# 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!

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 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. Commented Oct 19, 2020 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
Commented Oct 19, 2020 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. Commented Oct 19, 2020 at 2:51

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
Commented Oct 18, 2020 at 9:57
• Everything here is geocentric! Commented Oct 18, 2020 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
Commented Oct 18, 2020 at 10:07
• The "big picture" (from Horizons) i.sstatic.net/4BKDM.png
– uhoh
Commented Oct 18, 2020 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). Commented Oct 18, 2020 at 16:31