# Parabolic trajectory

In astrodynamics or celestial mechanics a parabolic trajectory is a Kepler orbit with the eccentricity equal to 1 and is an unbound orbit that is exactly on the border between elliptical and hyperbolic. When moving away from the source it is called an escape orbit, otherwise a capture orbit. It is also sometimes referred to as a C$_3$ = 0 orbit (see Characteristic energy).
Under standard assumptions a body traveling along an escape orbit will coast along a parabolic trajectory to infinity, with velocity relative to the central body tending to zero, and therefore will never return. Parabolic trajectories are minimum-energy escape trajectories, separating positive-energy hyperbolic trajectories from negative-energy elliptic orbits.

## Velocity

The orbital velocity ($v$) of a body travelling along a parabolic trajectory can be computed as:
    $v={\sqrt {2\mu \over r}}$
where:
* $r$ is the radial distance of the orbiting body from the central body,
* $\mu$ is the standard gravitational parameter.
At any position the orbiting body has the escape velocity for that position.
If a body has an escape velocity with respect to the Earth, this is not enough to escape the Solar System, so near the Earth the orbit resembles a parabola, but further away it bends into an elliptical orbit around the Sun.
This velocity ($v$) is closely related to the orbital velocity of a body in a circular orbit of the radius equal to the radial position of orbiting body on the parabolic trajectory:
    $v={\sqrt {2}}\,v_{o}$
where:
* $v_{o}$ is orbital velocity of a body in circular orbit.

## Equation of motion

For a body moving along this kind of trajectory the orbital equation is:
    $r={h^{2} \over \mu }{1 \over {1+\cos \nu }}$
where:
* $r\,$ is the radial distance of the orbiting body from the central body,
* $h\,$ is the specific angular momentum of the orbiting body,
* $\nu \,$ is the true anomaly of the orbiting body,
* $\mu \,$ is the standard gravitational parameter.

## Energy

Under standard assumptions, the specific orbital energy ($\epsilon$) of a parabolic trajectory is zero, so the orbital energy conservation equation for this trajectory takes the form:
    $\epsilon ={v^{2} \over 2}-{\mu \over r}=0$
where:
* $v\,$ is the orbital velocity of the orbiting body,
* $r\,$ is the radial distance of the orbiting body from the central body,
* $\mu \,$ is the standard gravitational parameter.
This is entirely equivalent to the characteristic energy (square of the speed at infinity) being 0:
    $C_{3}=0$

## Barker's equation

Barker's equation relates the time of flight $t$ to the true anomaly $\nu$ of a parabolic trajectory:
    $t-T={\frac {1}{2}}{\sqrt {\frac {p^{3}}{\mu }}}\left(D+{\frac {1}{3}}D^{3}\right)$
where:
* $D=\tan {\frac {\nu }{2}}$ is an auxiliary variable
* $T$ is the time of periapsis passage
* $\mu$ is the standard gravitational parameter
* $p$ is the semi-latus rectum of the trajectory ($p=h^{2}/\mu$ )
More generally, the time (epoch) between any two points on an orbit is
    $t_{f}-t_{0}={\frac {1}{2}}{\sqrt {\frac {p^{3}}{\mu }}}\left(D_{f}+{\frac {1}{3}}D_{f}^{3}-D_{0}-{\frac {1}{3}}D_{0}^{3}\right)$
Alternately, the equation can be expressed in terms of periapsis distance, in a parabolic orbit $r_{p}=p/2$:
    $t-T={\sqrt {\frac {2r_{p}^{3}}{\mu }}}\left(D+{\frac {1}{3}}D^{3}\right)$
Unlike Kepler's equation, which is used to solve for true anomalies in elliptical and hyperbolic trajectories, the true anomaly in Barker's equation can be solved directly for $t$. If the following substitutions are made
    ${\begin{aligned}A&={\frac {3}{2}}{\sqrt {\frac {\mu }{2r_{p}^{3}}}}(t-T)\\[3pt]B&={\sqrt[{3}]{A+{\sqrt {A^{2}+1}}}}\end{aligned}}$
then
    $\nu =2\arctan \left(B-{\frac {1}{B}}\right)$
With hyperbolic functions the solution can be also expressed as:
    $\nu =2\arctan \left(2\sinh {\frac {\mathrm {arcsinh} {\frac {3M}{2}}}{3}}\right)$
where
    $M={\sqrt {\frac {\mu }{2r_{p}^{3}}}}(t-T)$

## Radial parabolic trajectory

A radial parabolic trajectory is a non-periodic trajectory on a straight line where the relative velocity of the two objects is always the escape velocity. There are two cases: the bodies move away from each other or towards each other.
There is a rather simple expression for the position as function of time:
    $r={\sqrt[{3}]{{\frac {9}{2}}\mu t^{2}}}$
where
* μ is the standard gravitational parameter
* $t=0\!\,$ corresponds to the extrapolated time of the fictitious starting or ending at the center of the central body.
At any time the average speed from $t=0\!\,$ is 1.5 times the current speed, i.e. 1.5 times the local escape velocity.
To have $t=0\!\,$ at the surface, apply a time shift; for the Earth (and any other spherically symmetric body with the same average density) as central body this time shift is 6 minutes and 20 seconds; seven of these periods later the height above the surface is three times the radius, etc.