Final Answer:
Parabolic paths have an eccentricity precisely equal to 1, resulting in a trajectory at the boundary between bound and unbound paths, making them incompatible with the definition of a stable orbit. Thus the correct option is option (D).
Step-by-step explanation:
In celestial mechanics, a planet's orbit can be described by its eccentricity, a measure of how much its shape deviates from a perfect circle. A circular path (A) has an eccentricity of 0, and an elliptical path (B) has an eccentricity between 0 and 1. Hyperbolic paths (C) have eccentricities greater than 1, representing trajectories where the planet is not bound to the sun and will escape its gravitational influence. Parabolic paths (D) have an eccentricity precisely equal to 1.
To understand why a parabolic path is not a real orbit, we can refer to the specific energy (E) equation in orbital mechanics. For a bound orbit, the total energy must be negative (E < 0), indicating a closed trajectory. In the case of a parabolic orbit, the specific energy is zero (E = 0), resulting in a trajectory at the boundary between bound and unbound paths.
This signifies that the planet will approach the sun asymptotically but never complete a full orbit, making a parabolic path incompatible with the definition of a stable orbit.