Bi-elliptic transfer

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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.

The bi-elliptic transfer consists of two half elliptic orbits. From the initial orbit, a delta-v is applied boosting 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 delta-v is applied sending the spacecraft into the second elliptical orbit with periapsis at the radius of the final desired orbit where a third delta-v is performed injecting the spacecraft into the desired orbit.

While it requires one more burn than a Hohmann transfer and generally requires a greater period of time, the bi-elliptic transfer may require a lower amount of total delta-v than a Hohmann transfer in situations where the ratio of final to the initial semimajor axis is greater than 12.

[edit] Calculation

Utilizing the vis viva equation where,

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

where:

$v \,\!$ 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

The magnitude of the first delta-v at the initial circular orbit with radius $r 0$ is:

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

At $r b$ the delta-v is:

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

The final delta-v at the final circular orbit with radius $r f$ :

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

Where $a 1$ and $a 2$ are the semimajor axes of the two elliptical transfer orbits and are given by:

$a_1 = \frac{r_0+r_b}{2}$

$a_2 = \frac{r_f+r_b}{2}$

[edit] Example

For example, to transfer from circular low earth orbit with $r 0 = 6700$ km to a new circular orbit with $r 1 = 14 r 0 = 93800$ km using Hohmann transfer orbit requires delta-v of 2824.34+1308.38=4132.72 m/s. However if spaceship first accelerates 3060.31 m/s, thus getting in elliptic orbit with apogee at $r 2 = 40 r 0 = 268000$ km, then in apogee accelerates another 608.679 m/s, which places it in new orbit with perigee at $r 1 = 14 r 0 = 93800$ , and, finally, in perigee slows down by 447.554 m/s, placing itself in final circular orbit, then total delta-v will be only 4116.54, which is whopping 16 m/s less.