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 rb 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 [1].


Contents

[edit] Calculation

[edit] Delta-v

A bi-elliptic transfer from a low circular starting orbit (dark blue), to a higher circular orbit (red).
A bi-elliptic transfer from a low circular starting orbit (dark blue), to a higher circular orbit (red).

Utilizing the vis viva equation where,

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

where:

The magnitude of the first delta-v at the initial circular orbit with radius r0 is:

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

At rb 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 rf:

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

Where a1 and a2 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] 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 \;

[edit] Example

For example, to transfer from circular low earth orbit with r0 = 6700 km to a new circular orbit with r1 = 14r0 = 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 r2 = 40r0 = 268000 km, then in apogee accelerates another 608.679 m/s, which places it in new orbit with perigee at r1 = 14r0 = 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.

Burn Hohmanm ΔV (m/s) Bi-elliptic ΔV (m/s)
1 2824.34 3060.31
2 1308.38 608.679
3 - 447.554
Total 4132.72 4116.54
  • Forward applied ΔV
  • 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.

[edit] See also

[edit] References

  1. ^ Vallado, David Anthony (2001). Fundamentals of Astrodynamics and Applications. Springer, 317. ISBN 0792369033. 

[edit] External links

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