Bi-elliptic transfer

In astronautics and aerospace engineering, the bi-elliptic transfer is an orbital maneuver that moves a spacecraft from one orbit to another and may, in certain situations, require less delta-v than a Hohmann transfer maneuver.

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.

While they require one more engine burn than a Hohmann transfer and generally requires a greater travel time, some bi-elliptic transfers require a lower amount of total delta-v than a Hohmann transfer when the ratio of final to initial semi-major axis is 11.94 or greater, depending on the intermediate semi-major axis chosen.^[1]

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

Calculation

Delta-v

A bi-elliptic transfer from a low circular starting orbit (dark blue), to a higher circular orbit (red).

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

$v^2 = \mu \left( \frac{2}{r} - \frac{1}{a} \right)$

• $v \,\!$ 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:

$\Delta v_1 = \sqrt{ \frac{2 \mu}{r_0} - \frac{\mu}{a_1}} - \sqrt{\frac{\mu}{r_0}}$

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:

$\Delta v_2 = \sqrt{ \frac{2 \mu}{r_b} - \frac{\mu}{a_2}} - \sqrt{ \frac{2 \mu}{r_b} - \frac{\mu}{a_1}}$

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.

A bi-parabolic transfer from a low circular starting orbit (dark blue), to a higher circular orbit (red).

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

In this case one also speaks of a bi-parabolic transfer because the two transfer trajectories no longer are ellipses but parabola. The transfer time increases to infinity too.

Transfer time

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:

$T = 2 \pi \sqrt{\frac{a^3}{\mu}}$

The total transfer time $t$ is the sum of the time required for each half orbit. Therefore:

$t_1 = \pi \sqrt{\frac{a_1^3}{\mu}} \quad and \quad t_2 = \pi \sqrt{\frac{a_2^3}{\mu}}$

And finally:

$t = t_1 + t_2 \;$

Comparison with the Hohmann transfer

Delta-v

Delta-v requirements (normed with $v_1$) for four maneuvers between the two same circular orbits depending on the radius ratio $\tfrac{r_2}{r_1}$.

The figure on the right shows the Delta-v needed for a transfer between a circular orbit with radius $r_1$ and a circular orbit with radius $r_2$.

$\Delta v$ is normed with the initial velocity $v_1$ normiert, to make the comparison general. There are four curves: the Delta-v for a Hohmann transfer (blue), for a bi-elliptic transfer with $\alpha=\tfrac{r_b}{r_1}=20+\tfrac{r_2}{r_1}$ (red), for a bi-elliptic transfer with $\alpha=\tfrac{r_b}{r_1}=100+\tfrac{r_2}{r_1}$ (cyan) and for a bi-parabolic transfer $(r_b\rightarrow\infty)$ (green).^[3]

One sees that the Hohmann transfer is best as long as the radius ratio is smaller than 11.94. If the radius of the final orbit is more than 15,58 times larger than the radius of the initial orbit, then any bi-elliptic transfer requires less Delta-v than a Hohmann transfer as long as $r_b>r_2$.

The apoapsis distance $r_b$ is critical in between ratios of 11.94 and 15.58. The following table lists the minimum $\alpha=\tfrac{r_b}{r_1}$ (that is, the apoapsis distance with respect to the initial orbit radius) such that the bi-elliptic transfer is better in terms of energy for some cases.

Minimum $\alpha=\tfrac{r_b}{r_1}$, such that a bi-elliptic transfer needs less Delta-v.^[4]

Radius ratio $\tfrac{r_2}{r_1}$	Minimum $\alpha=\tfrac{r_b}{r_1}$	Comments
0 bis 11.94	–	Hohmann transfer is better
11.94	$\infty$	Bi-parabolic transfer
12	815.81
13	48.90
14	26.10
15	18.19
15.58	15.58
greater than 15.58	greater than $\tfrac{r_2}{r_1}$	Any bi-elliptic transfer is better

Transfer time

The long Transfer time of the bi-elliptic transfer

$t=\pi\sqrt{\frac{a_1^3}{\mu}}+\pi\sqrt{\frac{a_2^3}{\mu}}$

is a major drawback for this maneuver. It even becomes infinite for the pi-parabolic transfer limiting case.

The Hohmann transfer takes less than half of the time because there is just one half transfer ellipse, to be precise

$t=\pi\sqrt{\frac{a^3}{\mu}}$

Example

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

• Delta-v budget
• Oberth effect

References

1. Vallado, David Anthony (2001). Fundamentals of Astrodynamics and Applications. Springer. p. 318. ISBN 0-7923-6903-3.
2. Sternfeld, Ary J. [sic] (1934-02-12), "Sur les trajectoires permettant d'approcher d'un corps attractif central à partir d'une orbite keplérienne donnée" [On the allowed trajectories for approaching a central attractive body from a given Keplerian orbit], Comptes rendus de l'Académie des sciences (in French) (Paris) 198 (1): 711–713
3. Gobetz, F. W.; Doll, J. R. (May 1969). "A Survey of Impulsive Trajectories". AIAA Journal (American Institute of Aeronautics and Astronautics) 7 (5): 801–834. doi:10.2514/3.5231.
4. Escobal, Pedro R. (1968). Methods of Astrodynamics. New York: John Wiley & Sons. ISBN 978-0-471-24528-5.