4 ms·

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

• 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