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

Question

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?

Answer 1

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:

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

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