Bi-elliptical transfer

In space travel , the bi-elliptical transfer is a possible transition for a spacecraft between two orbits around the same central body (for example the earth or the sun).

Instead of going directly from the starting to the destination path as with the Hohmann transfer , the transfer takes place via two transfer ellipses. The first goes “beyond the goal”, the second leads to the desired target path. This may seem pointless at first, but if the target orbit is significantly higher than the starting orbit, the bi-elliptical transfer is energetically more favorable. In this article, only the case is considered in which the target path has a greater distance from the central body than the starting path. Also the simplifying assumption that the source and target track are circular and in the same plane that the speed changes instantaneously and that no perturbations present, for example by third body.

calculation

speed

A bi-elliptical transfer between a low starting orbit (blue) via the transfer ellipses (red and orange) to a high target orbit (green).

The fundamental equation for calculating coplanar transitions (such as the bi-elliptical transfer) is the Vis-Viva equation .

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

• ${\displaystyle r}$and are the current distance of the spacecraft from the central body and the current speed, ${\displaystyle v}$
• ${\displaystyle a}$is the major semi-axis of the railway,
• ${\displaystyle \mu =GM}$is the gravitational parameter of the central body (mass multiplied by the gravitational constant ). ${\displaystyle M}$ ${\displaystyle G}$

For a circular path the equation is simplified to ${\displaystyle (r=a)}$

${\displaystyle v={\sqrt {\frac {\mu }{r}}}}$

The figure on the right shows the course of the bi-elliptical transfer. The spacecraft is on a circular orbit (blue) with a radius . The speed is constant . The aim is to reach the higher circular path (green) with a radius . ${\displaystyle r_{1}}$ ${\displaystyle v_{1}={\sqrt {\frac {\mu }{r_{1}}}}}$ ${\displaystyle r_{2}}$

1. A momentary increase in speed brings the satellite to the first transfer ellipse (red), whose major semi-axis is. So the first change in speed is ${\displaystyle a_{1}}$

   ${\displaystyle \Delta v_{1}={\sqrt {\mu \left({\frac {2}{r_{1}}}-{\frac {1}{a_{1}}}\right)}}-{\sqrt {\frac {\mu }{r_{1}}}}}$
   It is to be applied tangentially in the direction of flight, since the starting path is a circle, the maneuver can start anywhere.

2. When the apoapsis is reached, the spacecraft is at a distance from the central body. The second instantaneous increase in speed takes place on the second transfer ellipse (orange), whose major semi-axis is. Again, the change in speed is tangential. The amount is ${\displaystyle r_{b}=2a_{1}-r_{1}}$ ${\displaystyle a_{2}={\tfrac {r_{b}+r_{2}}{2}}}$

   ${\displaystyle \Delta v_{2}={\sqrt {\mu \left({\frac {2}{r_{b}}}-{\frac {1}{a_{2}}}\right)}}-{\sqrt {\mu \left({\frac {2}{r_{b}}}-{\frac {1}{a_{1}}}\right)}}}$

3. When the periapsis of the second transfer ellipse is reached, the third speed change takes place. This time, however, it has to be reduced in size so that the satellite stays on the orbit

   ${\displaystyle \Delta v_{3}={\sqrt {\frac {\mu }{r_{2}}}}-{\sqrt {\mu \left({\frac {2}{r_{2}}}-{\frac {1}{a_{2}}}\right)}}}$

Overall, the fuel requirement (Delta v) is

${\displaystyle {\Delta v=\Delta v_{1}+\Delta v_{2}+\Delta v_{3}=\left({\sqrt {\mu \left({\frac {2}{r_{1}}}-{\frac {1}{a_{1}}}\right)}}-{\sqrt {\frac {\mu }{r_{1}}}}\,\right)+\left({\sqrt {\mu \left({\frac {2}{r_{b}}}-{\frac {1}{a_{2}}}\right)}}-{\sqrt {\mu \left({\frac {2}{r_{b}}}-{\frac {1}{a_{1}}}\right)}}\,\right)+\left({\sqrt {\frac {\mu }{r_{2}}}}-{\sqrt {\mu \left({\frac {2}{r_{2}}}-{\frac {1}{a_{2}}}\right)}}\,\right)}}$

If the radius of the target orbit is more than 15.58 times larger than the radius of the starting orbit, every bi-elliptical transfer is cheaper in terms of fuel than a Hohmann transfer, as long as it is. Below this value, a bi-elliptical transfer may require less Delta v (see section #Comparison with Hohmann transfer ). ${\displaystyle r_{b}>r_{2}}$

time

The transfer time can be calculated from half the cycle times of the transfer ellipses. The period of revolution is calculated for ellipses according to Kepler's third law ${\displaystyle T}$

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

The transfer time of a bi-elliptical transition is thus

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

This is considerably longer than with a Hohmann transfer, which is an important disadvantage of the bi-elliptical transfer (see section #Comparison with Hohmann transfer ).

Borderline case of Hohmann transfer

→ Main article : Hohmann transfer

For the borderline case , the bi-elliptical transfer goes over to the Hohmann transfer . ${\displaystyle r_{b}=r_{2}}$

Borderline case of bi-parabolic transfer

For the borderline case , the bi-elliptical transfer changes to the bi-parabolic transfer. ${\displaystyle r_{b}\rightarrow \infty }$

This case is purely theoretical, since the satellite is first brought infinitely far away from the central body. Firstly, it takes infinitely long, and secondly, one can then no longer apply the approximation of a two-body problem . Nevertheless, it is interesting to consider the comparison with the Hohmann transfer in the next section.

A bi-parabolic transfer between a lower starting orbit (blue) via two transfer parabolas (green and orange) to a higher-lying target orbit (red).

Point 1

The satellite is brought to an escape parabola (green).

${\displaystyle \qquad \Delta v_{1}=\left({\sqrt {2}}-1\right){\sqrt {\frac {\mu }{r_{1}}}}}$

At infinity its speed drops to 0. ${\displaystyle (\infty )}$

Point 2

An infinitesimally small thrust is now sufficient to bring the satellite to a new transfer parabola (orange).

${\displaystyle \qquad \Delta v_{2}=0}$

point 3

At the apex of the second parabola, you must now brake again to the target circular path.

${\displaystyle \qquad \Delta v_{3}=\left({\sqrt {2}}-1\right){\sqrt {\frac {\mu }{r_{2}}}}}$

Overall, the fuel requirement (Delta v) is

${\displaystyle \qquad \Delta v=\Delta v_{1}+\Delta v_{2}+\Delta v_{3}=\left({\sqrt {2}}-1\right)\left({\sqrt {\frac {\mu }{r_{1}}}}+{\sqrt {\frac {\mu }{r_{2}}}}\,\right)}$

This value is lower for all transitions than for a Hohmann transfer. The bi-parabolic transfer is the borderline case of a bi-elliptical transfer, for which the most delta v can be saved. ${\displaystyle r_{2}>11{,}94\,r_{1}}$

Comparison with the Hohmann transfer

speed

Δv requirement ( standardized to) for four maneuvers between two identical circular paths depending on the radius ratio . ${\displaystyle v_{1}}$ ${\displaystyle {\tfrac {r_{2}}{r_{1}}}}$

The figure on the right shows the required delta v, a measure for the fuel requirement and thus also for the energy when a transfer is made between a circular path with a radius and a circular path with a radius . ${\displaystyle r_{1}}$ ${\displaystyle r_{2}}$

${\displaystyle \Delta v}$is normalized with the initial speed so that the comparison is general. Four curves are shown: the fuel requirement for a Hohmann transfer (blue), for a bi-elliptical transfer with (red), for a bi-elliptical transfer with (cyan) and for a bi-parabolic transfer (green). ${\displaystyle v_{1}}$ ${\displaystyle \alpha ={\tfrac {r_{b}}{r_{1}}}=20+{\tfrac {r_{2}}{r_{1}}}}$ ${\displaystyle \alpha ={\tfrac {r_{b}}{r_{1}}}=100+{\tfrac {r_{2}}{r_{1}}}}$ ${\displaystyle (r_{b}\rightarrow \infty )}$

You can see that the Hohmann transfer is energetically the most favorable as long as the radius ratio is less than 11.94. If the radius of the target orbit is more than 15.58 times the radius of the starting orbit, any bi-elliptical transfer is cheaper in terms of fuel than a Hohmann transfer, as long as it is. ${\displaystyle r_{b}>r_{2}}$

For the area between 11.94 and 15.58, the distance between the common apoapsis of the two transfer ellipses (point 2 in the figures about the bi-elliptical transfer and the bi-parabolic transfer ) is decisive.

The following table lists some cases of how large (the distance of the apoapsis in relation to the radius of the starting orbit) must be at least so that the bi-elliptical transfer is energetically more favorable. ${\displaystyle \alpha ={\tfrac {r_{b}}{r_{1}}}}$

minimal for energetically more favorable bi-elliptical transfer ${\displaystyle \alpha ={r_{b}}/{r_{1}}}$

Radius ratio ${\displaystyle {r_{2}}/{r_{1}}}$	Minimal ${\displaystyle \alpha ={r_{b}}/{r_{1}}}$	Remarks
00 to 11.94	00,-	Hohmann transfer is cheaper
11.94	00 ${\displaystyle \infty }$	bi-parabolic transfer
12	815.81
13	048.90
14th	026.10
15th	018.19
15.58	015.58
greater than 15.58	greater than ${\displaystyle {\tfrac {r_{2}}{r_{1}}}}$	any bi-elliptical transfer is cheaper

This not necessarily intuitive connection can be explained by the Oberth Effect .

time

The long transfer time of a bi-elliptical transition

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

is a major disadvantage of this transfer maneuver. In the extreme case of the bi-parabolic transfer, the time is even infinitely long.

For comparison, a Hohmann transfer also needs

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

less than half the time because only half a transfer ellipse and not two half ellipses are driven.

example

different transfers required (km / s) ${\displaystyle \mathbf {\Delta v} }$

	Hohmann	bi-elliptical ${\displaystyle \alpha =100}$	bi-parabolic
${\displaystyle \Delta v_{1}}$	3.133	3.172	3.226
${\displaystyle \Delta v_{2}}$	0.833	0.559	not applicable
${\displaystyle \Delta v_{3}}$	not applicable	0.127	0.423
whole ${\displaystyle \mathbf {\Delta v} }$	3,966	3.858	3,649
Duration	118: 40: 49 h	782: 09: 27 h	${\displaystyle \infty }$

An example based on Example 6-2 illustrates the transfers:

A satellite that orbits the earth is to be brought from a circular runway onto the circular target orbit. The Hohmann transfer, the bi-elliptical transfer and the bi-parabolic transfer are compared in terms of speed and time. ${\displaystyle r_{1}=6569\,\mathrm {km} }$ ${\displaystyle r_{2}=382688\,\mathrm {km} }$

The ratio of the start to the finish radius is around 58.25. It can therefore be expected that the bi-elliptical and bi-parabolic transfer require less Delta v than the Hohmann transfer. For the bi-elliptical transfer a must be selected, for the example it is assumed. ${\displaystyle r_{b}>r_{2}}$ ${\displaystyle r_{b}=100\,r_{1}}$

Web links

• AFBA Prado, Journal of the Brazilian Society of Mechanical Sciences and Engineering , vol. 25, 2003, p. 122–128, doi : 10.1590 / S1678-58782003000200003 , online

Individual evidence

1. ↑ ^{a b c d e f} David A. Vallado: Fundamentals of Astrodynamics and Applications . Micorcosm Press, Hawthorne, CA 2013, ISBN 978-1-881883-18-0 , pp. 322-330 (English).
2. ^ ^{A b} Pedro R. Escobal: Methods of Astrodynamics . John Wiley & Sons, New York 1968, ISBN 978-0-471-24528-5 (English).
3. ↑ FW Gobetz, JR Doll: A Survey of Impulsive Trajectories . In: AIAA Journal . tape 7 , no. 5 , May 1969, p. 801-834 , doi : 10.2514 / 3.5231 (English).