Discontinuity in gravitational potential If you raise an object you increase it's (absolute value) potential energy.
But if you exceed an specific height suddenly the gravity can't pull over the object and it can be interpreted as the potential is turned to zero! (the intuition of potential energy is how much energy it can be released if you let the object go)
Does this mean that the potential energy's function is discontinuous? I guess not, but why? It's against my intuition!
 A: The potential is always defined as an integral $$U(\vec{x}) = \int^{\vec{x}}_{\vec{x}_0\:\:\Gamma} \vec{F}(\vec{r}) \cdot d\vec{r}\tag{1}$$ the path $\Gamma$ joins $\vec{x}_0$ to $\vec{x}$ and its form is irrelevent just because the vector field $\vec{F} = \vec{F}(\vec{r})$ is conservative.
It is an easy exercise to prove that the right-hand side of (1) as a function of $\vec{x}$ is always continuous if the function $\vec{F}$ is integrable (nomatter if one adopts the notion of eintegral of eithet Riemann or Lebesgue). If the function $\vec{F}$ is not integrable,  the potential does not exist at all.
Summing up: If the potential exists, it  is necessarily continuous.
Regarding your question, let us focus on your statement.
"But if you exceed an specific height suddenly the gravity can't pull over the object and it can be interpreted as the potential is turned to zero."
This is wrong, the correct statement would terminate like this
"...it can be interpreted as the potential is [continuously!] turned to a constant value."
since the force is the derivative of the potential and the derivative of a constant vanishes.
A: The problem here is that many introductory physics courses say that gravitation potential energy is
$$U = mgh.$$
This is an approximate expression that only works if your mass $m$ is much lighter than the Earth and relatively close to the Earth. However, it breaks down the second you start considering more complicated situations.
For example, suppose we had an object between the Earth and the moon. Then when it's close to the moon, we have
$$U \approx m g_{\text{moon}} h_{\text{moon}}$$
and when it's close to the Earth, we have
$$U \approx m g_{\text{Earth}} h_{\text{Earth}}.$$
However, these two expressions cannot be combined consistently everywhere. In introductory courses, they sometimes gloss over this point by saying that you should only consider the nearest / most important heavy body, like the one you would fall towards if you let the object go. But that implies that the potential discontinuously changes from the first option to the second option, as you move closer from the moon to the Earth, like the moon's potential suddenly "turns off".
This problem appears because both of the expressions above are only approximate. The true potential energy is
$$U = -\frac{GM_{\text{moon}}m}{r_{\text{moon}}} - \frac{GM_{\text{Earth}}m}{r_{\text{Earth}}}$$
where the $r$'s are the distances to the center of the moon and Earth. Using some calculus, you can show that this reduces to the previous two expressions when the object is very close to either the moon or the Earth.
