Astrophysics

Hohmann Transfer Orbit

The fuel-cheapest two-burn ellipse for moving between two circular orbits

The Hohmann transfer orbit is the fuel-cheapest two-burn elliptical path between two coplanar circular orbits: one prograde burn raises apoapsis to the target radius, a second circularizes there. For most radius ratios it minimizes total Δv.

  • BurnsTwo impulsive, both tangential (prograde)
  • Transfer semi-major axisa = (r₁ + r₂)/2
  • Speeds fromVis-viva: v = √(μ(2/r − 1/a))
  • Transfer timet = π√(a³/μ) (half the ellipse period)
  • Optimal up toRadius ratio ≈ 11.94 (then bi-elliptic wins)
  • Named forWalter Hohmann, 1925

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The intuition — coast, don't push

You're in a low circular orbit and you want to reach a higher one. The wasteful move is to point your engine "up" and climb radially. That fights gravity head-on and bends your velocity vector around, burning fuel the whole way. The Hohmann transfer does the opposite: it fires along the direction you're already moving, lets the ellipse do the climbing for free, and only fires again once — at the very top.

The trick rests on a single fact of orbital mechanics: a single prograde burn doesn't change where you are, it changes where you'll be half an orbit later. Speed up at the bottom of your orbit and you raise the opposite side. So one forward kick on the inner circle stretches it into an ellipse whose far point just kisses the target radius. You then coast — engine off — for half that ellipse, climbing all the way out for zero additional fuel. A second forward kick at the top circularizes you onto the destination orbit.

Two burns, one long unpowered coast in between. That is the entire maneuver, and for the great majority of orbit changes it is provably the cheapest two-impulse path that exists.

How the two burns work

Label the inner circular radius r₁ and the outer one r₂. The transfer ellipse is tangent to the inner orbit at its periapsis (closest point, radius r₁) and tangent to the outer orbit at its apoapsis (farthest point, radius r₂). That geometry fixes its semi-major axis:

a_transfer = (r₁ + r₂) / 2

Burn 1 (at r₁, periapsis): a prograde impulse Δv₁ that speeds you up from the inner circular speed to the (faster) periapsis speed of the transfer ellipse. You don't move yet — you've simply raised the apoapsis out to r₂.

Coast: engine off. You climb from r₁ to r₂ along the ellipse, trading kinetic energy for potential energy. By the time you reach apoapsis you've slowed down — and you're now moving slower than the outer circular orbit requires.

Burn 2 (at r₂, apoapsis): a second prograde impulse Δv₂ that speeds you up from the slow apoapsis speed to the outer circular speed. This raises the periapsis up to r₂ as well — the ellipse becomes a circle. You're done.

Going inward (high orbit to low), run it in reverse: both burns are retrograde (braking), but the magnitudes are identical.

The governing physics — vis-viva

Every speed in the maneuver comes from one equation, the vis-viva equation, which falls straight out of conservation of energy in a 1/r gravity field:

v² = μ (2/r − 1/a)

where μ = GM is the gravitational parameter of the central body (for Earth, μ ≈ 398,600 km³/s²), r is your current distance from the center, and a is the semi-major axis of the orbit you're on. For a circle, a = r, so it collapses to the circular speed:

v_circular = √(μ / r)

Now write the four speeds you need:

v_c1 = √(μ / r₁)                          inner circular speed
v_c2 = √(μ / r₂)                          outer circular speed
v_p  = √(μ (2/r₁ − 1/a_transfer))         transfer speed at periapsis
v_a  = √(μ (2/r₂ − 1/a_transfer))         transfer speed at apoapsis

The two burns and the total cost:

Δv₁ = v_p − v_c1        (prograde at r₁)
Δv₂ = v_c2 − v_a        (prograde at r₂)
Δv_total = |Δv₁| + |Δv₂|

And the transfer time is exactly half the period of the ellipse:

t_transfer = π √(a_transfer³ / μ)

Worked example — LEO to GEO

Lift a satellite from a 300 km low-Earth orbit to geostationary orbit. Using Earth's μ = 398,600 km³/s²:

  • r₁ = 6,378 + 300 = 6,678 km
  • r₂ = 42,164 km (geostationary radius)
  • a_transfer = (6,678 + 42,164)/2 = 24,421 km

Plug into vis-viva:

v_c1 = √(398600 / 6678)              =  7.726 km/s
v_p  = √(398600 (2/6678 − 1/24421))  = 10.152 km/s
Δv₁  = 10.152 − 7.726                =  2.426 km/s

v_c2 = √(398600 / 42164)             =  3.075 km/s
v_a  = √(398600 (2/42164 − 1/24421)) =  1.608 km/s
Δv₂  = 3.075 − 1.608                 =  1.467 km/s

Δv_total ≈ 3.89 km/s
t_transfer = π √(24421³ / 398600)    ≈ 5.27 hours

Notice Δv₁ is the bigger of the two — that's general: the first burn, deep in the gravity well where speeds are highest, almost always dominates. (This is also why the Oberth effect makes burns near periapsis so valuable.)

Hohmann vs. the alternatives

ManeuverBurnsTotal Δv (LEO→GEO)Transfer timeBest when
Hohmann transfer2 (both tangential)≈ 3.9 km/s≈ 5.3 hRadius ratio < ~11.94 — almost everything
Bi-elliptic transfer3More (≈ 4.2–4.5 km/s at this ratio of 6.3); less than Hohmann only past ratio ≈ 11.94Days to weeksExtreme radius ratios, time is cheap
One-tangent / fast transfer2 (one non-tangential)> 3.9 km/sShorter than HohmannWhen arrival timing beats fuel cost
Direct / brute-force climbContinuousMuch higherVariableNever optimal; only if forced
Low-thrust spiral (ion)Continuous thrustHigher Δv, but huge specific impulseWeeks to monthsElectric propulsion — fuel mass, not Δv, is what matters
Gravity assist (flyby)0 propulsive (uses a planet)Effectively free ΔvLong, geometry-lockedInterplanetary; trades patience for fuel

The headline number — Δv ≈ 3.9 km/s — is what every geostationary satellite operator pays. Multiply by the rocket equation and that single figure decides how much of a launch vehicle's mass can be payload versus propellant.

Real-world figures

  • Earth to Mars. A Hohmann transfer takes ≈ 259 days and needs about 5.6 km/s of total heliocentric Δv (perihelion departure plus aphelion arrival); the trans-Mars injection burn from low-Earth orbit itself is ≈ 3.6 km/s. The destination must be at the arrival point when you get there, so windows open only every 26 months (the Earth–Mars synodic period, ≈ 780 days).
  • Geostationary insertion. The ≈ 1.47 km/s apogee burn is the classic "apogee kick" performed by a satellite's onboard motor after the launcher drops it on a geostationary transfer orbit (GTO) — itself a Hohmann ellipse.
  • Earth to the Moon. A trans-lunar injection is essentially a Hohmann-like ellipse out to lunar distance, ≈ 3.1 km/s from LEO, ≈ 3 days of coast.
  • Cost of getting it wrong. Because Δv enters the rocket equation exponentially (m_fuel/m_dry = e^(Δv/v_e) − 1), an extra 1 km/s on a 3.9 km/s budget with a 300 s-Isp upper stage raises the propellant-to-dry-mass ratio by about half again (≈ 2.8 → ≈ 4.3). That's why nobody climbs radially.
  • Plane changes are murder. Rotating an orbit's plane by angle i costs Δv = 2·v·sin(i/2). At GEO speed (3.07 km/s) a 28.5° plane change alone costs ≈ 1.5 km/s — comparable to the entire apogee circularization burn.

When three burns beat two — the bi-elliptic surprise

It feels obvious that fewer burns means less fuel. It isn't always true. A bi-elliptic transfer fires you onto a huge ellipse whose apoapsis sails past the target out to some intermediate radius r_b, circularizes-ish at the top, then drops back down to r₂ with a third (braking) burn.

The reason it can win is the Oberth effect in reverse: when you climb very high, your speed at apoapsis is tiny, so the burn that raises periapsis there is extremely cheap. Push r_b to infinity and the second burn vanishes. The crossover is sharp and famous:

r₂/r₁ < 11.94          Hohmann always wins
11.94 < r₂/r₁ < 15.58  depends on the intermediate radius r_b
r₂/r₁ > 15.58          bi-elliptic always wins (with high enough r_b)

The catch is time. A bi-elliptic transfer to a high apoapsis can take many times longer than the Hohmann equivalent — sometimes weeks instead of hours — so in practice it's mostly a textbook curiosity and a tool for a few deep, slow maneuvers.

Assumptions and where they break

  • Coplanar orbits. The clean two-burn result assumes both orbits lie in the same plane. Real missions usually need a plane change too, and the cheapest move is to fold that into the high, slow apoapsis burn where it's least expensive.
  • Impulsive burns. The math treats each burn as instantaneous. Real engines thrust over minutes, smearing the impulse over an arc — a "finite-burn loss" that adds a few percent to the ideal Δv.
  • Two-body gravity only. No atmospheric drag, no solar pressure, no third bodies. For interplanetary work the simple Hohmann is just the patched-conic first guess; real trajectories add gravity assists and mid-course corrections.
  • Circular endpoints. If either orbit is already elliptical, the optimal transfer is no longer a textbook Hohmann and you solve a more general two-impulse optimization.

Common misconceptions

  • "You point the engine toward the target." No — you burn prograde, along your current velocity, both times. Pointing at the destination wastes Δv turning the velocity vector.
  • "Speeding up moves you outward immediately." A prograde burn raises the orbit on the far side, half an orbit later. You don't gain altitude where you fired; you gain it across the orbit.
  • "To go faster you thrust forward; to slow down you thrust back — so the higher orbit is faster." Counter-intuitively the higher circular orbit is slower (v = √(μ/r) shrinks with r). You added energy to climb, yet end up moving slower — the missing kinetic energy went into potential energy.
  • "Fewer burns is always cheaper." The bi-elliptic transfer disproves this for radius ratios above ≈ 11.94.
  • "Hohmann minimizes time." It minimizes fuel, and is actually the slowest useful two-burn route. Faster transfers exist; they just cost more Δv.
  • "It works for any launch date." For interplanetary transfers the destination has to be at the arrival point, which only happens once per synodic period — hence Mars windows every 26 months.

Frequently asked questions

Why is the Hohmann transfer the most fuel-efficient way between two orbits?

Because it touches both circular orbits exactly tangentially with only two impulsive burns, and both burns are aligned with the velocity vector — no Δv is wasted bending the velocity direction. Any path that crosses orbits at an angle, or uses three or more burns for a modest radius change, spends more total Δv. The Hohmann transfer is the minimum-energy two-impulse solution for radius ratios up to about 11.94; beyond that the bi-elliptic transfer wins.

What are the two burns in a Hohmann transfer?

Burn 1 is a prograde (forward) impulse on the inner circular orbit that raises the apoapsis to the target radius, placing the craft on an elliptical transfer orbit. The spacecraft then coasts for half an ellipse with the engine off. Burn 2 is a second prograde impulse at apoapsis that raises the periapsis up to the target radius, circularizing the orbit. Both burns add speed when going outward; both subtract speed when going inward.

How do you calculate Hohmann transfer Δv?

Use the vis-viva equation v = √(μ(2/r − 1/a)). The transfer ellipse has semi-major axis a = (r1 + r2)/2. Δv1 = v at r1 on the ellipse minus the inner circular speed √(μ/r1); Δv2 = the outer circular speed √(μ/r2) minus v at r2 on the ellipse. Total Δv = |Δv1| + |Δv2|. For a LEO (6,678 km) to GEO (42,164 km) transfer around Earth, that's roughly 2.44 + 1.47 ≈ 3.9 km/s.

How long does a Hohmann transfer take?

Exactly half the period of the transfer ellipse: t = π√(a³/μ), where a = (r1 + r2)/2. A LEO-to-GEO transfer takes about 5.3 hours. An Earth-to-Mars Hohmann transfer takes roughly 259 days (about 8.5 months), which is why Mars launch windows open only every 26 months.

When is a bi-elliptic transfer better than a Hohmann transfer?

When the ratio of final to initial radius exceeds about 11.94, a bi-elliptic transfer — which flings the craft far past the target before dropping back down — can use less total Δv than a Hohmann transfer, even though it adds a third burn. Between ratios of 11.94 and 15.58 it depends on the intermediate radius. The catch: bi-elliptic transfers can take many times longer, so they're rare in practice.

Why do Mars launch windows only open every 26 months?

A Hohmann transfer arrives at a fixed point on the target orbit after a fixed coast time (about 259 days for Mars). The destination planet must be at exactly that arrival point when the spacecraft gets there. Earth and Mars return to the same relative geometry only once per synodic period — about 780 days, or 26 months — so that's how often a minimum-energy departure lines up.