# Why is “delta-v + $\sqrt{{v_E}^2 + C_3}$ where ${v_E}^2 =$11.19 km/s” the correct way to calculate total propulsive delta-v? Please show all work

There is one answer to What spacecraft has had the greatest total propulsive delta-v? and I can not understand how its numbers have been calculated. Response to comments does not seem forthcoming so I'll ask separately for a good, clear, science and math-based explanation how this works. As we are told in school, please show all work!

From this answer:

Taking all of this in to account, the delta-v of each space craft defined as spacecraft only delta-v + $$\sqrt{{v_E}^2 + C_3}$$, where $${v_E}^2 = 11.19 km/s$$, the escape velocity from Earth.. The latter part converts the $$C_3$$ to the effective delta-v, when taking in to account losses from atmospheric drag, gravity drag, ineffective trajectories, etc. This seems to be the fairest way to calculate the effective delta-v. Taking all of this in to account, the following is the delta-v.

• Dawn- 22.89 km/s
• PSP- ~17.2 km/s
• New Horizons- 17.61 km/s
• Cassini- 15.69 km/s
• Juno- <14.5 km/s

The numbers changed from one edit to the next but have since stabilized.

Values for C3 and delta-v are scattered throughout the text, but if I understand correctly, if inserted in that equation result in those values.

I think they are meant to be geocentric C3 values rather than heliocentric (see this answer for examples of a heliocentric C3 and how to show one's work), and when quoted are actually the square roots of C3.

I can't understand the math;

• why velocities are added in quadrature
• why the units don't seem to work
• and how this produces the correct total propulsive delta-v for these spacecraft, either starting from Earth or from LEO.

Please explain in a clear, systematic way why this is the correct way to calculate total propulsive delta-v if it is, or how it should be done if it is not.

• I believe that what you're seeing here is a correction for the Oberth effect. When you conduct a burn in a gravity well it's effect is amplified as you climb out of the gravity well, the deeper the gravity well the more climb and the more benefit. – Loren Pechtel Oct 19 '20 at 2:26
• @LorenPechtel Interesting! That's a cool way to look at it. I'm still trying to figure out what the Oberth effect is and isn't. It should be easy but I'm still missing something. – uhoh Oct 19 '20 at 2:39
• Suppose you're heading away from Earth at 11.19 km/sec. Gravity keeps pulling on you, you get away but just barely, all your energy is spent on the escape. Now, lets try heading out at 12.19 km/sec. Gravity's pull is purely based on time, but you're moving faster, there's less time for gravity to act and it won't be able to claw away the whole 11.19 km/sec. The extra velocity that wasn't clawed away is the Oberth benefit. Note that this works both ways--do your capture burns as close to the planet as possible also. – Loren Pechtel Oct 19 '20 at 2:51

## 1 Answer

The calculation uses the following model for "total propulsive delta-v":

$$\Delta v_{total} = \Delta v_{spacecraft} + \Delta v_{launcher}$$

Here, $$\Delta v_{spacecraft}$$ is what propulsive capabilities the probe has by itself after leaving the Earth system entirely, and is presumed to be a known value that can be looked up.

$$\Delta v_{launcher}$$ is what's spent from starting still on the surface of the Earth, until the probe is sent on an escape trajectory away from the Earth.

For those escape trajectories, the quantity $$C_3$$ is known, and is defined as twice the excess energy after escape. The wikipedia page for characteristic energy has the following helpful formula to illustrate the relationship between orbital energy and $$C_3$$

$$\frac{1}{2} C_3 = \epsilon = \frac{1}{2} v^2 - \frac{\mu}{r} = constant$$

I would also like to expand on the $$\frac{1}{2} v^2 - \frac{\mu}{r}$$ part. When "escaped", $$r$$ is presumed to be some infinite, or at least very high number. The potential energy part thus goes towards zero.

We then have the following very handy relationship:

$$C_3 = v_{\infty}^2$$

$$C_3$$ is just the velocity "at infinity" squared.

Note the part about $$C_3$$ being constant along the trajectory. We can work from there:

$$\frac{1}{2} C_3 = \frac{1}{2} v^2 - \frac{\mu}{r}$$

$$C_3 = v^2 - \frac{2\mu}{r}$$

$$v^2 = \frac{2\mu}{r} + C_3$$

$$v = \sqrt{\frac{2\mu}{r} + C_3}$$

Now, by looking at the definition of escape velocity, $$v_e = \sqrt{\frac{2\mu}{r}}$$, or $$v_e^2 = \frac{2\mu}{r}$$. Which can then be substituted into the previous equation:

$$v = \sqrt{v_e^2 + C_3}$$

This is to be understood as the velocity of the escape trajectory when $$r$$ is the surface of the Earth, of which the launcher is presumed to supply everything since it's starting from zero:

$$\Delta v_{launcher} = \sqrt{v_e^2 + C_3}$$

Or to sum it up:

$$\Delta v_{total} = \Delta v_{spacecraft} + \Delta v_{launcher}$$

$$\Delta v_{total} = \Delta v_{spacecraft} + \sqrt{v_e^2 + C_3}$$

Exactly the equation in question.

• Is your $C_3$ geocentric or heliocentric? $\mu$ has to belong to some body, doesn't it? – uhoh Oct 18 '20 at 9:57
• Everything here is geocentric! – SE - stop firing the good guys Oct 18 '20 at 9:58
• Okay this is extremely helpful. So for the "velocities added in quadrature" part, it's really just energies being added normally, then "square-rooted" to obtain a velocity, in this case geocentric $v_{Earth, \infty}$? And that $C_3$ is sometimes called "injection energy"? – uhoh Oct 18 '20 at 10:07
• The "big picture" (from Horizons) i.stack.imgur.com/4BKDM.png – uhoh Oct 18 '20 at 16:27
• @uhoh That graph nicely illustrates three types of energy changes: 1) flybys (sudden jumps), 2) orbiting a different system (Juno and Cassini chaotic curbes) and 3) long ion engine burns (Dawn slopes). – SE - stop firing the good guys Oct 18 '20 at 16:31