4 ms·

> This is pretty good news if only that we will get more research into this phenomenon. Agreed. > Having a way to push your little probe along (even at a litt

> This is pretty good news if only that we will get more research into this phenomenon. Agreed. > Having a way to push your little probe along (even at a little bit of acceleration) without having to carry a lot of heavy propellant is a big deal. Maybe just some solar panels to generate the energy needed. I'm not sure that actually works. Most use cases are for exploring the outer solar system, or even outside of our solar system. As the probe gets away from the sun, the solar energy it gets decreases with the square of the distance, and so it probably will stop providing even the little bit of acceleration.

ajuc 10y ago
> As the probe gets away from the sun, the solar energy it gets decreases with the square of the distance, and so it probably will stop providing even the little bit of acceleration. 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 :)
- cperciva 10y ago
  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 ;).
GenerocUsername 10y ago
Nuclear/Batteries

ajuc 10y ago

> As the probe gets away from the sun, the solar energy it gets decreases with the square of the distance, and so it probably will stop providing even the little bit of acceleration. 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 :)

cperciva 10y ago
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 ;).

GenerocUsername 10y ago

Nuclear/Batteries