# What is the "launch energy" or C₃ measured in km² / s²

In some papers there is "launch energy" characteristic of spacecraft launched into space from Earth. Sometimes it is written as "C₃" (C3). For example, Ares V document lists "C₃" as equal or less than 35 km²/s².

What is the meaning of "C₃" (Characteristic energy?), and what is difference between 35 km²/s² (start from Earth), 10 km²/s² ("delivering 8 tonn to mars at a C₃ of 10 km²/s²"), 80 km²/s² (required for sample return from Jovian moons), and 110 km²/s² (needed to reach Saturn without gravity assists - in direct mission)

• Looks like C3 is V∞² - stereo.nrl.navy.mil/orig_stereo/PPA/… or books.google.com/books?id=oZfpYIUKDrUC&pg=PA57 - "The launch energy, C3, is defined as the square of the hyperbolic excess velocity (V∞) of the spacecraft with respect to the Earth, a measure related to how much velocity increase must be supplied to the spacecraft by the launch vehicle at launch."
– osgx
Mar 20, 2014 at 23:54
• I don't understand your question "what is the difference". Those are just different C3's for different target orbits to reach those bodies. Mar 21, 2014 at 0:18

It is what your velocity would be at a sufficient distance from Earth that its gravity doesn't matter, squared. That velocity is $v_\infty$, so $C_3=v_\infty^2$.

It can be calculated at any distance from Earth as your specific energy (energy per unit mass), times two:

$C_3 = v_\infty^2 = v^2-{2\mu\over r}$

So wherever you are, use the magnitude of your velocity, $v$, at that point, your distance from the center of the Earth or the barycenter of the Earth-Moon system, $r$, and the GM of Earth or the Earth-Moon system, $\mu$. Note that as $r$ goes to $\infty$, the expression goes to your velocity squared.

The term has an ancient derivation as the third integration constant when integrating the equation of motion.

• I've been reading about the recover(ies) of SOHO just now and thought I'd never be able to figure out what $C_3$ was, and suddenly this Q&A appear in the related sidebar to another Q&A. This is a really helpful answer - contains extra information which is also helpful!
– uhoh
Jun 4, 2016 at 8:06