Editing Escape velocity (section)

== Trajectory ==
If an object attains exactly escape velocity, but is not directed straight away from the planet, then it will follow a curved path or trajectory. Although this trajectory does not form a closed shape, it can be referred to as an orbit. Assuming that gravity is the only significant force in the system, this object's speed at any point in the trajectory will be equal to the escape velocity ''at that point'' due to the conservation of energy, its total energy must always be 0, which implies that it always has escape velocity; see the derivation above. The shape of the trajectory will be a [[parabola]] whose focus is located at the center of mass of the planet. An actual escape requires a course with a trajectory that does not intersect with the planet, or its atmosphere, since this would cause the object to crash. When moving away from the source, this path is called an [[escape orbit]]. Escape orbits are known as {{nowrap|1=''C''{{sub|3}} = 0}} orbits. ''C''{{sub|3}} is the [[characteristic energy]], −''GM''/2''a'', where ''a'' is the [[semi-major axis]] length, which is infinite for parabolic trajectories.

If the body has a velocity greater than escape velocity then its path will form a [[hyperbolic trajectory]] and it will have an excess hyperbolic velocity, equivalent to the extra energy the body has. A relatively small extra [[delta-v|delta-''v'']] above that needed to accelerate to the escape speed can result in a relatively large speed at infinity. [[bi-elliptic transfer|Some orbital manoeuvres]] make use of this fact. For example, at a place where escape speed is 11.2&nbsp;km/s, the addition of 0.4&nbsp;km/s yields a hyperbolic excess speed of 3.02&nbsp;km/s:
: <math>v_\infty = \sqrt{ V^2 - {v_\text{e}}^2 } = \sqrt{ (11.6 \text{ km/s})^2 - (11.2 \text{ km/s})^2 } \approx 3.02 \text{ km/s}.</math>

If a body in circular orbit (or at the [[periapsis]] of an elliptical orbit) accelerates along its direction of travel to escape velocity, the point of acceleration will form the periapsis of the escape trajectory. The eventual direction of travel will be at 90 degrees to the direction at the point of acceleration. If the body accelerates to beyond escape velocity the eventual direction of travel will be at a smaller angle, and indicated by one of the asymptotes of the hyperbolic trajectory it is now taking. This means the timing of the acceleration is critical if the intention is to escape in a particular direction.

If the speed at periapsis is {{mvar|v}}, then the [[Eccentricity vector|eccentricity]] of the trajectory is given by:
: <math>e=2(v/v_\text{e})^2-1</math>
This is valid for elliptical, parabolic, and hyperbolic trajectories. If the trajectory is hyperbolic or parabolic, it will [[asymptotically]] approach an angle <math>\theta</math> from the direction at periapsis, with
: <math>\sin\theta=1/e.</math>
The speed will asymptotically approach
: <math>\sqrt{v^2-{v_\text{e}}^2}.</math>