4 ms·

Yes, you move in ellipses; but you spend a lot of time far away and very little time close to the sun. I'm not sure that the average power is any higher than y

Yes, you move in ellipses; but you spend a lot of time far away and very little time close to the sun. I'm not sure that the average power is any higher than you would have for a circular orbit at the same energy... 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.)

novaleaf 10y ago
my attempt at a thought experiment that seems to suggest elliptical orbits are better in this regard: 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.
- cperciva 10y ago
  Ambient temperature is a red herring; the sun is hotter than the ambient temperature so you can always win by absorbing energy from the sun and then emitting it to the rest of space.
ajuc 10y ago
> 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 :)
- cperciva 10y ago
  Whoops, you're quite right. I was thinking that you wanted to add momentum when you were moving the slowest, but that's for when you want to get into a circular orbit (which is the opposite of what you're talking about). You should trust KSP more than you trust me. :-)
  - TeMPOraL 10y ago
    It's Oberth effect, basically? :). http://www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/ 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 ;).
    - photogrammetry 10y ago
      Does the Oberth effect really apply when your propellant mass is always zero?
      - TeMPOraL 10y ago
        It's about ∆v, so I'd say it does regardless of how you got that ∆v. I don't see a reason it couldn't work with solar sails, for example.
        MPSimmons 10y ago
        If anything, it should increase your inertia, since you're not sacrificing mass for ∆v.
      - repsilat 10y ago
        The simplest way (imo) to think about the Oberth effect is "you want to maximize the amount of time you're falling inwards (i.e. speeding up) and minimize the amount of time shooting outwards (i.e. slowing down.)" 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.

novaleaf 10y ago

my attempt at a thought experiment that seems to suggest elliptical orbits are better in this regard: 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.

cperciva 10y ago
Ambient temperature is a red herring; the sun is hotter than the ambient temperature so you can always win by absorbing energy from the sun and then emitting it to the rest of space.

ajuc 10y ago

> 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 :)

cperciva 10y ago
Whoops, you're quite right. I was thinking that you wanted to add momentum when you were moving the slowest, but that's for when you want to get into a circular orbit (which is the opposite of what you're talking about). You should trust KSP more than you trust me. :-)
- TeMPOraL 10y ago
  It's Oberth effect, basically? :). http://www.askamathematician.com/2013/01/q-how-does-the-oberth-effect-work-and-where-does-the-extra-energy-come-from-why-is-it-better-for-a-rocket-to-fire-at-the-lowest-point-in-its-orbit/ 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 ;).
  - photogrammetry 10y ago
    Does the Oberth effect really apply when your propellant mass is always zero?
    - TeMPOraL 10y ago
      It's about ∆v, so I'd say it does regardless of how you got that ∆v. I don't see a reason it couldn't work with solar sails, for example.
      - MPSimmons 10y ago
        If anything, it should increase your inertia, since you're not sacrificing mass for ∆v.
    - repsilat 10y ago
      The simplest way (imo) to think about the Oberth effect is "you want to maximize the amount of time you're falling inwards (i.e. speeding up) and minimize the amount of time shooting outwards (i.e. slowing down.)" 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.