# Why would a slow spiral from a C3 of zero take about 2.4 times as much ΔV as an impulsive maneuver?

I just read a fascinating comment that I don't understand. In part it says

...the Oberth effect does not depend on the mass of the object. A slow spiral in to a low orbit from a C3 of zero will take about 2.4 times as much ΔV as an impulsive maneuver to do the same thing, regardless of the GM.

The context is about a spacecraft (in this case DAWN) arriving at an asteroid and using low-thrust ion propulsion to enter into orbit around it and then lower the orbit.

But I don't understand the factor of 2.4 comment at all. In fact, while I think of C3 as excess v² above escape velocity of from the asteroid, I don't know where to "put" a spacecraft and how fast and moving in what direction in order to try to do verify this 2.4 with an ODE solver versus an impulsive solution.

Could someone walk me through what the assumptions are behind this comment and steps I'd need to do to arrive at the factor of ~2.4?

• I think “C3 of 0” just means the highest possible non-escape orbit — an orbit at the limit of sphere-of-influence. – Russell Borogove Jul 12 '18 at 2:14
• @RussellBorogove I'd thought that C3=0 just means any combination of speed and separation such that v²=2μ/r, which might or might not be a similar statement. I'm still confused if the Sun has to be considered at all to get to this ~2.4 or not. – uhoh Jul 12 '18 at 2:39

If you're in a circular orbit, your velocity is $\sqrt{\mu\over r}$. Escape velocity at that distance is $\sqrt{2\mu\over r}$. So the impulsive $\Delta V$ to reach escape velocity starting from that orbit is the difference of those two:

$$\Delta V_{ei}=\left(\sqrt 2-1\right)\sqrt{\mu\over r}$$

Now we escape by thrusting infinitesimal amounts in the orbit velocity direction. This keeps the orbit infinitesimally close to circular, simply increasing the radius over time. You can calculate the $dr$ as a function of $dv$. Then the trick is to integrate, in closed form, from your starting $r$ to $\infty$, which converges! That gives the integrated, gradual $\Delta V$ to escape of:

$$\Delta V_{eg}=\sqrt{\mu\over r}$$

The ratio of the gradual over the impulsive is:

$${1\over\sqrt 2-1}=\sqrt 2+1\approx 2.4$$

You can find more here.

• excellent, bingo, perfect, etc. I remember now what you explained there. Paraphrasing; if you are moving at a speed v in a circular orbit, then in the limit of very weak prograde propulsion, a delta v equal to that v will get you to infinity. – uhoh Jul 12 '18 at 4:23
• – uhoh Jul 12 '18 at 4:34
• And a simple explanation for why: When you burn a big rocket you do it low and get a big benefit from the Oberth effect. With an ion engine you don't. – Loren Pechtel Jul 12 '18 at 14:07
• @LorenPechtel You do the with ion engine. It's just that you chose to spiral out, so you are raising your radius throughout the maneuver, reducing the Oberth effect. You could choose to only operate your ion engine at only one point in the orbit, which becomes the periapsis, and leave it off the rest of the time. Depending on how short your firings are at periapsis, you will get all the same Oberth goodness. However it will take a really long time to escape. – Mark Adler Jul 12 '18 at 14:41
• @MarkAdler Yeah, if your probe didn't die before the burn was done. – Loren Pechtel Jul 12 '18 at 15:46