The equation in question is this one:
$$v^2 = v_e^2 + v_{\infty}^2$$
and it is showed in this answer.

The Escape velocity page on Wikipedia mentions a link (num. 4) to the book "Fundamentals of Astrodynamics", wherein the equation is derived from a hyperbolic escape trajectory.

In the answer the spacecraft is in LEO and has to be put into a Mars transfer orbit that is elliptic, but yet it uses the equation in question to calculate the needed velocity at the edge of Earth's SOI.
Already for a parabolic escape trajectory the $v_{\infty}$ would become zero, so how can this speed add to the escape velocity when a elliptic orbit is aimed ?

Is this (theoretically) correct, also since Wikipedia's Hohmann transfer orbit page shows the calculations of the delta-v's for entering and leaving the elliptical transfer orbit ?


2 Answers 2


In the example linked, the transfer orbit is elliptical about the Sun, but to get there you must hyperbolically escape the Earth. This is a patched conic approximation, where the multiple conic sections/orbits (about different gravitationally significant bodies) are 'patched' together at their intersections.

Also note that for the example linked the equation does not "calculate the needed velocity at the edge of Earth's SOI", this was done prior. The "needed velocity at the edge of Earth's SOI" is an input to the equation. The output is the velocity at a low Earth orbit height.

The equation is valid for all conic sections; however, with a negative sign convention, much like how the semi-major axis of a hyperbolic orbit is negative (but opposite). From Wikipedia (emphasis added):

The semi major axis ($a$) is not immediately visible with an hyperbolic trajectory but can be constructed as it is the distance from periapsis to the point where the two asymptotes cross. Usually, by convention, it is negative, to keep various equations consistent with elliptical orbits.

The semi major axis is directly linked to the specific orbital energy ($\epsilon$) or characteristic energy $C_{3}$ of the orbit, and to the velocity the body attains at as the distance tends to infinity, the hyperbolic excess velocity ($v_{\infty }$).

$v_{\infty }^{2}=2\epsilon =C_{3}=-\mu /a$, $v_{\infty }^{2}=2\epsilon > =C_{3}=-\mu /a$, or $a=-{\mu /{v_{\infty }^{2}}}$, $a=-{\mu /{v_{\infty }^{2}}}$

Also note that there are some shenanigans with the negative sign convention to make the equation(s) hold, mainly that squaring/square rooting a value does not affect its sign (i.e., $\sqrt{-x} \to -\sqrt{x}$).

For a circular 250 km low Earth orbit, the $v_{\infty}^2$ would be:

$$v^2=v_e^2+v_{\infty}^2 \to v_{\infty}^2=v^2-v_e^2=\frac{\mu}{r}-\frac{2\mu}{r} \to v_{\infty}^2=-60km^2/s^2$$

Negative, as expected for an elliptical (circle is a special case of an ellipse) orbit.

  • 1
    $\begingroup$ Thank you, the "patched conic approximation" helped me to understand that a hyperbolical escape would yet be possible and that the orbital velocity around the Sun could be the $v_{\infty}$ input for the equation. $\endgroup$
    – Cornelis
    Commented Jan 20, 2022 at 13:25
  • 1
    $\begingroup$ @Cornelis The orbital velocity about the Sun subtract the Earth's orbital velocity gives $v_{\infty}$, as the linked answer shows. $\endgroup$ Commented Jan 20, 2022 at 13:53
  • 1
    $\begingroup$ That's what I meant to say ! :) $\endgroup$
    – Cornelis
    Commented Jan 20, 2022 at 14:18

The velocities are used to express the kinetic energies.

If conservation of energy is valid, the sum of the kinetic energy and the gravitational potential energy is constant.

The parabolic, circular, elliptic and hyperbolic trajectories are all special cases of the transitions between these two forms of energy.

See wikipedia about orbital speed, escape speed and delta v budget

The first cosmic velocity is the minimal speed required for a circular orbit with zero height. The second cosmic velocity is the escape velocity.

If the speed is less than the first cosmic velocity, the rocket returns to the surface in a parabolic trajectory.

If the speed is greater than the first cosmic velocity and less than the second, the orbit is circular or elliptic.

If the speed is greater than the second cosmic velocity, the probe leaves on a hyperbolic trajectory.


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