Re: Bielliptic Transfer Orbits

From: Bill Woods (wwoods_at_popd.ix.netcom.com)
Date: 08/25/04 

To: sci-space-tech@moderators.isc.org
Date: Wed, 25 Aug 2004 08:32:44 GMT

Charles Talleyrand wrote:

> Can someone explain what a bieliptic transfer orbit is more
> efficient than a Hoffman transfer orbit when the ratio of
> initial to final orbits is large and there is no inclination
> change to make?
>
> This is counter-intuitive, and the obvious Google(tm) searches
> don't help.

You're in an orbit with radius R_1. Boost yourself up to escape
speed and coast out to infinity. Perform an infinitesimal orbital
correction to put yourself on the second transfer orbit, which has
periapsis at your target orbit at radius R_2. When you get there,
a third boost completes the maneuver.

Or, if you don't have infinite time to spend, boost yourself out
to a point somewhat closer, trading slightly smaller first and
third burns for a slightly larger second burn. The total delta V
will be greater, so use the biparabolic transfer as the limiting case.

If you work out the total delta V required this way, versus that
required by the Hohmann transfer, you'll find that for sufficiently
extreme values of R_1/R_2, the two-stage transfer is cheaper. 

--
  Bill Woods
 "If you examine my 16-year record in the Senate, you'll see
  that I'm just as effective when I'm not there as I was when
  I was there," said Mr. Kerry. "... I think it's disingenuous
  for Gov. Romney to suggest that my absence from the Senate
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 http://www.scrappleface.com/MT/archives/001737.html#001737

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