Bi-elliptic transfer

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

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

where

• ${\displaystyle v}$ is the speed of an orbiting body,
• ${\displaystyle \mu =GM}$ is the standard gravitational parameter of the primary body,
• ${\displaystyle r}$ is the distance of the orbiting body from the primary, i.e., the radius,
• ${\displaystyle a}$ is the semi-major axis of the body's orbit.

In what follows,

• ${\displaystyle r_{1}}$ is the radius of the initial circular orbit,
• ${\displaystyle r_{2}}$ is the radius of the final circular orbit,
• ${\displaystyle r_{b}}$ is the common apoapsis radius of the two transfer ellipses and is a free parameter of the maneuver,
• ${\displaystyle a_{1}}$ and ${\displaystyle a_{2}}$ are the semimajor axes of the two elliptical transfer orbits, which are given by

  ${\displaystyle a_{1}={\frac {r_{1}+r_{b}}{2}}}$,

  ${\displaystyle a_{2}={\frac {r_{2}+r_{b}}{2}}}$.

Starting from the initial circular orbit with radius ${\displaystyle r_{1}}$ (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

${\displaystyle \Delta v_{1}={\sqrt {{\frac {2\mu }{r_{1}}}-{\frac {\mu }{a_{1}}}}}-{\sqrt {\frac {\mu }{r_{1}}}}.}$

When the apoapsis of the first transfer ellipse is reached at a distance ${\displaystyle r_{b}}$ 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

${\displaystyle \Delta v_{2}={\sqrt {{\frac {2\mu }{r_{b}}}-{\frac {\mu }{a_{2}}}}}-{\sqrt {{\frac {2\mu }{r_{b}}}-{\frac {\mu }{a_{1}}}}}.}$

Lastly, when the final circular orbit with radius ${\displaystyle r_{2}}$ 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

${\displaystyle \Delta v_{3}={\sqrt {{\frac {2\mu }{r_{2}}}-{\frac {\mu }{a_{2}}}}}-{\sqrt {\frac {\mu }{r_{2}}}}.}$

If ${\displaystyle r_{b}=r_{2}}$, then the maneuver reduces to a Hohmann transfer (in that case ${\displaystyle \Delta v_{3}}$ 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 maximal possible savings can be computed by assuming that ${\displaystyle r_{b}=\infty }$, in which case the total ${\displaystyle \Delta v}$ simplifies to ${\displaystyle {\sqrt {\mu /r_{1}}}\left({\sqrt {2}}-1\right)\left(1+{\sqrt {r_{1}/r_{2}}}\right)}$. In this case, one also speaks of a bi-parabolic transfer because the two transfer trajectories are no longer ellipses but parabolas. The transfer time increases to infinity too.

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,

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

The total transfer time ${\displaystyle t}$ is the sum of the times required for each half-orbit. Therefore:

${\displaystyle t_{1}=\pi {\sqrt {\frac {a_{1}^{3}}{\mu }}}\quad {\text{and}}\quad t_{2}=\pi {\sqrt {\frac {a_{2}^{3}}{\mu }}},}$

and finally:

${\displaystyle t=t_{1}+t_{2}.}$

Delta-v required for Hohmann (thick black curve) and bi-elliptic transfers (colored curves) between two circular orbits as a function of the ratio of their radii

The figure shows the total ${\displaystyle \Delta v}$ required to transfer from a circular orbit of radius ${\displaystyle r_{1}}$ to another circular orbit of radius ${\displaystyle r_{2}}$. The ${\displaystyle \Delta v}$ is shown normalized to the orbital speed in the initial orbit, ${\displaystyle v_{1}}$, and is plotted as a function of the ratio of the radii of the final and initial orbits, ${\displaystyle R\equiv r_{2}/r_{1}}$; this is done so that the comparison is general (i.e. not dependent of the specific values of ${\displaystyle r_{1}}$ and ${\displaystyle r_{2}}$, only on their ratio).[2]

The thick black curve indicates the ${\displaystyle \Delta v}$ for the Hohmann transfer, while the thinner colored curves correspond to bi-elliptic transfers with varying values of the parameter ${\displaystyle \alpha \equiv r_{b}/r_{1}}$, defined as the apoapsis radius ${\displaystyle r_{b}}$ of the elliptic auxiliary orbit normalized to the radius of the initial orbit, and indicated next to the curves. The inset shows a close-up of the region where the bi-elliptic curves cross the Hohmann curve for the first time.

One sees that the Hohmann transfer is always more efficient if the ratio of radii ${\displaystyle R}$ is smaller than 11.94. On the other hand, 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, regardless of its apoapsis radius (as long as it's larger than the radius of the final orbit), requires less ${\displaystyle \Delta v}$ than a Hohmann transfer. Between the ratios of 11.94 and 15.58, which transfer is best depends on the apoapsis distance ${\displaystyle r_{b}}$. For any given ${\displaystyle R}$ in this range, there is a value of ${\displaystyle r_{b}}$ above which the bi-elliptic transfer is superior and below which the Hohmann transfer is better. The following table lists the value of ${\displaystyle \alpha \equiv r_{b}/r_{1}}$ that results in the bi-elliptic transfer being better for some selected cases.[4]

Minimal ${\displaystyle \alpha \equiv r_{b}/r_{1}}$ such that a bi-elliptic transfer needs less ${\displaystyle \Delta v}$[5]

Ratio of radii, ${\displaystyle {\frac {r_{2}}{r_{1}}}}$	Minimal ${\displaystyle \alpha \equiv {\frac {r_{b}}{r_{1}}}}$	Comments
<11.94	N/A	Hohmann transfer is always better
11.94	${\displaystyle \infty }$	Bi-parabolic transfer
12	815.81
13	48.90
14	26.10
15	18.19
15.58	15.58
>15.58	${\displaystyle >{\frac {r_{2}}{r_{1}}}}$	Any bi-elliptic transfer is better

The long transfer time of the bi-elliptic transfer,

${\displaystyle t=\pi {\sqrt {\frac {a_{1}^{3}}{\mu }}}+\pi {\sqrt {\frac {a_{2}^{3}}{\mu }}},}$

is a major drawback for this maneuver. It even becomes infinite for the 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,

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

While a bi-elliptic transfer has a small parameter window where it's strictly superior to a Hohmann Transfer in terms of delta V for a planar transfer between circular orbits, the savings is fairly small, and a bi-elliptic transfer is a far greater aid when used in combination with certain other maneuvers.

At apoapsis, the spacecraft is travelling at low orbital velocity, and significant changes in periapsis can be achieved for small delta V cost. Transfers that resemble a bi-elliptic but which incorporate a plane-change maneuver at apoapsis can dramatically save delta-V on missions where the plane needs to be adjusted as well as the altitude, versus making the plane change in low circular orbit on top of a Hohmann transfer.

Likewise, dropping periapsis all the way into the atmosphere of a planetary body for aerobreaking is inexpensive in velocity at apoapsis, but permits the use of "free" drag to aid in the final circularization burn to drop apoapsis; Though it adds an extra mission stage of periapsis-raising back out of the atmosphere, under some parameters this may cost significantly less delta V than simply dropping periapsis in one burn from circular orbit.