Hohmann Transfer Calculation: Two Burns, Delta-v and Travel Time

What does a Hohmann transfer connect?

An ideal Hohmann transfer connects two circular, coplanar orbits around the same central body with a tangent ellipse and two instantaneous burns. The first burn enters the ellipse. After coasting through half an orbit, the second burn matches the speed of the destination circle.

For an outward transfer with r₂ > r₁, both burns are prograde: each increases the spacecraft's speed at that location. The final circular speed is nevertheless lower than the initial circular speed. There is no contradiction, because the spacecraft slows down while moving outward between the burns.

This guide calculates an ideal two-body transfer. It is a reusable mathematical example, not a flight plan. It leaves out atmosphere, finite burn duration, plane changes, perturbations and the tracking needed to execute a real maneuver.

Define radii from the centre, not heights above the surface

All numerical examples use μ = 398600.4 km³/s², obtained from the IAU nominal terrestrial mass parameter 3.986004 × 10¹⁴ m³/s². A cubic kilometre contains 10⁹ cubic metres, so the conversion divides the SI value by 10⁹. This is a declared nominal constant for reproducibility.

The radii 7000 km and 14000 km are centre-to-centre distances. They are not altitudes. If the inputs are heights, add a consistently chosen central-body radius before using the equations. The IAU nominal Earth equatorial radius 6378.1 km would put a 7000 km orbit at a model height of 621.9 km.

Build the ellipse from the two circles

The transfer ellipse touches the inner circle at periapsis and the outer circle at apoapsis. Its semimajor axis is aₜ = (r₁ + r₂)/2, and its eccentricity is eₜ = (r₂ − r₁)/(r₂ + r₁) for the outward case. The central body is at a focus, not the centre of the ellipse.

With r₁ = 7000 km and r₂ = 14000 km, aₜ = 10500 km and eₜ = 1/3. Checking aₜ(1 − eₜ) and aₜ(1 + eₜ) returns the two endpoint radii. This is a useful first check before calculating any speed.

The burns occur on opposite sides of the central body. A line drawn radially outward from the departure point is not the spacecraft's path. Between the burns the spacecraft follows the curved ellipse under gravity, without continuous thrust in this ideal model.

Calculate four speeds before subtracting two burns

The circular speeds are v₁ = √(μ/r₁) and v₂ = √(μ/r₂). The transfer speeds follow vis-viva: vₜ₁ = √[μ(2/r₁ − 1/aₜ)] and vₜ₂ = √[μ(2/r₂ − 1/aₜ)]. Each subtraction must compare speeds at the same location.

The outward burns are Δv₁ = vₜ₁ − v₁ and Δv₂ = v₂ − vₜ₂. At these tangent points, the relevant velocity vectors lie along the same direction of motion, so the speed differences represent the tangential impulse magnitudes. This shortcut does not apply to an arbitrary maneuver with a direction change.

Total maneuver delta-v is the sum of the two impulse magnitudes. It is not the difference between the beginning and ending circular speeds, and it is not the vector sum of two burns applied at different points along the orbit.

A complete 7000-to-14000 km worksheet

The first burn raises the speed from 7.546053 to 8.713431 km/s. At the far side, gravity has reduced the transfer speed to 4.356716 km/s. The destination circle requires 5.335865 km/s, so a second increase of 0.979150 km/s is necessary.

The total is 2.146528 km/s. Subtracting the initial circular speed from the final one instead gives a negative change of about 2.210188 km/s; that quantity is a comparison of two endpoint speeds, not the required propulsion budget. The coast and its gravitational acceleration cannot be omitted from the story.

If the second burn is skipped, the spacecraft stays on the transfer ellipse in this model and later returns toward the inner radius. Merely reaching the destination radius does not place it in the destination circular orbit.

Transfer time is half the ellipse's period

Kepler's period law gives Pₜ = 2π√(aₜ³/μ). The coast from periapsis to apoapsis takes half of that, so tₜ = π√(aₜ³/μ). In the example, the result is 5353.834676 seconds, or 1.487176 hours.

Do not use the period of the initial circle or the final circle in place of the transfer ellipse's period. Do not divide the radial difference by one chosen speed either. The path is curved, the speed changes, and the radial component is zero at both apsides.

This ideal elapsed time starts immediately after the first impulse and ends at the second. A launch, departure preparation, finite engine firing, later phasing or arrival operations can add time to an actual mission.

Compare several destination radii

Each row changes the outer circular radius while keeping the initial radius and central parameter fixed. The 42164 km row is simply a numerical destination in this table; identifying a real geostationary orbit additionally requires the correct rotation period, equatorial plane and direction.

More distant destinations generally require a longer coast in this outward family because their transfer semimajor axes are larger. The two burns do not follow identical curves: the second burn's normalized magnitude eventually decreases even while the first continues to increase over the range shown.

See the two burns separately

Let R = r₂/r₁ and divide each burn by v₁ = √(μ/r₁). Then Δv₁/v₁ = √[2R/(1 + R)] − 1, while Δv₂/v₁ = [1 − √(2/(1 + R))]/√R. The graph contains dimensionless values; multiply by the actual initial circular speed to recover km/s.

The time column uses the initial circular period P₁ and equals ½[(1 + R)/2]³ᐟ². At R = 1 the two circles coincide and both burns vanish. The formula still returns half an unchanged circular period, but a zero-distance orbit change does not require waiting that long. Treat this as the limiting geometry, not a mandatory maneuver.

Lowering an orbit reverses the burns

For the inward transfer, start on the outer circle, reduce speed to enter the ellipse, coast to the inner radius, and reduce speed again to circularize. Positive signed values denote prograde changes here; negative values denote retrograde changes.

The same two radii produce the same total of magnitudes and the same half-ellipse coast time in either direction. The first and second magnitudes exchange roles. Adding the signed negative values and reporting a negative fuel requirement would confuse direction with the size of the maneuver.

Check the coast with energy and angular momentum

Between the burns, specific orbital energy ε = −μ/(2aₜ) and specific angular momentum h are constant. At the tangent endpoints h = r₁vₜ₁ = r₂vₜ₂. The worksheet has twice the radius at arrival and half the transfer speed, satisfying that equality.

The impulses change the energy and angular momentum from one orbit to the next. Consequently, demanding equal energy across all three table rows would be a mistake. Conservation applies during the unpowered two-body coast; the engines deliberately alter the orbital state at the endpoints.

How size and central mass change the answer

For a fixed radius ratio and unchanged μ, multiplying both radii by k multiplies every speed and delta-v by 1/√k, while the coast time grows by k³ᐟ². Doubling both radii therefore lowers the speed budget by √2 and lengthens the coast by 2√2.

With the geometry fixed, multiplying μ by four doubles the speeds and halves the coast time. The last row is a hypothetical different central parameter, not a claim that Earth's gravity changes. These scaling checks are useful for finding a misplaced square root or inconsistent distance unit.

Matching an orbit does not guarantee a rendezvous

A target object must also be at the arrival point when the transfer reaches it. In a circular coplanar outward model, place departure at angle zero and let the target's initial lead be φ. The target advances by n₂tₜ during the coast, where n₂ = √(μ/r₂³). Requiring arrival at π radians gives φ = π − n₂tₜ, modulo 2π.

For the 7000-to-14000 km example, the target's initial lead is 63.086570° in the chosen direction of motion. This is a synthetic phasing example. It does not include launch-site geometry, orbital inclination or a real spacecraft's current state.

NASA's interplanetary discussion illustrates why trajectory geometry and planet arrival must be considered together. Extending a solar transfer calculation to a complete Earth-to-Mars mission also needs departure from Earth's neighborhood and arrival in Mars' neighborhood; a single heliocentric delta-v table is not that complete budget.

What this calculation does and does not optimize

The Hohmann construction is an important reference for transfers between circular coplanar orbits with impulsive tangential burns. It is not a universal statement that two burns are always the best answer for every mission. Other transfer families, time constraints, plane changes and low-thrust propulsion can change the comparison.

Delta-v is not propellant mass. Converting a maneuver budget to propellant requires a propulsion model, including exhaust performance and vehicle mass. Likewise, a high target radius does not by itself imply an escape trajectory: the transfer ellipse here remains bound.

Review the vis-viva derivation for the speed equation and Kepler's third law for period and units. For a separate Earth-based observing task, the sunrise and sunset tool gives local solar times; it is not an orbital-transfer planner.

Frequently asked questions