Escape orbit

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 parabolic trajectory can be computed as:
:$v = \sqrt$
where:
*$r$ is the radial distance of orbiting body from 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:
:$v = \sqrt\, v_o$
where:
*$v_o$ is orbital velocity of a body in circular orbit.

Equation of motion

For a body moving along this kind of trajectory an orbital equation becomes:
:$r =$
where:
*$r\,$ is radial distance of orbiting body from central body,
*$h\,$ is specific angular momentum of the orbiting body,
*$\nu\,$ is a 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:
:$\epsilon = - = 0$
where:
*$v\,$ is orbital velocity of orbiting body,
*$r\,$ is radial distance of orbiting body from 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:
:$t - T = \frac \sqrt \left(D + \frac D^3 \right)$

where:
*$D = \tan \frac$ 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
:$t_f - t_0 = \frac \sqrt \left(D_f + \frac D_f^3 - D_0 - \frac D_0^3\right)$

Alternately, the equation can be expressed in terms of periapsis distance, in a parabolic orbit $r_p = p/2$:
:$t - T = \sqrt \left(D + \frac D^3\right)$

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
A &= \frac \sqrt (t - T) \\ pt B &= \sqrt \end

then
: $\nu = 2\arctan\left(B - \frac\right)$

With hyperbolic functions the solution can be also expressed as: Eq.(40) and Appendix C.

: $\nu = 2\arctan\left(2\sinh\frac\right)$

where

: $M = \sqrt (t - T)$

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 /math>

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.

See also

* Kepler orbit
* Parabola

References

{{DEFAULTSORT:Parabolic Trajectory
Orbits