Can someone help Scott Manley explain this Snoopy-capturing maneuver?

Question

In his recent video Apollo 10's Lunar Module Snoopy Is Lost In Space - Could We Bring it Home? Scott Manley uses some fancy supplemental techniques in KSP to simulate and explain a possible mission to capture Snoopy (Apollo 10 Lunar Module) "...and returning him safely to the Earth."

But at about 09:38 Manley surprisingly finds himself at a bit of a loss for words. This explanation says little more than "it takes less Delta V because you square it and so it takes less Delta V":

Now when you’re doing capture from deep space, you want to get down close to the Earth, because that way you’re actually using Earth’s gravity to reduce the amount of Delta V you need to perform the capture.

If Earth wasn’t there, you’d actually need more Delta V to match the orbit, because you fall down into it you get this square law where you’re adding your velocity squared, and… listen, it’s just a simple thing to do. You drop down into the gravity well and then you perform your burn there and then it becomes… it means you need less Delta V to perform this.

Question: Can someone help Scott Manley out and explain more clearly what it is that he's trying to explain?

cued at 09:38

Answer 1

Despite watching a few times it looks like I linked the use of the term "capture" at this point in the video with the use of the term "capture" at 08:20 in the video. But that's about the mechanical capture of Snoopy during a docking maneuver and this is about the gravitational capture of the combined pair in the Earth's gravitational field.

What Manley is trying to say is that:

If you want to transition from a gravitationally unbound to gravitationally bound status in the Earth's field, this is a manipulation of the energy of the spacecraft's orbit, which depends on position and $v^2$, and a $\Delta v$ impulse can have the largest effect on $v^2$ when

1. the $\Delta \mathbf{v}$ vector is parallel or antiparallel with $\mathbf{v}$, and
2. the speed $v$ is maximum.

And for some further mathematical insight into why this is true, and some intriguing graphical illustrations of this as well, one can read the several excellent answers to Oberth effect for Earth vehicles.