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Wikipedia, In astronautics and aerospace engineering, the bielliptic transfer is an orbital maneuver that moves a spacecraft from one orbit to another and may, in certain situations, require less deltav than a Hohmann transfer. The bielliptic transfer consists of two half elliptic orbits. From the initial orbit, a deltav is applied boosting the spacecraft into the first transfer orbit with an apoapsis at some point r_{b} away from the central body. At this point, a second deltav is applied sending the spacecraft into the second elliptical orbit with periapsis at the radius of the final desired orbit where a third deltav is performed injecting the spacecraft into the desired orbit. While it requires one more burn than a Hohmann transfer and generally requires a greater period of time, the bielliptic transfer may require a lower amount of total deltav than a Hohmann transfer in situations where the ratio of final to the initial semimajor axis is greater than 11.94 . CalculationDeltav
Utilizing the vis viva equation where, where:
The magnitude of the first deltav at the initial circular orbit with radius r_{0} is: At r_{b} the deltav is: The final deltav at the final circular orbit with radius r_{f}: Where a_{1} and a_{2} are the semimajor axes of the two elliptical transfer orbits and are given by: Transfer timeLike the Hohmann transfer, both transfer orbits used in the bielliptic transfer constitute exactly one half of an elliptic orbit. This means that the time required to execute each phase of the transfer is simply half the orbital period of each transfer ellipse. Using the equation for the orbital period and the notation from above, we have: The total transfer time t is simply the sum of the time required for each half orbit Therefore we have: and finally: ExampleFor example, to transfer from circular low earth orbit with r_{0} = 6700 km to a new circular orbit with r_{1} = 14r_{0} = 93800 km using Hohmann transfer orbit requires deltav of 2824.34+1308.38=4132.72 m/s. However if spaceship first accelerates 3060.31 m/s, thus getting in elliptic orbit with apogee at r_{2} = 40r_{0} = 268000 km, then in apogee accelerates another 608.679 m/s, which places it in new orbit with perigee at r_{1} = 14r_{0} = 93800, and, finally, in perigee slows down by 447.554 m/s, placing itself in final circular orbit, then total deltav will be only 4116.54, which is 16.18 m/s less.
Evidently, the bielliptic orbit spends more of its deltaV early on (in the first burn). This yields a higher contribution to the specific orbital energy and, due to the Oberth effect, is responsible for the net reduction in required deltaV. See alsoExternal links
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Published  July 2009


