How does possible orbital eccentricity vary with distance from the Sun?

Question

I was pondering a recent comment by Henning Makholm to this answer, he was addressing whether the distance from us of objects beyond Uranus are more affected by our orbit, or their orbits, depending on their orbital eccentricity:

Assuming that our hypothetical object orbits (very) roughly in the ecliptic plane, I don't think eccentricity matters... if the object moved fast enough to get away from Earth faster than Earth goes around half its orbit, it would have more than solar escape velocity!

I had to chew on that for a while.

What is the relationship between an object's distance from its primary, and how eccentric its orbit can be before it will escape the primary? Does this depend on how deep the gravity well is? Is there a term for this, and a formula?

Answer 1

A periodic orbit always has eccentricity in the interval $[0,1)$.

An eccentricity of $1$ denotes a parabolic orbit, in which the object always has exactly the solar escape velocity for its distance from the sun. Such an orbit has no aphelion; after perihelion the object will escape to infinity, in an ultimate direction in direct opposition to its perihelion, while its speed drops off to $0$.

An eccentricity greater than $1$ is a hyperbolic orbit, corresponding to an interstellar trajectory that is deflected to some degree by the sun. Here the object also escapes to infinity, but instead of converging to $0$ its speed (relative to the sun) will drop towards a finite limit as it proceeds ever farther away.

(All of this is unless the object meets another planet or star that pertubes its path away from the ideal Keplerian trajectories we're talking about here).

Thus, by definition (more or less), an eccentricity of $1$ marks the boundary between periodic and escaping orbits. This is independent of how close to the sun the object comes; the eccentricity measures the shape of the orbit rather than its size.