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This question already has an answer here:

The velocity required to escape a gravitational body is termed escape velocity. But how is it possible for any object to escape a body if it's gravitational field acts over an infinite range, so the velocity of the escaping object should always be decreasing until it reaches zero then steadily moving back towards the body?

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marked as duplicate by sammy gerbil, ZeroTheHero, Jon Custer, Buzz, John Rennie newtonian-mechanics Dec 31 '18 at 8:54

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The escape velocity is the velocity you need so that you are always traveling away from the body, no matter how far away you get — your speed never reaches zero, so you never turn around and start moving back. If we choose the gravitational potential energy to vanish at infinity, $V(\infty) = 0$, then we find the escape speed by stipulating that our speed should only reach 0 when we're infinitely far away, \begin{equation} E_0 = E_f\\ \rightarrow \frac{1}{2}mv_{\text{esc}}^2 - \frac{GmM}{r_0} = V(\infty) + \frac{1}{2}mv_{\infty}^2 = 0, \end{equation} You might say that we can never reach infinity, but that's the point: as long as you're a finite distance away, your velocity away from the body will be non-zero.

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You are right, but at infinity the gravitational effect is zero, that is because the potential decreases as $\Phi(r) \sim r^{-1}$ so $\lim_{r\to\infty}\Phi(r) = 0$. With this in mind, the escape velocity is defined as the minimum velocity you have to imprint a particle so that it reaches infinity with velocity $0$

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The escape velocity is an idealization. But you will notice that the escape velocity is a finite value and is derived by assuming that we will get to infinitely far away from the body with a velocity of $0$. So if we move away faster than this velocity, we will be even better off.

Another thing to consider is that you can still be far enough away from a body to where the gravitational effects are negligible.

So if you were to leave your body with the escape velocity or faster, then you are going to get far enough away so that the effects of gravity are negligible.

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The gravitational field of a body (let's call it body 1) has an infinite range, but its magnitude or strength decreases with distance. In other words, the gravitational field is strongest near the body and weakens with distance. Mathematically, the gravitational field is proportional to $\frac{1}{r^2}$, where $r$ is the distance from the centre of the body. So the velocity of the body (let's call it body 2) will decrease but if it is large enough to begin with, it will go on moving away from body 1. If it is not large enough, it will do what you say and fall back towards body 1.

To be more precise, the velocity of body 2 should be high enough for its kinetic energy to be greater than the potential energy it has due to the attraction from body 1.

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To escape from a body means to move in a hyperbolic orbital. This has an unbounded distance to that object. In non-relativistic mechanics and Newtonian gravity this means that the kinetic energy should exceed the gravitational potential at any position. At infinity where the gravitational potential tends to zero the kinetic energy should then tend to a finite - nonzero - value. At the surface of the body, assuming it has one, the kinetic energy should be at least as large as the potential energy, which allows the derivation of a simple expression for the escape velocity.

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Well, consider the function $log(x)$ - its first derivative is always "slowing down" but it never reverses and starts going down.

In the case of escape velocity we can exploit conservation of energy. The kinetic energy of the escaping body is $\frac{1}{2} m v^2$. The gravitational potential energy is $-1 * \frac{K_1}{r}$ where $K_1$ is a positive constant and $r$ is the distance between the escaping body and the larger body. So, $\frac{1}{2} m v^2 - \frac{K_1}{r}$ must stay constant. Let's say $\frac{1}{2} m v^2 - \frac{K_1}{r} = K_2 $ where $K_2$ is another constant. As $r$ increases the velocity must fall - but it need not fall as low as zero.

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