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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?

Is there a more accurate definition of escape velocity?

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 $rrightarrowinfty$.

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 $vecv_e$ 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.

The object doesn't escape the gravitational influence; as you have noted, $1/r^2$ is never equal to $0$. However, what we mean by "escape velocity"is that the object has enough energy (large enough velocity) so that its path would have to "turn around at infinity". In other words, the object's velocity is high enough and decreasing slow enough to where the object will never turn around.

You are right. The notion of "escape" from a gravitational field is something that only exists in an idealized mathematical model of the situation. In that idealized model, we have only one gravitator, in a space-time of infinite extent. With that, we can state the idea of the escape velocity as thus:

It is the minimum speed that an object must be thrown away from the gravitator, in order that it will reach infinite distance, after an infinite lapse of time.

If you throw the object with less speed than this, it will come back: either it will strike the gravitator's surface, or it will go into a repeating orbit around it that nonetheless stays within some maximum distance from the object.

In formal language, if $mathbfr(t)$ is a vector pointing from the gravitator toward the thrown object and which traces its trajectory, and at $t = 0$ we throw the object, the escape velocity is the speed it must begin with, i.e. the minimum speed $v_mathrmesc$ where that if

$$|mathbfr'(0)| = v_mathrmesc$$

then

$$lim_t rightarrow infty |mathbfr(t)| = infty$$

. In reality, of course, there are two limitations: for one, we can never have an infinite time go by, which means that the object will, at all times, still be theoretically under the influence of the gravitator, even if that influence is extremely small. The other limitation is that in reality, there will be other gravity sources present at distance, which will then change the trajectory in other ways once their influence begins to dominate control over the motion of the thrown object and this, in turn, will also prevent reaching that infinite distance. For example, if we are talking about the Solar System, and we are "escaping" the Earth, if we launch a craft away with no more than Earth's escape velocity, then it will only "escape" until the Sun's gravity begins to dominate it, at which point it will now follow a more complicated trajectory based on mutual influence between the Earth and Sun (mostly, it will orbit the Sun, since the Earth will move away in the interim, but it may encounter the Earth again), and also, to some extent, the other planets.

Nonetheless, it is useful because we can take it as a ballpark for the minimum speed, and hence minimum effort which our engine must expend, to "get away" from the gravity source and get to a more remote destination like, say, Mars. Of course, actually travelling there will require more speed change (so-called "delta vee"), because we will also need to navigate to intercept that destination and then settle onto its surface safely.

The important part here is the speed of the orbiting object.

Every distance of the orbiting object has its own orbiting speed (orbiting along the circle). If you increase the speed, the orbiting path changes to an ellipse; if you increase the speed more, the ellipse becomes more and more prolonged.

If you continue to increase speed, there will be the moment in which the path becomes parabolic — and parabola is not closed curve anymore, so the object escapes — not from the gravitational force, but from the orbiting around the object.

Throw a stone up, and it will fall back on Earth. But if you give it very, very large speed, it will go away, into space. Surely, it's always under the gravitational influence of the planet but it has speed great enough that it never returns back.

The minimum speed needed for this is called escape velocity. And at that speed (or greater) the path of the object is an open curve; this means the curve extends to infinity and the object, hence, has no bounds on how far it can go.