed an object must be launched from a planet's surface with, in order to break completely free of that planet's gravity and never fall back, even after travelling forever.

Derivation. An object of mass $M$ launched from Earth's surface with speed $v_i$ has total energy $E_i=\tfrac12Mv_i^2-\dfrac{GMM_e}{R_e}$. For the minimum possible escape speed, the object should just barely reach an infinite distance with zero kinetic energy left over (both $U$ and $K.E.$ tend to zero as $r\to\infty$), so $E_f=0$. Energy conservation ($E_i=E_f$) then gives

using $g=GM_e/R_e^2$. Numerically, with $g=9.8\ \text{m s}^{-2}$ and $R_e=6400\ \text{km}$, Earth's escape speed works out to about $11.2\ \text{km s}^{-1}$. …