


By
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 maneuver. The bielliptic transfer consists of two halfelliptic orbits. From the initial orbit, a first burn expends deltav to boost the spacecraft into the first transfer orbit with an apoapsis at some point away from the central body. At this point a second burn sends the spacecraft into the second elliptical orbit with periapsis at the radius of the final desired orbit, where a third burn is performed, injecting the spacecraft into the desired orbit.^{} While they require one more engine burn than a Hohmann transfer and generally requires a greater travel time, some bielliptic transfers require a lower amount of total deltav than a Hohmann transfer when the ratio of final to initial semimajor axis is 11.94 or greater, depending on the intermediate semimajor axis chosen.^{} The idea of the bielliptical transfer trajectory was first published by Ary Sternfeld in 1934.^{} CalculationDeltavThe three required changes in velocity can be obtained directly from the visviva equation where
In what follows,
Starting from the initial circular orbit with radius r_{1} (dark blue circle in the figure to the right), a prograde burn (mark 1 in the figure) puts the spacecraft on the first elliptical transfer orbit (aqua halfellipse). The magnitude of the required deltav for this burn is When the apoapsis of the first transfer ellipse is reached at a distance r_{b} from the primary, a second prograde burn (mark 2) raises the periapsis to match the radius of the target circular orbit, putting the spacecraft on a second elliptic trajectory (orange halfellipse). The magnitude of the required deltav for the second burn is Lastly, when the final circular orbit with radius r_{2} is reached, a retrograde burn (mark 3) circularizes the trajectory into the final target orbit (red circle). The final retrograde burn requires a deltav of magnitude If r_{b} = r_{2}, then the maneuver reduces to a Hohmann transfer (in that case can be verified to become zero). Thus the bielliptic transfer constitutes a more general class of orbital transfers, of which the Hohmann transfer is a special twoimpulse case. The maximal possible savings can be computed by assuming that r_{b} = ∞, in which case the total simplifies to . In this case, one also speaks of a biparabolic transfer because the two transfer trajectories are no longer ellipses but parabolas. The transfer time increases to infinity too. 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 half the orbital period of each transfer ellipse. Using the equation for the orbital period and the notation from above, The total transfer time t is the sum of the times required for each halforbit. Therefore: and finally: Comparison with the Hohmann transferDeltavThe figure shows the total required to transfer from a circular orbit of radius r_{1} to another circular orbit of radius r_{2}. The is shown normalized to the orbital speed in the initial orbit, , and is plotted as a function of the ratio of the radii of the final and initial orbits, ; this is done so that the comparison is general (i.e. not dependent of the specific values of r_{1} and r_{2}, only on their ratio).^{} The thick black curve indicates the for the Hohmann transfer, while the thinner colored curves correspond to bielliptic transfers with varying values of the parameter , defined as the apoapsis radius r_{b} of the elliptic auxiliary orbit normalized to the radius of the initial orbit, and indicated next to the curves. The inset shows a closeup of the region where the bielliptic curves cross the Hohmann curve for the first time. One sees that the Hohmann transfer is always more efficient if the ratio of radii R is smaller than 11.94. On the other hand, if the radius of the final orbit is more than 15.58 times larger than the radius of the initial orbit, then any bielliptic transfer, regardless of its apoapsis radius (as long as it’s larger than the radius of the final orbit), requires less than a Hohmann transfer. Between the ratios of 11.94 and 15.58, which transfer is best depends on the apoapsis distance r_{b}. For any given R in this range, there is a value of r_{b} above which the bielliptic transfer is superior and below which the Hohmann transfer is better. The following table lists the value of that results in the bielliptic transfer being better for some selected cases.^{}
Transfer timeThe long transfer time of the bielliptic transfer, is a major drawback for this maneuver. It even becomes infinite for the biparabolic transfer limiting case. The Hohmann transfer takes less than half of the time because there is just one transfer halfellipse, to be precise, ExampleTo transfer from a circular low Earth orbit with r_{0} = 6700 km to a new circular orbit with r_{1} = 93 800 km using a Hohmann transfer orbit requires a Δv of 2825.02 + 1308.70 = 4133.72 m/s. However, because r_{1} = 14r_{0} > 11.94r_{0}, it is possible to do better with a bielliptic transfer. If the spaceship first accelerated 3061.04 m/s, thus achieving an elliptic orbit with apogee at r_{2} = 40r_{0} = 268 000 km, then at apogee accelerated another 608.825 m/s to a new orbit with perigee at r_{1} = 93 800 km, and finally at perigee of this second transfer orbit decelerated by 447.662 m/s, entering the final circular orbit, then the total Δv would be only 4117.53 m/s, which is 16.19 m/s (0.4%) less. The Δv saving could be further improved by increasing the intermediate apogee, at the expense of longer transfer time. For example, an apogee of 75.8r_{0} = 507 688 km (1.3 times the distance to the Moon) would result in a 1% Δv saving over a Hohmann transfer, but require a transit time of 17 days. As an impractical extreme example, an apogee of 1757r_{0} = 11 770 000 km (30 times the distance to the Moon) would result in a 2% Δv saving over a Hohmann transfer, but the transfer would require 4.5 years (and, in practice, be perturbed by the gravitational effects of other Solar system bodies). For comparison, the Hohmann transfer requires 15 hours and 34 minutes.
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.
Text from Wikipedia is available under the Creative Commons Attribution/ShareAlike License; additional terms may apply. Published in April 2019.
Please see some ads intermixed with other content from this site: 

