Parabolic trajectory

Velocity

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

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

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.

Equation of motion

For a body moving along this kind of trajectory the orbital equation is:

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

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.

Energy

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

${\displaystyle ϵ=\frac{v^2}{2}-\frac{\mu }{r}=0}$

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

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

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

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, given by ${\displaystyle p=h^2/\mu }$

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

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

then

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

With hyperbolic functions the solution can be also expressed as:

${\displaystyle \nu =2\arctan \left(2\sinh \frac{\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{h}\frac{3M}{2}}{3}\right)}$

where

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

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:

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

where

• ${\displaystyle \mu }$ 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.