How this instrument works
A Hohmann transfer swaps one circular orbit for another by coasting along half of an ellipse that touches both circles: periapsis at the inner radius, apoapsis at the outer one. Because the ellipse must touch both circular orbits exactly, its semi-major axis is fixed by geometry alone — the arithmetic mean of the two radii, a = (r₁ + r₂) ⁄ 2. No other ellipse connects the two orbits with a single tangent burn at each end.
Kepler's third law ties an orbit's period to its semi-major axis cubed: a full lap of that ellipse would take 2π√(a³⁄GM), with GM the central body's standard gravitational parameter. A transfer, though, never completes the lap — the spacecraft boards at periapsis and steps off at apoapsis, exactly half the ellipse, which is why the transfer-time formula carries a lone π rather than 2π. The same cubed-radius relationship governs the Moon around Earth and Earth around the Sun; only GM changes.
The formula treats both burns as instantaneous velocity changes and both orbits as circular and coplanar, with the transferring spacecraft's own mass too small to perturb the primary. Real hardware burns for tens of seconds to a few minutes rather than zero, and if the destination orbit sits in a different plane, a separate inclination change with its own fuel bill is not accounted for here — that cost has to be added, or folded into one combined burn, outside this instrument.
- Enter the Central body mass — Earth is 5.9722 × 10²⁴ kg; this value sets GM and scales the entire transfer time.
- Enter the Departure orbit radius, measured from the body's center rather than its surface — for a 400 km-altitude Earth orbit that is 6,771 km.
- Enter the Destination orbit radius the same way, from the center; geostationary orbit around Earth sits at 42,164 km.
- Read the Transfer time in seconds, or switch its unit to hours — the figure is coast time only, from the departure burn to the arrival burn.
Worked example — 400 km low Earth orbit to geostationary orbit
A satellite sits in a 400 km-altitude parking orbit around Earth, radius r₁ = 6,771,000 m once the planet's own 6,371 km bulk is added in, and needs to reach geostationary orbit at r₂ = 42,164,000 m. With Earth's mass M = 5.9722 × 10²⁴ kg, the semi-major axis of the connecting ellipse is a = (6,771,000 + 42,164,000) ⁄ 2 = 24,467,500 m.
Feeding that into T = π√(a³⁄GM) returns 19,044.2658217 seconds — about 5 hours and 17 minutes of coasting between the departure burn at perigee and the circularization burn at apogee. That duration, not the fuel cost, is often the number that decides a ground-station handover schedule or a crew's on-orbit timeline, since the two burns themselves last only seconds by comparison.
Questions
Why is the transfer ellipse's semi-major axis just the average of the two radii?
Because the ellipse is defined by touching both circular orbits exactly once: its closest point, periapsis, sits at the inner radius r₁ and its farthest point, apoapsis, sits at the outer radius r₂. A semi-major axis is half the sum of periapsis and apoapsis distances by definition, so a = (r₁ + r₂) ⁄ 2 falls straight out of the geometry, no extra physics required.
Why does the formula use π instead of the 2π in Kepler's third law?
Kepler's third law gives the period of one full lap of an orbit, 2π√(a³⁄GM). A Hohmann transfer only rides the ellipse from periapsis to apoapsis — exactly half a lap — so the transfer time is half the Keplerian period, and the 2π becomes a lone π.
Does the transfer time include the two rocket burns?
No. Both burns are modeled as instantaneous velocity changes, so T is pure coast time between them. Real burns take anywhere from seconds to several minutes depending on the engine's thrust-to-weight ratio, which adds a little to the true door-to-door time but rarely changes the answer by more than a rounding error.
Is a Hohmann transfer the fastest way between two orbits?
No, it is the cheapest, not the quickest. It minimizes the total velocity change, delta-v, needed for two coplanar circular orbits, which is why it is the default baseline, but a higher-thrust direct burn reaches the destination sooner at the cost of much more propellant.
What happens if the departure and destination orbits are not in the same plane?
This formula does not price that in. A plane change needs its own delta-v, ideally spent where orbital speed is lowest, usually combined with the apoapsis burn, and for large-angle changes it can cost more fuel than the transfer itself. Treat T here as the in-plane coast time only.
Who actually calculates Hohmann transfer times like this?
Mission designers sizing the coast arc between a launch vehicle's parking orbit and a satellite's operational orbit, satellite operators scheduling an orbit-raising campaign from low Earth orbit to geostationary orbit, and students checking Walter Hohmann's 1925 result against a modern orbital mechanics textbook.