Escape velocity is the minimum speed needed for an object to escape from the gravitational influence of a massive body. However, gravity has infinite range. Object $A$ is always getting pulled by the gravity of object $B$ no matter the distance between object A and B. (Maybe the force becomes extremely small with increasing distance, but still not zero)
So, how can an object possibly “escape the gravitational influence of another object”? Is there a more accurate definition of escape velocity? Won't the orbital path just become infinitely large with increasing initial velocity?
Best Answer
First, consider an object with a radial orbit (zero angular momentum orbit) in a $1/r^2$ central force field. The total energy $E$ of the object (which is constant) is the sum of the (negative) potential energy $U(r)$ due to the force field and the kinetic energy $T(v)$ due to the radial speed.
There are three cases to consider:
$E \lt 0$: The orbit is bounded, i.e., there is a maximum, finite distance $r = r_{max}$ where the speed (and thus kinetic energy) is zero, and the object has maximum (least negative) potential energy
$E \ge 0$: The orbit is unbounded, i.e., the object never has zero speed. As $r \rightarrow \infty$, the kinetic energy $T$ asymptotically approaches $E$ from above.
$E = 0$: A special case of the unbounded orbit in that $T \rightarrow 0$ as $r\rightarrow\infty$.
It is this special case that is relevant to the definition of escape velocity. At any radius $r_0$ in the central force field, there is a speed $v_e$ such that
$$T(v_e) = |U(r_0)|$$
Thus, an object starting at $r = r_0$ with outward radial velocity $\vec{v}_e$ has just enough kinetic energy to, ahem, 'coast to a stop at $r = \infty$'.
More precisely, the object will coast arbitrarily far away with speed arbitrarily close to zero. In this sense, the object 'escapes' the central force field, but just so.