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.

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 semi-major axis is greater than 11.94 [Cite book | last = Vallado | first = David Anthony | title = Fundamentals of Astrodynamics and Applications | page = 317 | publisher = Springer | date = 2001 | isbn = 0792369033 | url = http://books.google.com/books?id=PJLlWzMBKjkC&printsec] .

Calculation

Delta-v

Utilizing the "vis viva" equation where,:$v^2\; =\; mu\; left(\; frac\{2\}\{r\}\; -\; frac\{1\}\{a\}\; ight)$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::$Delta\; v\_1\; =\; sqrt\{\; frac\{2\; mu\}\{r\_0\}\; -\; frac\{mu\}\{a\_1\; -\; sqrt\{frac\{mu\}\{r\_0$

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 $r\_f$::$Delta\; v\_3\; =\; sqrt\{frac\{mu\}\{r\_f\; -\; sqrt\{\; frac\{2\; mu\}\{r\_f\}\; -\; frac\{mu\}\{a\_2$

Where $a\_1$ and $a\_2$ are the semimajor axes of the two elliptical transfer orbits and are given by::$a\_1\; =\; frac\{r\_0+r\_b\}\{2\}$:$a\_2\; =\; frac\{r\_f+r\_b\}\{2\}$

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 simply half the orbital period of each transfer ellipse.

Using the equation for the orbital period and the notation from above, we have:

:$T\; =\; 2\; pi\; sqrt\{frac\{a^3\}\{mu$

The total transfer time $t$ is simply the sum of the time required for each half orbit Therefore we have:

:$t\_1\; =\; pi\; sqrt\{frac\{a\_1^3\}\{mu\; quad\; and\; quad\; t\_2\; =\; pi\; sqrt\{frac\{a\_2^3\}\{mu$

and finally:

:$t\; =\; t\_1\; +\; t\_2\; ;$

Example

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 16.18 m/s less.

*"green|Forward applied ΔV"
*"red|Reverse applied ΔV"

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.

ee also

*Delta-v budget
*The Oberth effect

References

External links