Bi-elliptic Transfer Calculator
Compute three-burn bi-elliptic transfer delta-v, time of flight, and compare with Hohmann for any orbit ratio.
🌌 What is a Bi-elliptic Transfer?
A bi-elliptic transfer is an orbital maneuver that moves a spacecraft between two circular orbits using three engine burns and two intermediate ellipses, rather than the single ellipse of a Hohmann transfer. The spacecraft first burns to enter a large transfer ellipse that reaches far beyond the target orbit, then burns at apoapsis to reshape the trajectory, and finally burns to circularize at the destination.
The key advantage of bi-elliptic transfers appears at large orbit ratios. When the target orbit radius is more than about 11.94 times the initial orbit radius, a bi-elliptic path can consume less total delta-v than the classic two-burn Hohmann transfer. At ratios above 15.58 the bi-elliptic approach always wins on propellant efficiency, regardless of the intermediate altitude chosen. In the range from 11.94 to 15.58, the outcome depends on exactly how large the intermediate orbit is made.
Bi-elliptic transfers are used in mission planning for high-orbit satellites, lunar missions, and deep space probes where the large orbit ratio makes the extra complexity worthwhile. A common application is repositioning geostationary satellites from a graveyard orbit or planning transfers to very high Earth orbits used by navigation constellations. The main drawback is time: because the spacecraft coasts far from Earth before returning, the total transfer time can be many times longer than a Hohmann trajectory covering the same altitude change.
This calculator supports five central bodies (Earth, Moon, Mars, Venus, Jupiter) and shows the complete delta-v budget broken into the three individual burns, the semi-major axes of both transfer ellipses, and the total time of flight. The comparison tab places the bi-elliptic result alongside the equivalent Hohmann result and identifies which regime the orbit ratio falls into.
📐 Formula
Δvtotal = Δv1 + Δv2 + Δv3
a1 = (r1 + rb) / 2 — semi-major axis of first transfer ellipse
a2 = (rb + r2) / 2 — semi-major axis of second transfer ellipse
Δv1 = |√(μ (2/r1 − 1/a1)) − √(μ/r1)| — departure burn from initial orbit
Δv2 = |√(μ (2/rb − 1/a2)) − √(μ (2/rb − 1/a1))| — apoapsis burn
Δv3 = |√(μ/r2) − √(μ (2/r2 − 1/a2))| — arrival circularization burn
TOF = π√(a13/μ) + π√(a23/μ) — total transfer time
r1 = central body radius + initial altitude (m)
r2 = central body radius + target altitude (m)
rb = central body radius + intermediate altitude (m)
μ = gravitational parameter of the central body (m³/s²)
Example: Earth LEO 400 km to 200,000 km with rb at 800,000 km gives total Δv ≈ 3.854 km/s, saving about 192 m/s over the Hohmann path.
📖 How to Use This Calculator
Steps
1
Select central body -- Choose the central body (Earth, Moon, Mars, Venus, or Jupiter) from the dropdown. This sets the gravitational parameter and surface radius used in all calculations.
2
Enter initial orbit altitude -- Type the altitude of your starting circular orbit in kilometers above the surface. Use the slider for quick exploration or type directly for precision.
3
Enter target orbit altitude -- Type the altitude of the destination circular orbit. The target must differ from the initial altitude by at least 1 km.
4
Set intermediate altitude -- Enter the bi-elliptic intermediate altitude. This must exceed both the initial and target altitudes. A larger intermediate orbit generally reduces total delta-v at the cost of longer transfer time.
5
Read results or compare with Hohmann -- The Bi-elliptic Transfer tab shows the three-burn breakdown and time of flight. Switch to the vs Hohmann Comparison tab to see which maneuver is more efficient for your orbit ratio.
💡 Example Calculations
Example 1 -- LEO to High Earth Orbit (Bi-elliptic Zone)
Earth: 400 km circular orbit to 200,000 km, intermediate at 800,000 km
1
Orbit ratio: r2/r1 = (6378.1 + 200,000) / (6378.1 + 400) = 206,378.1 / 6,778.1 ≈ 30.4 -- well above 15.58, so bi-elliptic always wins.
2
First transfer ellipse: a1 = (6,778.1 + 806,378.1) / 2 ≈ 406,578 km. Departure burn: Δv1 ≈ 3.133 km/s.
3
Apoapsis burn at 800,000 km: Δv2 ≈ 0.358 km/s. Second transfer ellipse: a2 = (806,378.1 + 206,378.1) / 2 ≈ 506,378 km.
4
Circularization at 200,000 km: Δv3 ≈ 0.364 km/s. Total Δv ≈ 3.854 km/s. Equivalent Hohmann: ≈ 4.046 km/s. Savings: ≈ 192 m/s.
Total Δv = 3.854 km/s | Transfer time ≈ 35.7 days
Try this example →Example 2 -- LEO to GEO (Hohmann Zone)
Earth: 400 km circular orbit to geostationary orbit (35,786 km), intermediate at 100,000 km
1
Orbit ratio: r2/r1 = (6378.1 + 35,786) / (6378.1 + 400) = 42,164.1 / 6,778.1 ≈ 6.22 -- well below 11.94, so Hohmann always wins.
2
Hohmann transfer: Δv ≈ 3.859 km/s (two burns via GTO semi-major axis 24,471 km).
3
Bi-elliptic via 100,000 km intermediate: Δv ≈ 4.245 km/s. Hohmann saves about 386 m/s of propellant.
Hohmann Δv = 3.859 km/s | Bi-elliptic penalty: +386 m/s
Try this example →Example 3 -- Mars Low Orbit to High Mars Orbit (Bi-elliptic Zone)
Mars: 400 km orbit to 100,000 km, intermediate at 500,000 km
1
Mars radii: initial r1 = 3,789.5 km, target r2 = 103,389.5 km. Orbit ratio r2/r1 ≈ 27.3 -- deep in the bi-elliptic zone.
2
Mars gravitational parameter: μ = 4.283 × 1013 m3/s2. Transfer ellipse 1: a1 = (3,789.5 + 503,389.5) / 2 ≈ 253,590 km.
3
Three burns calculated using Mars gravity. Bi-elliptic total Δv ≈ 2.67 km/s, compared to Hohmann ≈ 2.92 km/s. Savings ≈ 250 m/s.
Total Δv ≈ 2.67 km/s | Bi-elliptic saves ≈ 250 m/s over Hohmann
Try this example →❓ Frequently Asked Questions
What is a bi-elliptic transfer and how does it differ from Hohmann?
A bi-elliptic transfer uses three engine burns and two transfer ellipses to move between circular orbits. A Hohmann transfer uses only two burns and one ellipse. The extra burn in bi-elliptic allows the spacecraft to reach a large intermediate orbit where it can change energy more efficiently, reducing total delta-v when the orbit ratio is high.
When is a bi-elliptic transfer more efficient than Hohmann?
Bi-elliptic is always more efficient when the orbit ratio r2/r1 exceeds 15.58. Between 11.94 and 15.58 the answer depends on how large the intermediate orbit is chosen. Below 11.94 Hohmann is always more efficient regardless of intermediate altitude.
Why does bi-elliptic save delta-v only at high orbit ratios?
At high orbit ratios the spacecraft moves very slowly near the apoapsis of the large intermediate ellipse. Applying a small velocity change at this low-speed point produces a large change in orbital energy. This efficiency gain is large enough to offset the cost of the extra burn, resulting in a net delta-v saving over the direct Hohmann path.
What are the three burns in a bi-elliptic transfer?
The first burn departs from the initial circular orbit and inserts into the first transfer ellipse, raising apoapsis to the intermediate radius rb. The second burn at apoapsis transfers to the second ellipse, which has its periapsis at the target radius. The third burn at periapsis circularizes the orbit at the target altitude.
How do I choose the intermediate orbit altitude?
A larger intermediate orbit generally saves more delta-v when the orbit ratio is above 11.94, but it also increases transfer time significantly. The optimal choice is theoretically infinity, but practical missions balance propellant savings against mission duration, communication blackout periods, and radiation belt exposure.
Can bi-elliptic transfers be used for GEO insertion from LEO?
Not efficiently. The LEO-to-GEO orbit ratio is about 6.6, well below the 11.94 Hohmann threshold. Using a bi-elliptic path would cost hundreds of meters per second more delta-v than a standard GTO Hohmann transfer and significantly extend the mission timeline without any benefit.
How long does a bi-elliptic transfer take compared to Hohmann?
A bi-elliptic transfer always takes longer than an equivalent Hohmann transfer because the spacecraft must coast to a much farther intermediate point and back. For a LEO-to-high-orbit transfer with an intermediate orbit at 800,000 km the transfer time can exceed 35 days, compared to a few hours for a Hohmann path covering the same altitude change.
What is the orbit ratio transition zone between 11.94 and 15.58?
In the transition zone the bi-elliptic can be more or less efficient than Hohmann depending on the intermediate altitude. A higher intermediate orbit favors bi-elliptic. A lower intermediate orbit (close to the target altitude) pushes the result toward Hohmann efficiency. The exact crossover point within this zone must be calculated for each specific intermediate orbit choice.
What are real missions that use bi-elliptic-like transfers?
Missions to very high Earth orbits, some lunar trajectories, and certain deep space probes use bi-elliptic or bi-parabolic inspired trajectories. Satellite repositioning maneuvers in geostationary arc sometimes use three-burn strategies when large longitude changes are required, exploiting the efficiency of changing orbital energy at high altitude.
What is the semi-major axis of each bi-elliptic transfer ellipse?
The semi-major axis of the first transfer ellipse is a1 = (r1 + rb) / 2, where r1 is the initial orbit radius and rb is the intermediate radius. The second ellipse has semi-major axis a2 = (rb + r2) / 2, where r2 is the target orbit radius. Both values are shown in the calculator results.
Does this calculator account for inclination changes?
No. This calculator assumes coplanar circular orbits with no inclination change. Adding an inclination change to a bi-elliptic transfer is most efficient when performed at the high-altitude apoapsis burn, since the spacecraft velocity is lowest there. Combined plane change and bi-elliptic maneuvers are covered in advanced astrodynamics texts.
Why is the apoapsis burn delta-v so small compared to the other two burns?
At the intermediate apoapsis the spacecraft is moving very slowly because it has traded most of its kinetic energy for potential energy. A small velocity change at this low-speed point is all that is needed to reshape the second ellipse. This is exactly the reason bi-elliptic transfers are efficient: the expensive orbital energy exchange happens at the cheapest possible velocity point.