Benutzer:Stormregion0/BielliptischerOrbit

Der bi-elliptische Transfer ist ein Transferorbit der in bestimmten Fällen energetisch günstiger ist als der Hohmann-Transfer. Er besteht dabei aus zwei halben elliptischen Bahnen. Im Ausgangsorbit wird dem Flugobjekt ein Kraftstoß versetzt so das Delta-v ausreicht um in die erste elliptische Transferbahn zu kommen. Wenn das Flugobjekt die Apoapsis erreicht hat wird ein zweiter Kraftstoß versetzt um in die zweite elliptische Transferbahn zu kommen dessen Periapsis dem Radius des Zielorbits entspricht. Dort wird dann ein dritter Kraftstoß ausgeführt um in dem Zielorbit zu bleiben.

Ein solcher Transfer brauch ein Kraftstoß mehr als ein Hohmann-Tranfer und grundsätzlich ist die Reisezeit größer, trotzdem können bi-elliptische Transfers energetisch günstiger sein wenn das Verhältnis der

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.

Berechnung

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 of the orbiting body from 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) 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:

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

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 retrograde 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.

The maximum savings possible can be computed by assuming that $r_{b}=\infty$, in which case the total $\Delta v$ simplifies to $\left({\sqrt {2}}-1\right)\left({\sqrt {\mu /r_{0}}}+{\sqrt {\mu /r_{f}}}\right)$.

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 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 required for each half orbit. Therefore:

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

And finally:

$t=t_{1}+t_{2}\;$

Example

To transfer from a circular low Earth orbit with r₀=6700 km to a new circular orbit with r₁=93800 km using a Hohmann transfer orbit requires a Δv of 2825.02+1308.70=4133.72 m/s. However, because r₁=14r₀ >11.94r₀, 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 r₂=40r₀=268000 km, then at apogee accelerated another 608.825 m/s to a new orbit with perigee at r₁=93800 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 Δv saving could be further improved by increasing the intermediate apogee, at the expense of longer transfer time. For example, an apogee of 75.8r₀=507,688 km (1.3 times the distance to the Moon) would result in a 1% Δv saving over a Hohmann transfer, but a transit time of 17 days. As an impractical extreme example, an apogee of 1757r₀=11,770,000 km (30 times the distance to the Moon) would result in a 2% Δv 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.

Δv for various orbital transfers

Type	Hohmann	Bi-elliptic
Apogee (km)	93800	268000	507688	11770000	∞
Burn 1 (m/s)	Vorlage:Green	Vorlage:Green	Vorlage:Green	Vorlage:Green	Vorlage:Green
Burn 2 (m/s)	Vorlage:Green	Vorlage:Green	Vorlage:Green	Vorlage:Green	0
Burn 3 (m/s)	0	Vorlage:Red	Vorlage:Red	Vorlage:Red	Vorlage:Red
Total (m/s)	4133.72	4117.53	4092.38	4051.04	4048.76
Percentage	100%	99.6%	99.0%	98.0%	97.94%

• Vorlage:Green
• Vorlage:Red

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

• Delta-v budget
• Oberth effect