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 C3 = 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 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 $$r_p = p/2$$:
 * $$t - T = \sqrt{\frac{2 r_p^3}{\mu}} \left(D + \frac{1}{3} D^3\right)$$

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{align}

A &= \frac{3}{2} \sqrt{\frac{\mu}{2r_p^3}} (t - T) \\[3pt] B &= \sqrt[3]{A + \sqrt{A^{2}+1}} \end{align}$$

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.