Bi-elliptic transfer

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The bi-elliptic transfer consists of two half elliptic orbits. From the initial orbit, a first burn expends delta-v to boost 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 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.[citation needed]


The idea of the bi-elliptical transfer trajectory was first[citation needed] published by Ary Sternfeld in 1934.[2]




The three required changes in velocity can be obtained directly from the vis-viva equation,

  •  v \,\! is the speed of an orbiting body
  • \mu = GM\,\! is the standard gravitational parameter of the primary body
  •  r \,\! is the distance of the orbiting body from the primary
  •  a \,\! is the semi-major axis of the body's orbit
  • r_b is the common apoapsis distance of the two transfer ellipses and is a free parameter of the maneuver.
  • a_1 and a_2 are the semimajor axes of the two elliptical transfer orbits, which are given by
    a_1 = \frac{r_0+r_b}{2}
    a_2 = \frac{r_f+r_b}{2}

Starting from the initial circular orbit with radius r_0 (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 half ellipse). The magnitude of the required delta-v 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 half ellipse). The magnitude of the required delta-v for the second burn is:


Lastly, when the final circular orbit with radius r_f is reached, a retrograde burn (mark 3) circularizes the trajectory into the final target orbit (red circle). The final retrograde burn requires a delta-v of magnitude:

\Delta v_3 = \sqrt{ \frac{2 \mu}{r_f} - \frac{\mu}{a_2}} - \sqrt{\frac{\mu}{r_f}}

If r_b=r_f, then the maneuver reduces to a Hohmann transfer (in that case \Delta v_3 can be verified to become zero). Thus the bi-elliptic transfer constitutes a more general class of orbital transfers, of which the Hohmann transfer is a special two-impulse case.

The maximum savings possible can be computed by assuming that r_b=\infty, in which case the total \Delta v simplifies to \left(\sqrt 2 - 1\right) \left(\sqrt{\mu/r_0} + \sqrt{\mu/r_f}\right).

Transfer time[edit]

Like the Hohmann transfer, both transfer orbits used in the bi-elliptic 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 time required for each half orbit. Therefore:


And finally:

t = t_1 + t_2 \;


To transfer from a circular low Earth orbit with r0=6700 km to a new circular orbit with r1=93800 km using a Hohmann transfer orbit requires a Δv of 2825.02+1308.70=4133.72 m/s. However, because r1=14r0 >11.94r0, it is possible to do better with a bi-elliptic transfer. If the spaceship first accelerated 3061.04 m/s, thus achieving an elliptic orbit with apogee at r2=40r0=268000 km, then at apogee accelerated another 608.825 m/s to a new orbit with perigee at r1=93800 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 75.8r0=507,688 km (1.3 times the distance to the Moon) would result in a 1% Δv saving over a Hohmann transfer, but a transit time of 17 days. As an impractical extreme example, an apogee of 1757r0=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.

Δv for various orbital transfers
Type Hohmann Bi-elliptic
Apogee (km) 93800 268000 507688 11770000
Burn 1 (m/s) 2825.02 3061.04 3123.62 3191.79 3194.89
Burn 2 (m/s) 1308.70 608.825 351.836 16.9336 0
Burn 3 (m/s) 0 -447.662 -616.926 -842.322 -853.870
Total (m/s) 4133.72 4117.53 4092.38 4051.04 4048.76
Percentage 100% 99.6% 99.0% 98.0% 97.94%
  • Δv applied prograde
  • (negative) Δv applied retrograde

Evidently, the bi-elliptic orbit spends more of its delta-v 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 delta-v.

See also[edit]