Plan an Orbital Maneuver
Plan an Orbital Maneuver — Project Guide

🧑‍🔧 Plan an Orbital Maneuver

Field: Aerospace

Written and maintained by the PhDino author · Last reviewed 21 September 2026 · Every calculator used here is tested against independent reference values · how PhDino checks its numbers

Work out the delta-v an orbital transfer needs, then how much of your rocket's mass has to be propellant to deliver it, and how that changes with the engine.

Every orbital maneuver starts with the same question: how much delta-v (total velocity change) does the maneuver actually require? Once that number exists, the rocket equation translates it into a hard mass requirement: for a chosen engine efficiency and starting mass, exactly how much of that mass has to be propellant.

This guide runs those two steps in the order a real mission plan does: figure out what the maneuver costs in delta-v first, then check whether a given vehicle can actually deliver it, then ask how the answer moves when you change the engine or the geometry.

Delta-v is the currency of spaceflight

In orbit you cannot steer with a wheel; you change orbit by changing your velocity, and the total velocity change a mission needs is its delta-v budget. Every burn adds to the bill. What makes the budget matter is the rocket equation, which converts delta-v into mass: the more velocity you must add, the larger the share of your starting mass that has to be propellant, and that share grows exponentially rather than in a straight line.

That exponential is why the same mission looks affordable or hopeless depending on the engine. The rocket equation links the delta-v to the engine's specific impulse (a measure of how much velocity change a kilogram of propellant delivers) and to the ratio of the starting mass to the final mass, so it answers questions in both directions.

What a Hohmann transfer assumes

A Hohmann transfer moves a spacecraft between two circular orbits in the same plane with two burns. The first burn, along the direction of motion, stretches the orbit into an ellipse whose lowest point is the starting orbit and whose highest point touches the target orbit. The second burn, at that high point, raises the low point and circularizes. It is the cheapest two-burn transfer between coplanar circular orbits, and it takes half an orbit of the transfer ellipse.

It also assumes that each burn is instantaneous, that only one body's gravity acts, and that the two orbits share a plane. Real burns take time and lose a little efficiency to gravity, other bodies perturb the orbit, and the orbits usually differ in inclination, which the example handles with a plane change.

Steps

  1. 1🌍Hohmann Transfer Orbit
    Pick a central body and the initial and final orbit altitudes for the transfer. The classic case is Earth low orbit to geostationary. This returns the delta-v for each of the two burns and the total. Hit "📋 Copy Values" once you have a result: the total delta-v carries directly into the rocket equation next.
    🧮 Open the Hohmann Transfer Orbit calculator
  2. 2🚀Rocket Equation
    "📥 Paste Values" fills in Delta-V with the total from Step 1. Enter your engine's specific impulse and the vehicle's starting (wet) mass, and leave Final (Dry) Mass at 0 to solve for it. The result is the mass left after the burns, so the starting mass minus that is exactly the propellant the maneuver takes. Run it again with a different specific impulse, or with a larger delta-v, to see how the answer moves.
    🧮 Open the Rocket Equation calculator

Worked example: a 10 tonne spacecraft from a 400 km orbit to geostationary orbit

A spacecraft with a fully fueled mass of 10,000 kg sits in a circular orbit 400 km above the Earth. It must move to a geostationary orbit 35,786 km up, using a chemical engine. The first question is the delta-v, the second is the propellant, and the third is how the answers move when the engine or the geometry changes.

Step 1: The delta-v budget

You enterValue
Central BodyEarth
Initial Orbit Altitude400 km
Final Orbit Altitude35,786 km
The calculator returnsValue
Burn 1 (Depart Initial Orbit)2.399 km/s
Burn 2 (Circularize at Final Orbit)1.457 km/s
Total Delta-V3.857 km/s

The first burn, at the low point, adds 2.399 km/s, and the second, at the high point of the transfer ellipse, adds 1.457 km/s, for 3.857 km/s in all. The transfer coasts for half an orbit of the ellipse, about 5.3 hours.

The spacecraft starts at 7.67 km/s and ends at 3.07 km/s. It has been accelerated twice and finishes slower than it began, because the energy went into altitude, and at the top of the ellipse it is moving at only 1.62 km/s, which is why the second burn is the smaller one.

Step 2: Propellant with a 320 s engine

You enterValue
Specific Impulse, Isp (0 = solve for it)320 sec
Initial (Wet) Mass (0 = solve for it)10,000 kg
Final (Dry) Mass (0 = solve for it)0 kg
Delta-V (0 = solve for it)3.857 km/s
The calculator returnsValue
Specific Impulse (Isp)320 sec
Initial (Wet) Mass10,000 kg
Final (Dry) Mass2,926 kg
Delta-V3.857 km/s

The effective exhaust velocity is Isp × g₀ = 320 × 9.80665 = 3,138 m/s, and the delta-v in the same units is 3,857 m/s, so the mass ratio is exp(3,857 ÷ 3,138) = 3.42. Leaving Final Mass at 0 tells the calculator to solve for it: 2,926 kg.

That means the burns consume 7,074 kg of propellant, 70.7 percent of the starting mass. Everything that is not propellant, meaning the satellite, the engine, the tanks and the structure, must fit in the remaining 2,926 kg, which makes this a demanding stage.

Step 3: The same maneuver with a 450 s engine

You enterValue
Specific Impulse, Isp (0 = solve for it)450 sec
Initial (Wet) Mass (0 = solve for it)10,000 kg
Final (Dry) Mass (0 = solve for it)0 kg
Delta-V (0 = solve for it)3.857 km/s
The calculator returnsValue
Specific Impulse (Isp)450 sec
Initial (Wet) Mass10,000 kg
Final (Dry) Mass4,173 kg
Delta-V3.857 km/s

A hydrogen-and-oxygen engine has a specific impulse around 450 s. With it the mass ratio falls and the final mass rises to 4,173 kg, so the propellant is 5,827 kg, 58.3 percent of the total. That is 1,247 kg more for the satellite and its structure from the engine alone, with the same starting mass and the same maneuver.

Step 4: Adding a 28.5° plane change

You enterValue
Specific Impulse, Isp (0 = solve for it)320 sec
Initial (Wet) Mass (0 = solve for it)10,000 kg
Final (Dry) Mass (0 = solve for it)0 kg
Delta-V (0 = solve for it)4.224 km/s
The calculator returnsValue
Specific Impulse (Isp)320 sec
Initial (Wet) Mass10,000 kg
Final (Dry) Mass2,603 kg
Delta-V4.224 km/s

Both orbits in a Hohmann transfer share a plane. A launch from latitude 28.5° leaves the parking orbit inclined 28.5° to the equator, while geostationary orbit is at 0°. The cheapest place to turn is the slow high point, and the turn is folded into the circularizing burn: Δv = √(va² + vc² − 2·va·vc·cos i), with va = 1.618 km/s the transfer-orbit speed there and vc = 3.075 km/s the circular speed.

That gives 1.825 km/s for the second burn instead of 1.457, an extra 0.37 km/s, and 4.224 km/s in total. The rocket equation returns 2,603 kg at the end, with 7,397 kg, 74.0 percent of the vehicle, going into propellant.

When the two speeds are equal, as when you only turn the plane of a circular orbit and change nothing else, the formula collapses to Δv = 2v·sin(Δi/2). Turning 28.5° on its own at geostationary speed would cost 2 × 3.075 × sin(14.25°) = 1.51 km/s, far more than the 0.37 km/s extra paid by folding the turn into the burn at the slow high point.

What it adds up to

Two changes of similar size matter very differently. The plane change adds about a tenth to the delta-v and costs 3.2 percentage points of the vehicle's mass in extra propellant. Moving from a 320 s to a 450 s engine removes 12.5 points of propellant. The exponential in the rocket equation is the reason: every additional kilometer per second costs more mass than the one before it, and better efficiency is the lever that acts on all of it at once.

Notes & limitations

  • The transfer assumes both orbits share the same plane and that each burn is instantaneous. A real transfer between differently inclined orbits needs a plane change with its own delta-v cost, which the worked example adds by hand (see the Hohmann Transfer article).
  • Leaving a different field at 0 lets the Rocket Equation solve for something else entirely. For instance, leaving Specific Impulse at 0 (with wet mass, dry mass and delta-v all filled in) answers "what engine efficiency would I need" instead.

Where Hohmann numbers stop being true

The budget above is a clean starting figure, and real missions add to it.

  • Finite burns. A real engine burns for minutes, not instantly, and gravity works against it during the burn, so the actual delta-v is somewhat higher.
  • Low-thrust propulsion. Electric engines have very high specific impulse but tiny thrust, so they spiral out over weeks instead of following an ellipse, and the delta-v of a spiral is larger than a Hohmann transfer's.
  • Margins and reserves. Real budgets add delta-v for navigation corrections, phasing and, in low orbit, end-of-life disposal, often a few percent or more.
  • Staging. The final mass includes every tank and engine you carry to the end. Dropping empty tanks along the way, staging, is how vehicles reach the high delta-v values of a launch.
  • Other bodies. Interplanetary transfers need departure and arrival hyperbolas at each planet as well as the transfer ellipse, which this calculator does not include.

What the rocket equation does not know

The equation treats the final mass as one number. It does not know how much of it is structure and how much is payload, which is why mission planners work backward from a structural fraction, the share of a stage that is tanks and engine, and check that the answer is buildable at all.

Frequently asked questions

What does specific impulse mean in plain terms?

It is how efficiently an engine turns propellant into velocity, measured in seconds. It equals the effective exhaust velocity divided by standard gravity, so an engine at 450 s pushes its exhaust out faster than one at 320 s and gets more velocity change from each kilogram.

Why does the second burn cost less than the first?

Because the speed to be gained is smaller at the top. The first burn lifts the spacecraft from 7.67 to 10.07 km/s, a gain of 2.40 km/s. At the high point the transfer orbit is moving at 1.62 km/s while the circular speed there is 3.07 km/s, a shortfall of only 1.46 km/s.

How much delta-v does a plane change cost?

For a circular orbit where only the plane turns and the speed stays the same, Δv = 2v·sin(Δi/2), with v the orbital speed and Δi the change in inclination. At geostationary speed, 3.075 km/s, a 28.5° turn costs 1.51 km/s. Because the cost scales with speed, the turn is cheapest where the spacecraft is slowest, which is why the worked example folds it into the burn at the high point of the transfer, where the extra is only 0.37 km/s. When the speeds before and after differ, use the full law-of-cosines form in the example.

Can I use this for a trip to Mars?

Only partly. Choosing the Sun as the central body gives the two heliocentric burns of an interplanetary Hohmann transfer, but leaving and reaching a planet takes additional delta-v to escape or be captured, which is not included. Treat the number as the transfer's share of the total, not the whole of it.

Why does a 10 tonne spacecraft end up at under 3 tonnes?

Because most of its mass was propellant, 10,000 kg to start and 2,926 kg to finish. At a mass ratio of about 3.4, roughly 71 percent of the vehicle is fuel, and the rest has to hold everything else.

How long does the transfer take?

Half an orbit of the transfer ellipse, about 5.3 hours for this LEO-to-geostationary case. The rocket equation ignores time, so timing questions such as launch windows and phasing are separate problems.

Further reading

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Educational tool — not a substitute for a licensed engineer or the official code text.