# 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 sometimes requires 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]

Ary Sternfeld in 1934 first published the idea of the bi-elliptical transfer trajectory.[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 from the orbiting body to 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) so raises the periapsis to the radius of the target circular orbit, thus 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 retograde 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.

### Transfer time

Like the Hohmann transfer, both transfer orbits of the bi-elliptic transfer constitute half an elliptic orbit. The time required to execute each phase of the transfer therefore 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 that each half orbit requires:

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

## 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 2824.34+1308.38=4132.72 m/s. However, because r1=14r0 >11.94r0, a bi-elliptic transfer is better. If the spaceship first accelerates 3060.31 m/s, thus achieving an elliptic orbit with apogee at r2=40r0=268000 km, then in apogee accelerates another 608.679 m/s to a new orbit with perigee at r1=93800 km, and finally in perigee decelerates by 447.554 m/s, entering final circular orbit, then the total delta-v will be only 4116.54, which is 16.18 m/s (0.4%) less.

Burn Hohmann ΔV (m/s) Bi-elliptic ΔV (m/s)
1 2824.34 3060.31
2 1308.38 608.679
3 - 447.554
Total 4132.72 4116.54