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Isn't G a continuous function and although you leave the immediate vicinity of the earth with an escape velocity won't it always exert a force, however small it may be. Won't that force eventually pull the object back to the earth (assuming the absence of other objects)

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It's easiest to think of this problem in terms of energy. At launch the object of mass m has certain amount of kinetic energy corresponding to its velocity v

\begin{equation} E_k = \frac{mv^2}{2} \end{equation}

and certain amount of gravitational potential energy corresponding to its separation r from the central body of mass M

\begin{equation} E_p = -\frac{GMm}{r} \end{equation}

As the object moves away from the central body, it slows down under the influence of the gravitational field, so its kinetic energy depletes while its gravitational potential energy increases since it is now further away from the central body. Kinetic energy is transformed into gravitational potential energy and the total mechanical energy is conserved.

The critical realization is that the amount of kinetic energy ΔEk the object loses as it moves a unit of distance further away from the central body becomes smaller as r grows. This is due to the fact that decrease in kinetic energy ΔEk is equal to the increase in gravitational potential energy ΔEp and Ep's growth slows down considerably as r increases as can easily be seen on the chart of Ep against r. At infinity Ep reaches 0, which means that as r makes infinite change, Ep grows only by a finite value. This fact explains why it is possible to depart to infinity using a finite amount of kinetic energy.

If an object's initial kinetic energy is smaller than that finite amount then it will fall back. If it is equal or larger then it will depart to infinity.

Let's consider some concrete numbers. Imagine a 1kg spacecraft 10,000km away from the center of the Earth speeding away with velocity 8.929km/s. Its kinetic energy is 39.8MJ and its gravitational potential energy is -39.8MJ. When the object reaches 11,000km distance from Earth's center, its velocity will have decreased to 8.513km/s due to gravitational pull of the Earth and its kinetic energy has dropped to 36.2MJ. At the same time its separation from the center of the Earth grew to 11,000km so its gravitational potential energy is now -36.2MJ. Thus on the first 1,000km the drop in kinetic energy (and increase in gravitational potential energy) was 3.6MJ. As the object continues on to 12,000km its velocity drops to 8.151km/s and its kinetic energy to 33.2MJ. At the same time its gravitational potential energy increases to -33.2MJ. This time, the drop in kinetic energy (and corresponding increase in gravitational potential energy) is only 3MJ. For each subsequent 1,000km the drop in kinetic energy becomes smaller and smaller. If you continue this way summing up all infinite decreases in kinetic energy that the body will experience as it departs from Earth, you'll find that the total kinetic energy dropped by 39.8MJ to 0J while the gravitational potential energy increased from -39.8MJ to 0J. At the same time though, distance increased by ∞km.

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No, because the fall-off of gravity is $1/r^2$, the force becomes weaker and weaker as the object moves away, at such a rate that it will never pull the object back in, even though the force is never actually zero. If the gravitational force only fell off as $1/r$ then you would be right, it would always pull the object back in eventually.

My guess is your discomfort is equivalent to some form of Zeno's Paradox.

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It's true that it doesn't matter how far away you get from the Earth, there will always be an attractive force, and that if you are at rest gravity will pull you towards the Earth.

However, if you start at Earth's surface with that speed, you can show that your trajectory will never take you back to Earth. It will be a hyperbola. The reason is that your speed in the direction opposite to the Earth never goes down to zero. It gets smaller and smaller as you get further (because gravity pulls you back), but it never gets to zero.

A vary nice way to see it (if you are familiar with potential energy) is the definition from wikipedia: the escape velocity is the speed you need so your kinetic energy plus the gravitational potential energy add up to zero. This means that the object has enough energy to get out of the gravitational well. It doesn't matter if the gravitational well extends to infinity, the object will not return to the Earth because it will always have nonzero speed.

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Yes the Earth will always exert a force, but that force will get smaller and smaller as the object get further and further. Even if we assume Earth and the object are the only 2 masses in the universe, the gravity from earth will always slow down the object, forever (ie until its distance is infinite), but the velocity of the object will always have a radial positive component, even if it tends towards 0 at infinity.

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Well, whether the radial component of velocity remains positive will depend on its initial value. – David Z Dec 10 '11 at 2:58
I would say the important bit is that the velocity will tend toward some constant value as t->infinity. Like an asymptote. It doesn't need to go to zero. – Steven Lu Dec 10 '11 at 3:56
@DavidZaslavsky I meant radial in the referential of Earth. Surely the distance from the earth will keep on increasing for an object with an initial velocity higher than the escape velocity no? – FrenchKheldar Dec 10 '11 at 5:10
I'm pretty sure the infinity limit would be zero, even if it takes a LONG time to reach that. The only way it would converge towards a finite velocity would be if there was another force to balance gravity. – FrenchKheldar Dec 10 '11 at 5:17

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