EDIT: Ok, let's think through the calculus here. Kepler's 2nd law says that the area swept by the line between a planet and the Sun is constant, i.e., r^2 dθ/dt = 2 π a b / P where a and b are the semi-major and semi-minor axes and P is the orbital period. Thus dt / r^2 = P dθ / (2 π a b); and the former is the instantaneous power absorbed. Integrating over an orbit, we get E = P / (a b); or the average power absorbed is proportional to 1 / (a b).
Since the energy of an orbit is determined solely by the semi-major axis a, this means the power absorbed for any given orbital energy is maximized for b ≪ a; so you're right, that highly eccentric orbits are a better way to acquire energy for the purpose of orbital escape. (Subject of course to practical considerations of energy storage -- you'll absorb the most power at exactly the time you don't want to be changing your momentum.)
Thanks for doing the math. But I'm still confused, I thought you do want to accelerate when near the sun, as that makes the far end of the orbit go further? Again, no math behind, just how it worked in KSP :)
You should trust KSP more than you trust me. :-)
http://www.askamathematician.com/2013/01/q-how-does-the-ober...
By the way, I only know about it because I played KSP a lot ;).
It doesn't just work with orbits and rockets and propellants, it works just as well with an oscillating weight on a spring that you flick with your fingers.
if you have an efficient insulator, you could absorb an arbitrarily large amount of heat at the perigee, and retain it for the remainder of the orbit.
if you were in a circular orbit, you'd only be able to absorb up to the ambient temp.
In space you move in elipses. It's possible to choose an elongated orbit that moves the probe closer and then further from the sun repeatedly and each time it is close to the sun it accelerates forward, moving the "far" side of orbit further away, until it gains enough speed to escape sun gravity.
Or at least it's possible in Kerbal Space Program :)