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.

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.^{[2]}

The idea of the bi-elliptical transfer trajectory was first published by Ary Sternfeld in 1934.^{[3]}

# Calculation

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

where

- is the speed of an orbiting body,
- is the standard gravitational parameter of the primary body,
- is the distance of the orbiting body from the primary, i.e., the radius,
- is the semi-major axis of the body's orbit.

In what follows,

- is the radius of the initial circular orbit,
- is the radius of the final circular orbit,
- is the common apoapsis radius of the two transfer ellipses and is a free parameter of the maneuver,
- and are the semimajor axes of the two elliptical transfer orbits, which are given by , .

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,

and finally:

# Comparison with the Hohmann transfer

The long transfer time of the bi-elliptic transfer,

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

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

# Example

To 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 Δ*v* of 2825.02 + 1308.70 = 4133.72 m/s. However, because *r*1 = 14*r*0 > 11.94*r*0, 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 *r*2 = 40*r*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 Δ*v* saving could be further improved by increasing the intermediate apogee, at the expense of longer transfer time. For example, an apogee of 75.8*r*0 = 507 688 km (1.3 times the distance to the Moon) would result in a 1% Δ*v* saving over a Hohmann transfer, but require a transit time of 17 days. As an impractical extreme example, an apogee of 1757*r*0 = 11 770 000 km (30 times the distance to the Moon) would result in a 2% Δ*v* 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 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