Parabolic trajectory

The orbital velocity (${\displaystyle v}$) of a body travelling along parabolic trajectory can be computed as:

${\displaystyle v={\sqrt {2\mu \over r}}}$

where:

• ${\displaystyle r}$ is the radial distance of orbiting body from central body,
• ${\displaystyle \mu }$ 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 (${\displaystyle v}$) 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:

${\displaystyle v={\sqrt {2}}\,v_{o}}$

where:

• ${\displaystyle v_{o}}$ is orbital velocity of a body in circular orbit.

For a body moving along this kind of trajectory an orbital equation becomes:

${\displaystyle r={h^{2} \over \mu }{1 \over {1+\cos \nu }}}$

where:

• ${\displaystyle r\,}$ is radial distance of orbiting body from central body,
• ${\displaystyle h\,}$ is specific angular momentum of the orbiting body,
• ${\displaystyle \nu \,}$ is a true anomaly of the orbiting body,
• ${\displaystyle \mu \,}$ is the standard gravitational parameter.

Under standard assumptions, the specific orbital energy (${\displaystyle \epsilon }$) of a parabolic trajectory is zero, so the orbital energy conservation equation for this trajectory takes the form:

${\displaystyle \epsilon ={v^{2} \over 2}-{\mu \over r}=0}$

where:

• ${\displaystyle v\,}$ is orbital velocity of orbiting body,
• ${\displaystyle r\,}$ is radial distance of orbiting body from central body,
• ${\displaystyle \mu \,}$ is the standard gravitational parameter.

This is entirely equivalent to the characteristic energy (square of the speed at infinity) being 0:

${\displaystyle C_{3}=0}$

Barker's equation relates the time of flight ${\displaystyle t}$ to the true anomaly ${\displaystyle \nu }$ of a parabolic trajectory:[1]

${\displaystyle t-T={\frac {1}{2}}{\sqrt {\frac {p^{3}}{\mu }}}\left(D+{\frac {1}{3}}D^{3}\right)}$

where:

• ${\displaystyle D=\tan {\frac {\nu }{2}}}$ is an auxiliary variable
• ${\displaystyle T}$ is the time of periapsis passage
• ${\displaystyle \mu }$ is the standard gravitational parameter
• ${\displaystyle p}$ is the semi-latus rectum of the trajectory (${\displaystyle p=h^{2}/\mu }$ )

More generally, the time between any two points on an orbit is

${\displaystyle t_{f}-t_{0}={\frac {1}{2}}{\sqrt {\frac {p^{3}}{\mu }}}\left(D_{f}+{\frac {1}{3}}D_{f}^{3}-D_{0}-{\frac {1}{3}}D_{0}^{3}\right)}$

Alternately, the equation can be expressed in terms of periapsis distance, in a parabolic orbit ${\displaystyle r_{p}=p/2}$:

${\displaystyle t-T={\sqrt {\frac {2r_{p}^{3}}{\mu }}}\left(D+{\frac {1}{3}}D^{3}\right)}$

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 ${\displaystyle t}$. If the following substitutions are made

${\displaystyle {\begin{aligned}A&={\frac {3}{2}}{\sqrt {\frac {\mu }{2r_{p}^{3}}}}(t-T)\\[3pt]B&={\sqrt[{3}]{A+{\sqrt {A^{2}+1}}}}\end{aligned}}}$

then

${\displaystyle \nu =2\arctan \left(B-{\frac {1}{B}}\right)}$

With hyperbolic functions the solution can be also expressed as:[2]

${\displaystyle \nu =2\arctan \left(2\sinh {\frac {\mathrm {arcsinh} {\frac {3M}{2}}}{3}}\right)}$

where

${\displaystyle M={\sqrt {\frac {\mu }{2r_{p}^{3}}}}(t-T)}$

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:

${\displaystyle r={\sqrt[{3}]{4.5\mu t^{2}}}}$

where

• μ is the standard gravitational parameter
• ${\displaystyle t=0\!\,}$ 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 ${\displaystyle t=0\!\,}$ is 1.5 times the current speed, i.e. 1.5 times the local escape velocity.

To have ${\displaystyle t=0\!\,}$ 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.

• Kepler orbit
• Parabola