It seems it is very difficult to have e=1 perfectly in nature. The final state (being captured or running away) of a celestial body with a parabolic trajectory, is determined by minor perturbation?
No, but nor do elliptical or hyperbolic trajectories. These are features of a model we have for gravity, not part of nature.
The modelling process involves constructing a mathematical framework that describes a natural system and allows for predictions. It is often said that "no models are correct, but some models are useful". For gravity, one common model is the Newtonian model. In the Newtonian model, you can set up two particles in such a way that one follows a parabolic curve (in some inertial frame) Parabolic trajectories exist in the Newtonian model.
In nature, there are more than two particles in the universe and so there are no parabolic, elliptical nor hyperbolic trajectories. Moreover, the Newtonian model is known to only approximate gravity (even General Relativity is probably an approximation, gravity is probably some kind of quantum interaction. But the details are not yet known)
Nevertheless. The Newtonian model is a very useful model for solar system dynamics. Long-period comets, falling from the distant Oort cloud, will have orbits that are very long and are well approximated by parabolas. (but only approximated, as comets, in particular, have significant non-gravitational accelerations due to reaction forces from outgassing, in addition to gravitational perturbations of the planets, and effects from the non-spherical shape of the sun)
Does a parabolic trajectory really exist in nature?
This is a great question! But let's ask two simpler questions first, then move back to yours.
- Do Keplerian orbits (circular, elliptical, parabolic, hyperbolic) exist in nature?
- If they did, what fraction would be parabolic?
The answer to #1 is no. If there were only two spherical objects in the universe, each would be affected only by the other. In that case their trajectory would be Keplerian.
The answer to #2 is zero, and that's for the reason you suspected; eccentricity would have to be exactly unity, and for any finite number of eccentricities that can occur, there's no finite probability that any exact value would come up. There's be zero chance of an eccentricity of 0, 0.5 1, 2 or any specific number.
But to your question about a population of real-world orbits, Those that have eccentricities of say 0.01 to 0.99 we can safely call them elliptical, because in the short run at least they will not be affected by the gravity of other solar system objects enough to push them into exactly zero, or to and beyond 1.
For orbits so close to the hairy edge like an eccentricity of 0.99999999, or 1.00000001, we might call them "parabolic" sometimes, but in those cases we would have an understanding that there is no such thing as an exactly parabolic orbit and that the obit may easily change between bound and unbound states due to a weak perturbing force by another body.
Short answer: Yes, orbits with a value $e=1$ certainly exist. An object perturbed out of a bound state from $e<1$ to $e>1$ must at some time have a value of $e=1$.
Long answer: The intermediate value theorem states that for a continuous function $f$, if $f(t_1)=x_1$ and $f(t_2)=x_2$, then for any value $x$ in the interval $(x_1,x_2)$, there exists a value $t$ in the interval $(t_1,t_2)$ such that $f(t)=x$.
For real bodies, we can assume that eccentricity is a continuous function of time since position and velocity are continuous functions of time (we assume no teleportation or infinite acceleration engines).
If we apply the intermediate value theorem to the eccentricity of an orbit as it passes through a bound state at $t_1$ where eccentricity $e_1 = 0.999$ to an unbound state at $t_2$ where eccentricity $e_2=1.001$, then there must be at least one value $t$, $t_1<t<t_2$ such that the eccentricity is $1$ at time $t$.
- The above proof works for bodies perturbed out of or into a gravitationally bound state (with obligatory XKCD comic):
The resulting orbital path won't appear exactly parabolic in most coordinate systems since it will have a value of $e=1$ for an infinitely short period of time.
We can also define a coordinate system in which an orbit is perfectly parabolic using the same methods I've used here.
Sure. Go to the moon, and throw a rock.
When you're on a planet with no atmosphere, like the Moon, gravity can be considered as a field oriented in a constant direction, straight down, and its force is a effectively constant value in a constant direction. When this occurs, and there is no atmosphere to cause wind resistance, then the trajectory of an object thrown into space and experiencing no other forces will be parabolic in shape as it goes up and then comes back down, as long as its initial speed doesn't exceed the escape velocity of the planet, and it doesn't travel enough distance (vertically or horizontally) for the force of gravity to notably change.