In this answer I've interpreted the x axis of the plot in the question (also shown below) labeled as $C_3$ (characteristic energy) to be the "excess $v^2$ above escape velocity from Earth", and so I'd written the relation.

$$C_3 = v^2 - v_{esc}^2.$$

Could I have some confirmation that this is a correct way to use the $C_3$ used in the plot? Am I missing a factor of 2 or are there other subtleties nor not-so-subtleties that I've overlooked?

The expression in @MarkAdler's answer is cast in terms of potential energy (per unit mass) but I've assumed conservation of total energy and applied it to kinetic energy (per unit mass).

The figure below is from the linked question, there is currently no original source listed for the image.

enter image description here


1 Answer 1


That is the formula. Wikipedia shows it a bit different:

$\tfrac{1}{2} v^2 - \mu/r = \text{constant} = \tfrac{1}{2} C_3$

Multiply by two, and the only question becomes the $2\mu/r$ term. Escape velocity is $\sqrt{\frac{2GM}{R}}$ per this source. If you square that, as you did in your initial formula, you come out with exactly the same thing as the Wikipedia article. Thus your formulation is correct.

As for the chart, it comes from Nasa's Performance Vehicle website.

  • 1
    $\begingroup$ yay! Thank you for the validation, now I can sleep soundly. $\endgroup$
    – uhoh
    Mar 19, 2018 at 18:25

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