# 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.[citation needed]

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[citation needed] 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.

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

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:

$t = t_1 + t_2 \;$

## Example

To transfer from circular low earth orbit with r0=6700 km to a new circular orbit with r1=93800 km using Hohmann transfer orbit requires delta-v 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 in apogee accelerated another 608.825 m/s to a new orbit with perigee at r1=93800 km, and finally in perigee decelerated by 447.662 m/s, entering final circular orbit, then the total delta-v would be only 4117.53, which is 16.19 m/s (0.4%) less.

The Δv ratio 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, 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 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%