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Orbital mechanics · Lesson 4

Changing an orbit

A short engine burn changes velocity at the spacecraft's current location. The new velocity determines the path followed afterward; the burn does not instantly move the spacecraft to another altitude.

We will transfer from a circular Earth orbit of radius 7,000 km to one of radius 14,000 km. Both circles lie in the same plane and run in the same direction.

Separate the impulse from the coast

Δv, pronounced delta-v, is the change in the velocity vector. In an impulsive model, a burn takes negligible time compared with the orbital period. Position stays fixed while velocity changes. This approximation works best for burns short enough that the craft does not move far during thrust.

A prograde burn points along the current velocity; a retrograde burn points against it. From a circular orbit, a modest prograde impulse makes the burn point the periapsis of a larger ellipse. The opposite side rises. A modest retrograde impulse makes the burn point apoapsis and lowers the opposite side.

These rules have conditions. They assume tangential burns from a circle that do not reverse velocity or produce escape. A burn in a different direction, or on an existing ellipse, can change the orbit differently. To predict the result, use position and the complete new velocity vector.

Update the velocity at the point where the burn occurs

An impulse changes the spacecraft's state in a specific way. Its position immediately before and after the ideal burn is the same. Its velocity vector changes by the commanded delta-v. Starting from that updated state, gravity determines the next coast. A diagram that instantly moves the spacecraft from the lower circle to the higher circle would skip the transfer entirely and misrepresent what the engine does.

For a tangential prograde burn, the old velocity and the added velocity point in the same direction. Their magnitudes can therefore be added. For a tangential retrograde burn that does not reverse the craft, subtract the magnitude of the impulse. For an impulse at another angle, resolve the velocity change into components and add those components. Adding scalar speeds would lose the direction information that determines the new path.

At the instant of the burn, radius is unchanged, so gravitational potential energy per unit mass is unchanged. A speed increase raises kinetic energy and therefore total orbital energy. On a bound ellipse, less negative specific energy means a larger semi-major axis. This connects a local engine action to the size of the subsequent orbit without suggesting that the engine physically pushes the distant side upward.

Why does a modest tangential prograde burn from a circle make this point periapsis? Radial velocity is initially zero, but the new transverse speed exceeds the circular value for the existing radius. The path cannot remain that same circle under the available gravity. It begins moving outward, so the burn point is the nearest radius on the new bound ellipse. A tangential speed reduction instead starts inward motion and makes the original point apoapsis.

This reasoning has limits that should be tested before applying the rule. An outward radial impulse gives nonzero radial speed, so the burn point is not an apsis. A large prograde impulse can make the trajectory unbound, leaving no farthest apoapsis to raise. A retrograde impulse large enough to reverse motion needs a separate vector analysis. The familiar rule is a shortcut for the restricted case, not a replacement for the state update.

Take the exercise's speed reduction to 7.0 km/s at radius 7,000 km. The new kinetic term is 24.5 MJ/kg, while the potential term is about negative 56.943 MJ/kg. Their sum is about negative 32.443 MJ/kg. Solving the semi-major-axis energy relation gives a near 6,143 km. Because the burn point is apoapsis, the other extreme radius is twice a minus 7,000 km, about 5,286 km.

That last result is not a lower circular orbit. It is the predicted near end of a mathematical ellipse, below Earth's mean radius of 6,371 km. The craft would encounter the atmosphere and ground before tracing the complete ideal curve. The calculation therefore diagnoses an unsafe geometric intersection within the teaching model. It is not a maneuver recommendation or an atmospheric-entry calculation.

A finite-duration burn changes position while thrust is acting. Its velocity change accumulates along a moving trajectory, so the instantaneous approximation becomes less accurate as burn duration becomes significant relative to orbital time. An actual low-thrust transfer can require many revolutions. The two-burn model isolates the state changes and coasting geometry; it does not claim that every propulsion system can execute them instantaneously.

Make an ellipse tangent to both circles

A Hohmann transfer connects the two coplanar circles with half an ellipse. Its periapsis is at r₁ and its apoapsis at r₂, so aₜ = (r₁ + r₂)/2. Burn once at the inner circle to enter the ellipse, coast to its apoapsis, then burn again to match the outer circle.

Two burns, with a coast between them

Target circular radius14,000 km

Before burn 1: circular parking orbit.

Speed Step 0 / 10

Calculated maneuver, independent of playback
First prograde burn	Second prograde burn	Coast time
1.167 km/s	0.979 km/s	89.2 minutes

Total Δv = 2.147 km/s. The final circular speed is 5.336 km/s, less than the initial 7.546 km/s despite two prograde burns.

Initial radius is 7,000 km, measured from Earth's center. All three paths share one distance scale. The first burn changes velocity at the right-hand point; the second occurs at the left-hand point. The coast follows Kepler's equation in physical time, compressed into eight teaching steps. Changing the target resets playback. Earth is a symbolic marker.

For r₁ = 7,000 km and r₂ = 14,000 km, aₜ = 10,500 km. Initial circular speed is 7.546 km/s. Vis-viva gives transfer departure speed 8.713 km/s, requiring a first prograde burn of 1.167 km/s.

During the outward coast, speed falls to 4.357 km/s at the far end. The target circle needs 5.336 km/s, so the second burn adds 0.979 km/s. Without it, the craft simply follows the ellipse back inward.

1. Enter the transfer ellipse

{Δv}_{1} = \sqrt{μ (\frac{2}{r_{1}} - \frac{1}{a_{t}})} - \sqrt{\frac{μ}{r_{1}}}

2. Match the outer circular speed

{Δv}_{2} = \sqrt{\frac{μ}{r_{2}}} - \sqrt{μ (\frac{2}{r_{2}} - \frac{1}{a_{t}})}

Time between the burns

t_{coast} = π \sqrt{\frac{{a_{t}}^{3}}{μ}}

aₜ is the transfer ellipse's semi-major axis. The coast covers half its period.

The coast takes about 89.2 minutes. Total Δv is 1.167 + 0.979 = 2.147 km/s using unrounded values. The final circular speed is lower than the initial circular speed even though both burns increase speed locally. The outward coast accounts for the larger intervening decrease.

Read the transfer as two state changes and one coast

Select Nearby orbit, then Double radius. The latter restores the 14,000 km target and resets the demonstration. The three paths remain visible on the same distance scale: the original circle, the transfer ellipse, and the target circle. Sharing a scale lets you check that the transfer touches each circle at the stated radius. The central Earth marker is symbolic rather than a scaled surface boundary.

At Reset, the spacecraft is on the right-hand departure point before burn one. The first Next step applies the conceptual departure burn without moving the point to a different radius. Eight subsequent teaching intervals cover the outward coast. At step nine, the spacecraft reaches the target radius but has not circularized. Step ten represents the second burn. Its position stays at the meeting point while its velocity changes to match the outer circle.

The readout table describes the complete calculated maneuver at every playback position. It is not a running total of fuel already used. You can inspect both burns and the coast time before pressing Play or Next step. This separation makes it possible to compare scenarios without having to remember values that appeared at different moments in an animation. The animation depicts sequence; the table supplies the numerical budget.

Derive the transfer ellipse from its endpoints. Adding 7,000 and 14,000 km and dividing by two gives 10,500 km. The difference divided by the sum gives eccentricity one third. At departure, vis-viva uses current radius 7,000 km and semi-major axis 10,500 km. The calculated transfer speed is about 8.713 km/s. Subtract the original circular speed, about 7.546 km/s, because the burn supplies only the difference.

During the coast, keep the same semi-major axis and change the current radius. At 14,000 km the transfer speed is about 4.357 km/s. The target circle requires about 5.336 km/s. Subtract transfer speed from target speed for the second burn, giving about 0.979 km/s. Reversing this subtraction would incorrectly suggest a retrograde burn. The transfer arrives too slowly to stay on the outer circle, so it needs a local speed increase.

The flight time comes from half the period of the transfer ellipse, not half the period of either circle. Use 10,500 km in the period formula, then divide by two. The result is approximately 5,354 seconds, or 89.2 minutes. The spacecraft moves through half an ellipse, but does not cover equal arc lengths in equal times. The coast animation uses equal time progression rather than equal geometric angles.

Now select Triple radius. The target moves farther away, the transfer ellipse grows, and the computed coast takes longer. The same sequence of two tangential prograde burns still applies within this widget's range. Compare the separate burn values instead of guessing that each scales directly with radius. The circular and elliptical speed formulas contain square roots and reciprocal radii, so the changes are not linear.

Reaching the outer circle without the second burn is a useful failure case. The spacecraft has the right radius at apoapsis but the wrong velocity for the circle. With no new impulse it continues on the same ellipse and returns inward. The visual contact between paths is therefore necessary for an instantaneous transfer, but does not itself complete one. The shared position is where a velocity change can connect the two trajectories.

Δv is not fuel mass

A maneuver budget adds the magnitudes of separate velocity changes. Do not subtract one burn from another because they point in different directions at different places. Each burn costs propellant.

Converting Δv to propellant requires engine exhaust speed and spacecraft mass. For an ideal rocket:

Ideal rocket equation, optional

Δv = v_{e} \ln (\frac{m_{0}}{m_{f}})

Here vₑ is effective exhaust speed, m₀ is mass before a burn, mf is mass after, and ln is the natural logarithm. You do not need this formula for the exercises. It explains why equal Δv budgets need not mean equal fuel masses for different vehicles.

A Hohmann transfer is a useful low-Δv option between circular coplanar orbits, but it is not universally best. Faster transfers can cost more; for sufficiently large radius ratios, a three-burn bi-elliptic transfer can use less Δv than a two-burn Hohmann transfer. Plane changes and rendezvous timing add other constraints.

Real engines have finite thrust. Very low-thrust propulsion can gradually change an orbit over many revolutions instead of tracing this two-impulse ellipse. Real missions must also avoid atmospheric entry and collisions. Our diagrams are not a numerical prediction of a complete mission.

Separate velocity cost, energy change, and propellant

The maneuver budget adds the magnitudes of the two impulses because each requires propulsion. The local prograde directions at the two opposite sides of the orbit point in opposite directions in the Earth-centered drawing. That does not make the propellant expenditures cancel. Nor does the vector sum of engine impulses alone describe the entire velocity change over the journey, because gravity also changes velocity throughout the coast.

Keep four speeds in chronological order when checking the result. The initial circle is slower than the transfer at departure. The coast then reduces speed substantially. The second burn increases speed from the transfer-arrival value to the outer circular value. The final speed can be lower than the initial speed despite two local increases. Omitting the coast from the account produces the apparent paradox.

Energy tells the same story using different quantities. The initial circle has specific energy about negative 28.471 MJ/kg, the transfer ellipse about negative 18.981 MJ/kg, and the outer circle about negative 14.236 MJ/kg. Both burns raise the energy; the coast keeps the middle value constant. These energy increments are not numerically equal to delta-v. They have different units and depend on the speed at which a burn occurs.

For a tangential speed increment, expand the difference between the new and old kinetic terms. The energy gain per unit mass equals old speed times the speed increment, plus half the square of the increment. At fixed delta-v, a larger starting speed therefore produces a larger energy change. This is one reason maneuver location matters. It does not mean that an arbitrarily chosen high-speed burn will meet a mission's geometry and timing constraints.

The optional rocket equation connects a delta-v budget to a mass ratio under ideal propulsion assumptions. Suppose exhaust speed is 3 km/s and initial mass is twice final mass. The natural logarithm of two is approximately 0.693, so the ideal delta-v is about 2.079 km/s. A 1,000 kg vehicle ending at 500 kg and a 2,000 kg vehicle ending at 1,000 kg have the same ratio and ideal delta-v, but consume different propellant masses.

For consecutive burns using the same effective exhaust speed, mass ratios multiply. The logarithm of their product equals the sum of their logarithms, which is consistent with adding impulse magnitudes for the ideal budget. Real missions also carry reserves and account for changes in vehicle configuration, engine performance, and losses. The teaching table does not include those factors, so it should not be interpreted as a complete fuel loading calculation.

Compare transfer options against an explicit objective. A low-delta-v transfer may take longer than a mission allows. A faster route can require larger burns or a more difficult arrival. A plane change, collision constraint, or moving target can alter the feasible choices. The Hohmann construction is valuable because it has clear assumptions and a calculable baseline, not because every real mission should follow it.

Finally, preserve precision until the reporting step. Adding displayed values of 1.167 and 0.979 gives 2.146, while the model's unrounded sum rounds to 2.147 km/s. That difference is not missing propulsion or a violation of the budget. Each displayed input has already been rounded independently. Use unrounded intermediate quantities for calculations and explain rounding when a reader compares the visible numbers by hand.

Exercises

Account for two prograde burns

How can two prograde burns take a spacecraft from a 7.546 km/s circle to a 5.336 km/s circle?

Hint

Track four speeds: before and after each burn. Include the coast between them.

Worked answer

Speed rises at each burn, then falls during the coast

Initial circle: 7.55 km/s

After burn 1: 8.71 km/s

Before burn 2: 4.36 km/s

Final circle: 5.34 km/s

The four speeds share one scale. The large decrease happens during the outward coast, not during either prograde burn.

Speed goes from 7.546 to 8.713 km/s at burn 1, falls to 4.357 km/s during the coast, and rises to 5.336 km/s at burn 2. Both impulses are prograde. The final orbit is higher and has a lower circular speed, but reaching it required increased orbital energy. A claim that the engine directly slowed the craft into the higher circle confuses a local burn with the whole journey.

Omit the second burn

A controller stops after the first Hohmann burn because the trajectory already touches the desired altitude. What happens at the far end?

Hint

Compare the transfer speed at 14,000 km with the circular speed there.

Worked answer

At the target radius, two speeds are different

Transfer arrival: 4.36 km/s

Required circular speed: 5.34 km/s

Without the second burn, the craft remains on the transfer ellipse. Touching the target circle at one point does not put it on that circle.

The craft reaches the radius of the target circle but has only 4.357 km/s instead of 5.336 km/s. With no second impulse it stays on the transfer ellipse and falls inward again. To circularize it needs about 0.979 km/s of additional prograde velocity. Sharing one position with a circular path is not enough to share the orbit.

Brake from a circular orbit

A craft in a 7,000 km circular Earth orbit reduces its tangential speed to 7.0 km/s. Does it immediately fall to a lower circle? Describe the new ideal path and its physical problem.

Hint

Position cannot jump during an impulse. Ask whether the new tangential speed can support a circle at the unchanged radius.

Worked answer

The lowered ellipse intersects Earth

Unlike the symbolic planet markers elsewhere, Earth's radius and both paths here share a distance scale. The burn occurs at the left-hand apoapsis, 7,000 km from the center. The ideal ellipse enters Earth; atmosphere would affect it before the marked surface crossing. The hidden part inside Earth cannot be flown.

The burn point remains at r = 7,000 km but becomes apoapsis because the new tangential speed is below circular speed. Energy becomes more negative, the semi-major axis decreases, and periapsis moves inward. At a chosen post-burn speed of 7.0 km/s, a is about 6,143 km and periapsis about 5,286 km, below Earth's surface. The mathematical ellipse therefore intersects the planet; the craft would encounter the atmosphere and ground rather than complete it.

Sources and model limits

The calculations use Newtonian two-body motion and ideal instantaneous burns. They omit atmospheric drag, an uneven gravity field, other attracting bodies, and finite engine thrust unless the text explicitly discusses them. They are teaching calculations, not flight instructions.

NASA, Basics of Space Flight: Trajectories. Transfer trajectories, timing, and the reference-frame distinction in gravity assists.
Bryan Weber, Orbital Mechanics & Astrodynamics: Hohmann transfer. Two impulsive burns between circular coplanar orbits; limits of the transfer model.

Earth examples use μ = 398,600.4418 km³/s² and a rounded mean radius of 6,371 km. Reported answers are rounded, not measurements. The Sun-centered example uses astronomical units and years with a one-year circular Earth orbit.

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