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

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

where:

• ${\displaystyle r}$  is the radial distance of the orbiting body from the 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 the orbital equation is:

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

where:

• ${\displaystyle r\,}$  is the radial distance of the orbiting body from the central body,
• ${\displaystyle h\,}$  is the specific angular momentum of the orbiting body,
• ${\displaystyle \nu \,}$  is the 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 the orbital velocity of the orbiting body,
• ${\displaystyle r\,}$  is the radial distance of the orbiting body from the 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 (epoch) 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:

${\displaystyle r={\sqrt[{3}]{{\frac {9}{2}}\mu t^{2}}}}$ 

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