6 ms·

> To reach the Sun you need to subtract 100 percent of Earth’s orbital velocity; to reach solar escape velocity you need only add 41 percent to it. You could s

• joe_the_user 12y ago
  Your "trick" doesn't change any of the conservation of angular momentum considerations in the article. To "get rid of the remaining kinetic energy" would only result in the object getting escape velocity relative to earth. The object would never have "stopped" relative to it's orbit around the sun but even at it's further point would be traveling with fairly close to the same angular velocity as the earth (as described in the article etc).
  • CompelTechnic 12y ago
    I'm about 80% sure leni536 is right, from playing Kerbal Space Program.
    • TeMPOraL 12y ago
      Let me guess: the wonderful experience of putting a satellite orbiting in the precisely opposite direction to the one required by contract, noticing that your orbital velocity at apoapsis is like ~200m/s and realizing that you have enough delta-v to just turn around and reverse the orbit on the spot?
    • jheriko 12y ago
      i think you are right based on a similar feeling from that game, but be careful with that intuition. how much delta-v do you spend lifting the orbit so that lowering the periapse is cheap?
      • TheLoneWolfling 12y ago
        For large orbital changes, substantially less than the niave transfer.
  • benjoffe 12y ago
    That "trick" would indeed work and is a well known orbital manoeuvre: http://en.wikipedia.org/wiki/Bi-elliptic_transfer http://en.wikipedia.org/wiki/Bi-elliptic_transfer (without the final circularising burn). This manoeuvre would be preferable to the article's alternative (a Hohmann transfer reducing periapsis to the sun's radius without the final circularising burn) in all cases where the initial semi-major axis (relative to the sun) is roughly greater than 11.94 solar radii (ignoring minor factors that make this case slightly different to the wikipedia example). edit: to make it clear, I'm talking about a heliocentric bi-elliptic transfer after Earth escape, where the initial orbit is equivalent to Earth's orbit, and the final inner orbit can be imagined as one scraping the surface of the sun, but since the sun would destroy the craft, the final circularising burn is not required (in sloppy terms: hitting the sun on the edge would be easier than hitting it straight on).
    • teraflop 12y ago
      The downside, of course, is that your nuclear waste spends an awfully long time in an orbit that crosses, or at least passes very close to, Earth's. Better not lose control of your rocket, or you'll have a politically inconvenient situation on your hands.
      • jsprogrammer 12y ago
        More important than politically inconvenient: a technically inconvenient situation, as that waste could be on a trajectory back to earth. https://youtu.be/YIqXql6LLz0?t=6m50s https://youtu.be/YIqXql6LLz0?t=6m50s
      • jheriko 12y ago
        thats not how this manoeuvre works. the burn to fall into the sun happens way before any possible return to earth if that orbit were not altered... all that happens is that you start near the earth - thats unavoidable because the rocket comes form the earth. it will of course cross the earth's orbit on the way into the sun, but, with the exception of very small ranges of orbits the chance of hitting the earth on return is tiny, and easily avoided with some timing.
• kllrnohj 12y ago
  A straight Hohmann transfer requires ~0.5-1% more dV than a bi-elliptic transfer. Calling this "by far" not optimal is by far not correct. So instead of 32 km/s you need 31.68 km/s. This changes nothing.
  • benjoffe 12y ago
    Where did you get the "0.5-1%" figure from? That may be true in some cases, but not in this extreme case, here it's closer to half the delta-v (which, due to the rocket equation is enormously less difficult to achieve). Solar escape velocity is about 16.5km/s* , and since ~99% of this manoeuvre's delta-v budget is used at launch the entire delta-v cost is asymptotically this value. *http://en.wikipedia.org/wiki/New_Horizons#Launch http://en.wikipedia.org/wiki/New_Horizons#Launch (see paragraph 4)
    • pdonis 12y ago
      > Solar escape velocity is about 16.5km/s No, it's about 42 km/s, i.e., sqrt(2) times the Earth's orbital velocity around the Sun, which is 30 km/s. New Horizons was launched to take as much advantage of the Earth's orbital velocity as it could, so that the rocket didn't need to provide all of the 42 km/s of delta-v needed for solar escape. (You could use the same trick for your scheme, of course. The main issue I see with your scheme is that it greatly lengthens the time that the waste is out in space.)
  • leni536 12y ago
    It's not a full bi-elliptic transfer, it's missing the third "retrograde" burn. In this situation the target orbit would be inside the Sun, the third burn is completely unnecessary. Your rule of thumb comparison of the two strategy doesn't work here. [1] https://en.wikipedia.org/wiki/Bi-elliptic_transfer https://en.wikipedia.org/wiki/Bi-elliptic_transfer
• InclinedPlane 12y ago
  To what end? To spend literally decades, perhaps centuries, with nuclear waste floating around the solar system? Why not just leave it in deep space to start with? Why bother sending it into space in the first place? Meanwhile, if you succeed in sending nuclear waste into the Sun what do you think happens to it? It doesn't go away, it just gets vaporized and then scattered into the solar wind.
  • tvon 12y ago
    I think leni536 is just addressing the kind of interesting problem of what it takes to shoot anything into the Sun.
  • Cerium 12y ago
    Solar energy is simply a nuclear power system with a sensible waste disposal strategy.
    • elcct 12y ago
      But it also has not been approved by any authority. It is just like sending waste to some shady dude few blocks away "to take care of it"
      • DonHopkins 12y ago
        The sun isn't the shadiest dude in the solar system.