What's the point of characteristic energy?

Question

My understanding of orbital mechanics is that delta-v is the most useful notion of ability to change orbits in space, akin to energy being the most useful notion of the ability to do work. But if delta-v is the standard "space currency" when it comes to the physics of designing rockets, when and why is characteristic energy used?

One can easily find characteristic energy defined as the energy per unit mass over that required to escape from a gravity well. However, those definitions don't teach why this would be a useful metric to have. We could just as easily track the total kinetic energy of all propellant on the rocket, which is a fairly useless metric. We could also track the average distance traveled by a particle of exhaust before reaching a deflection of 45 degrees, which is similarly useless.

My question is, given there are infinitely many metrics we could keep track of, why is C3, characteristic energy, useful? Further, what does it tell us that my current "space currency" delta-v does not?

Answer 1

Partial answer, waiting for a rocket scientist to chime in.

This is a cool question!

Celta-v calculated from exhaust velocity using the Tsiolkovsky rocket equation for each stage would be a huge overestimate because it doesn't account for things like atmospheric drag or gravity.

So you'd have to numerically integrate over a specific trajectory for a final delta-v relative to the launch site taking those into effect explicitly, then subtract off Earth's rotation and escape velocity to get a useful number; speed in space at a given distance from Earth after SECOx.

That turns out to have pretty much the same launch capability information as a geocentric C3 figure.

So I guess C3 is just the bottom-line figure that mission planners need to know when selecting rockets from the "launch vehicles 2021" catalog, and it contains the results of the numerical (or real-life) consequences of all the staging, drag and gravitational effects for a given trajectory.

But there are short cuts!

If Tsiolkovsky gives you say 15 km/s using masses and exhaust velocities of a staged launch...

• Specific impulse and delta-v in Tsiolkovsky's rocket equation
• How does the Rocket Equation work when you use boosters?

then folks will just subtract off Earth's escape velocity of about 11.2 km/s and a fudge factor for gravity loss (and a bit for aerodynamic loss):

Gravity losses depend on the time over which thrust is applied as well the direction the thrust is applied in. Gravity losses as a proportion of delta-v are minimised if maximum thrust is applied for a short time, or if thrust is applied in a direction perpendicular to the local gravitational field. During the launch and ascent phase, however, thrust must be applied over a long period with a major component of thrust in the opposite direction to gravity, so gravity losses become significant. For example, to reach a speed of 7.8 km/s in low Earth orbit requires a delta-v of between 9 and 10 km/s. The additional 1.5 to 2 km/s delta-v is due to gravity losses, steering losses and atmospheric drag

So your 15 km/s Tsiolkovsky delta-v gets you 15 - 11.2 - 1.5 = 2.3 km/s after leaving Earth's influence.

That's a C3 or characteristic energy of 5.3 km^2/s^2, since the potential term is zero.

See also:

• Modeling a rocket using Tsiolkovsky's equation and ordinary differential equations

Just don't get too excited and tweet the wrong C3!

• How and/or why did Elon Musk mistweet Roadster's C3 so badly?