A vehicle is in one orbit and needs to be in another. A communications satellite is released in a low parking orbit and belongs in geostationary orbit, 42,164 km from the centre of the Earth. A lunar stack sits at 200 km and has to reach the Moon. The question is the same each time: which path between the two costs the least propellant, and where along it should the engine burn?

It matters because the answer is most of the budget. Getting to low orbit is the expensive part of spaceflight, but everything after it — the trans-lunar injection, the capture at the far end, the burns that raise a satellite to its station — is paid for by an upper stage whose every kilogram of propellant was itself lifted from the ground. A transfer that is a few hundred metres per second cheaper is a payload that is tonnes heavier.

The last lesson ended with vis-viva, $v^{2} = μ (2/ r - 1/ a)$ : the speed on any orbit from the distance $r$ , the semi-major axis $a$ and the body's gravitational parameter $μ$ . With it, every burn becomes a change of $a$ at a known $r$ , and this lesson is mostly that equation, applied carefully.

To go up, speed up

Start with a paradox. A higher circular orbit is slower: circular speed is $μ / r$ , so it falls as the orbit grows. Yet the way to reach a higher orbit is to burn prograde, twice, and speed up both times.

The resolution is the coast in between. A prograde burn raises the far side of the orbit. The vehicle then climbs towards it, and climbing trades kinetic energy for potential: it arrives at the top slower than it left the bottom, slower even than the circular speed up there. A second prograde burn at the top makes up the difference and raises the near side to match. Both burns add speed; the coast takes more away than they added; and the rocket equation charges for the burns, not for the speed you end up with.

The Hohmann transfer

Take two circular orbits in the same plane, radii $r_{1}$ inside and $r_{2}$ outside. Treat the burns as impulsive — instantaneous changes of velocity. That is a good approximation whenever a burn is short compared with the orbital period, and it is the assumption the whole of this section rests on.

The transfer orbit is the ellipse that just touches both circles: periapsis on the inner one, apoapsis on the outer. Its semi-major axis is half the sum of its apsides:

a_{t} = \frac{r _{1} + r _{2}}{2}

At the start, vis-viva gives the speed the vehicle needs at periapsis of that ellipse, and the first burn is the difference from circular speed there:

v_{p} = μ (\frac{2}{r _{1}} - \frac{1}{a _{t}}) = \frac{μ}{r _{1}} \frac{2 r _{2}}{r _{1} + r _{2}}, Δ v_{1} = \frac{μ}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1)

At apoapsis the same equation, with $r = r_{2}$ , gives the arrival speed $v_{a}$ . (So does angular momentum: at both apsides the velocity is perpendicular to the radius, so $r_{1} v_{p} = r_{2} v_{a}$ .) The second burn raises it to circular speed at $r_{2}$ :

v_{a} = \frac{μ}{r _{2}} \frac{2 r _{1}}{r _{1} + r _{2}}, Δ v_{2} = \frac{μ}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}})

The coast between them is half the transfer ellipse, so by Kepler's third law it takes

t = π \frac{a _{t}^{3}}{μ}

Why this ellipse and not some other? Two reasons, one about direction and one about place. A small burn $Δ v$ changes the specific energy $ε = v^{2} /2 - μ / r$ by $v \cdot Δ v$ , because the position does not move during an impulse. For a given size of burn that dot product is largest when the burn is along the velocity, so any other direction spends propellant on turning rather than on energy. And at an apsis the velocity is perpendicular to the radius, so a burn there changes the far side of the orbit and leaves the near side alone — exactly what is needed to touch the other circle and nothing more. Among transfers with two burns between coplanar circles, the Hohmann transfer is the cheapest. With three, as the next section shows, it sometimes is not.

Worked example: the landing page

The diagram on the landing page flies a Hohmann transfer from $r_{1} = 15, 178$ km to $r_{2} = 42, 164$ km, with $μ = 398, 600.4418$ km³/s². Circular speed on the inner orbit is $μ / r_{1} = 5.125$ km/s. The transfer ellipse has $a_{t} = 28, 671$ km and eccentricity $(r_{2} - r_{1}) / (r_{2} + r_{1}) = 0.471$ , and needs 6.215 km/s at periapsis, so

$Δ v_{1} = 6.215 - 5.125 = 1.090$ km/s, prograde, on the inner orbit.
At apoapsis the vehicle arrives at 2.237 km/s. Circular speed at $r_{2}$ is 3.075 km/s, so $Δ v_{2} = 0.838$ km/s, prograde again.
Total: 1.927 km/s. The coast takes $π a_{t}^{3} / μ = 24, 157$ s, 6.71 hours.

These are the numbers the landing page's readout shows as the burns fire.

Worked example: low orbit to geostationary

From a 200 km parking orbit ( $r_{1} = 6, 578$ km, 7.784 km/s) to geostationary radius, the transfer ellipse is the geostationary transfer orbit, GTO: $a_{t} = 24, 371$ km, $e = 0.730$ . The first burn takes the vehicle from 7.784 to 10.239 km/s, 2.455 km/s. It reaches apogee five and a quarter hours later at 1.597 km/s, and the second burn takes it to 3.075 km/s, 1.477 km/s. Total, 3.932 km/s.

That total is for an equatorial launch. From Cape Canaveral, at 28.5° north, the transfer orbit is inclined 28.5° to the equator, and a geostationary satellite must end up in the equatorial plane. Rotating a velocity $v$ through an angle $Δ i$ costs $2 v sin (Δ i /2)$ on its own: 3.83 km/s in the parking orbit, 1.51 km/s once circular at geostationary radius. But the circularisation and the plane change can be one burn at apogee, where the vehicle is slowest. The single burn has to turn a 1.597 km/s velocity into a 3.075 km/s velocity 28.5° away, and the law of cosines gives its size:

Δ v_{2} = v_{a}^{2} + v_{GEO}^{2} - 2 v_{a} v_{GEO} cos Δ i = 1.837 km/s

— against 2.99 km/s for the same two changes made one after the other. Combined burns are cheaper whenever two changes are made at the same place, because the two corrections partly share one direction. Hold on to where this burn is made: the plane change is placed where the vehicle is slow, which is the opposite of the rule this lesson is about to arrive at.

When one ellipse is not the cheapest

A bi-elliptic transfer uses three burns and two half-ellipses. The first burn sends the vehicle out on an ellipse whose apoapsis $r_{b}$ lies beyond the target. At $r_{b}$ a second, small prograde burn raises the periapsis to $r_{2}$ . The vehicle falls back inwards and, at $r_{2}$ , a third burn — retrograde this time — rounds the orbit off. Each burn comes straight from vis-viva: the first is the speed at periapsis of an $(r_{1}, r_{b})$ ellipse less circular speed at $r_{1}$ ; the second is the difference between the speeds at apoapsis of the $(r_{2}, r_{b})$ ellipse and of the $(r_{1}, r_{b})$ one; the third is the speed at periapsis of the $(r_{2}, r_{b})$ ellipse less circular speed at $r_{2}$ .

It looks wasteful: it overshoots and comes back. Yet for large enough ratios $r_{2} / r_{1}$ it is cheaper. The reason is the subject of the second half of this lesson. The bi-elliptic makes more of its energy change in the first burn, deep in the well where the vehicle is fast and each metre per second buys the most energy. The second burn, which fixes the angular momentum, is made far out where the vehicle is barely moving and a small push changes the far periapsis a great deal. The price is the third burn and a very long coast.

The crossovers are fixed numbers. Below $r_{2} / r_{1} = 11.94$ no bi-elliptic transfer beats the Hohmann transfer, however far out $r_{b}$ is. Above $15.58$ every bi-elliptic transfer does, for any $r_{b}$ beyond $r_{2}$ . In between it depends on how far out the middle burn is made. And the limit $r_{b} \to \infty$ , the bi-parabolic transfer, costs exactly $(2 - 1) (v_{1} + v_{2})$ — escape, turn round at infinity for nothing, fall back and circularise — which is the least any of them can do. Curiously, 15.58 is also the ratio at which the Hohmann transfer itself is most expensive in units of the starting circular speed: 0.536 $v_{1}$ .

Take a 400 km start ( $r_{1} = 6, 778$ km, $v_{1} = 7.669$ km/s) and a target twenty times further out, $r_{2} = 135, 563$ km. The Hohmann transfer costs 4.101 km/s and takes 1.1 days. A bi-elliptic transfer with its far apoapsis at twice the target radius, 271,125 km, costs 4.031 km/s and takes 8.3 days: 70 m/s saved for a week of extra coasting. The bi-parabolic limit is 3.887 km/s and takes forever.

Figure · one ellipse or two

TARGET RADIUS r₂ / r₁20.0

FAR APOAPSIS r_b / r₂2.00

TARGET r₂: 135,563 km
HOHMANN Δv: 4.101 km/s
HOHMANN TIME: 1.1 d
BI-ELLIPTIC Δv: 4.031 km/s
BI-ELLIPTIC TIME: 8.3 d
BI-ELLIPTIC SAVES: 70 m/s

Hohmann: 2,915 + 1,186 m/s. Bi-elliptic: 3,043 out to 271,125 km, 722 there, then 265 retrograde at the target.

Start orbit 400 km up (r₁ = 6,778 km, 7.669 km/s). The curve is what a bi-elliptic transfer saves over a Hohmann transfer to each target radius, with its far apoapsis at the chosen multiple of the target. The grey curve is the bi-parabolic limit. Nothing beats the Hohmann transfer below r₂/r₁ = 11.94; above 15.58 any bi-elliptic does.

In practice the bi-elliptic transfer is rarely flown in its pure form: the ratios where it pays are beyond most destinations, and the time is long. Its idea is flown all the time, though. Many launches to geostationary orbit use a supersynchronous transfer orbit, whose apogee is well above geostationary altitude, because the plane change made up there is cheaper still — the same trade of a longer coast for a burn made where the vehicle is slow.

The Oberth effect

Here is the rule the bi-elliptic transfer was exploiting, stated plainly: the same burn, from the same engine, adds more orbital energy the faster the vehicle is moving when it makes it. In a gravity well that means the deeper it is, because on any orbit the vehicle is fastest at periapsis.

The energy argument

Burn prograde, impulsively, by $Δ v$ at a point where the speed is $v$ . The position does not change during the impulse, so the potential energy does not either, and the whole change in specific energy is kinetic:

Δ ε = \frac{1}{2} (v + Δ v)^{2} - \frac{1}{2} v^{2} = v Δ v + \frac{1}{2} Δ v^{2}

The cost of the burn, in propellant, is set by the rocket equation from $Δ v$ alone. It does not depend on $v$ . The energy it buys grows in proportion to $v$ . Energy is what matters to the orbit: it fixes the semi-major axis, and on an escape trajectory it fixes the speed left over far away, $v_{\infty} = 2 ε$ . So a burn at 10 km/s is worth nearly twice a burn at 5 km/s, for the same propellant.

Where the energy comes from

That looks like something for nothing, and it is worth checking the bookkeeping. Take one short moment of the burn, in the frame of the planet. The vehicle, of mass $m$ , is moving at $v$ and expels a small mass $d m$ of propellant backwards at the exhaust speed $v_{e}$ relative to itself. Momentum is conserved, so the vehicle gains $d v$ with

m d v = v_{e} d m

Now count kinetic energy. The vehicle gains $\frac{1}{2} m (v + d v)^{2} - \frac{1}{2} m v^{2} \approx m v d v = v v_{e} d m$ . The propellant was moving at $v$ with the vehicle and leaves at $v - v_{e}$ , so its kinetic energy changes by $\frac{1}{2} d m [(v - v_{e})^{2} - v^{2}] = \frac{1}{2} d m v_{e}^{2} - v v_{e} d m$ . The two together:

v v_{e} d m + (\frac{1}{2} d m v_{e}^{2} - v v_{e} d m) = \frac{1}{2} d m v_{e}^{2}

which is exactly the energy the combustion released, since in the vehicle's own frame it threw $d m$ backwards at $v_{e}$ . The books balance. But look at how they balance. The vehicle's share, $v v_{e} d m$ , exceeds the chemical energy whenever $v > v_{e} /2$ , and the excess comes out of the propellant's own kinetic energy. The propellant was carrying kinetic energy because it had been riding along at $v$ ; burning it at speed hands most of that over to the vehicle and leaves the exhaust nearly stationary, or even still moving forward.

Put numbers on it. An upper stage in a 200 km orbit moves at 7.784 km/s. A Merlin Vacuum, with a vacuum specific impulse of 348 s, has $v_{e} = 3.413$ km/s. Its exhaust leaves the engine backwards and is still travelling forward at 4.371 km/s in the Earth's frame. Each kilogram of it has given up 20.7 MJ of kinetic energy on top of the 5.8 MJ its combustion released, and the vehicle receives 26.6 MJ per kilogram burned — four and a half times what the chemistry alone supplied. Nothing is free: that kinetic energy was put into the propellant on the way up, by burning other propellant. The Oberth effect is the rule for where it is best spent, and the answer is wherever it is largest.

Leaving a planet

The effect is largest on escape. A hyperbolic trajectory with excess speed $v_{\infty}$ has energy $v_{\infty}^{2} /2$ , so at periapsis radius $r_{p}$ vis-viva gives

v_{p}^{2} = v_{\infty}^{2} + \frac{2 μ}{r _{p}} = v_{\infty}^{2} + v_{esc}^{2}

The escape speed adds in quadrature. Differentiate at fixed $r_{p}$ and $v_{\infty} d v_{\infty} = v_{p} d v_{p}$ : each metre per second added at periapsis comes out as $v_{p} / v_{\infty}$ metres per second at infinity. That ratio is the gearing, and deep in the well it is large.

The late-2026 opportunity for Mars, as the simulator's window search finds it, needs a departure $C_{3}$ — the energy measure mission planners use, $v_{\infty}^{2}$ — of 9.26 km²/s², so $v_{\infty} = 3.043$ km/s. From a 200 km parking orbit, where escape speed is 11.009 km/s, the periapsis speed has to be $3.04 3^{2} + 11.00 9^{2} = 11.421$ km/s. The injection burn is the difference from circular speed, $11.421 - 7.784 = 3.637$ km/s, and at that point each extra metre per second becomes 3.75 m/s of excess speed. The alternative — spend just enough to escape, 3.224 km/s, coast out to where the Earth's pull no longer matters, and then burn the 3.043 km/s there — costs 6.267 km/s. Burning deep saves 2.63 km/s, and that is why every interplanetary departure is a single burn from a low parking orbit.

Figure · the same burn, somewhere else

BURN AT

POSITION ON THE ORBIT (TRUE ANOMALY)0 °

BURN SIZE Δv0.50 km/s

BURN: 0.50 km/s
SPEED AT THE BURN: 6.215 km/s
ENERGY GAINED: 3.23 MJ/kg
NEW APOAPSIS r: 92,002 km
VERSUS THE SAME BURN AT APOAPSIS: ×2.60

The landing page's transfer ellipse, drawn to scale: 15,178 × 42,164 km from the centre of the Earth. The burn is prograde and impulsive. Its energy is v·Δv + Δv²/2: the lower panel is that gain around the whole orbit, and the dashed line is the 6.95 MJ/kg this orbit needs to escape.

Set the burn to 0.5 km/s and move it round the orbit: the energy it buys ranges from 3.23 MJ/kg at periapsis to 1.24 MJ/kg at apoapsis, a factor of 2.6 for the same propellant. At periapsis it throws the apoapsis out to 92,000 km; at apoapsis it raises the periapsis only to 27,700 km. Then increase the burn. A little over 1 km/s fired at periapsis is enough to leave the Earth altogether — escape from periapsis needs 1.033 km/s — while at apoapsis the same burn leaves the vehicle bound, and escape from there would take 2.111 km/s.

Arriving

The same arithmetic runs backwards on arrival. A vehicle falling towards a planet on a hyperbola is fastest at periapsis, and a braking burn there removes the most energy for its size. Capture into orbit round the Moon or Mars is always a periapsis burn, and the approach is steered so that periapsis is as low as is safe.

When the rule breaks

The Oberth effect is a statement about energy. It says nothing about the other elements, and it assumes an impulse. Where either of those fails, so does the rule.

Not every change is a change of energy. Raising a periapsis is done at apoapsis, and circularising a transfer orbit is done at the top, not the bottom. That is not a failure of the rule; it is a different job. Burning at periapsis would move the wrong side of the orbit.
Plane changes reverse it. Turning the velocity through an angle costs $2 v sin (Δ i /2)$ , which is proportional to the speed. The cheapest place to change a plane is where the vehicle is slowest — which is why GTO's plane change is made at apogee, and why a supersynchronous transfer makes it higher still.
Real burns take time. The effect is strongest at a point, and a real burn is spread along an arc. Part of it happens before periapsis and part after, where the vehicle is slower, and the thrust is pointed along a velocity that is turning under it. A trans-lunar injection lasting a few minutes out of a 90-minute orbit loses little. An ion engine spiralling out over months loses most of it. The usual practice is to centre the burn on the point it is meant for, starting it half its length early, or to split one long burn into several shorter ones on successive periapsis passes.
Periapsis has a floor. The deepest point available is limited by the atmosphere and, in the end, by the surface. A burn cannot be made lower than the vehicle can safely fly.
Far from any body there is nothing to exploit. Out on the asymptote of a hyperbola, $v$ is just $v_{\infty}$ and the gearing has fallen to one. A correction made there is worth what it costs and no more.

In Vivapse

The transfer on the landing page is components/hero-transfer.tsx: two radii, vis-viva for the four speeds, $Δ v_{1} = v_{p} - v_{1}$ and $Δ v_{2} = v_{2} - v_{a}$ , and Kepler's equation for the coast between. The simulator itself does the same arithmetic wherever it plans a departure. hyperbolicInjection(rp, mu, vInf) in src/sim/lambert.ts is the escape equation above in two lines — the periapsis speed $v_{\infty}^{2} + 2 μ / r_{p}$ and the burn as that minus circular speed — and hyperbolicCapture is its mirror for arrival. For Mars, the launch-window search in src/sim/windows.ts uses it to turn a heliocentric Lambert solution's $v_{\infty}$ into the injection $Δ v$ that a program reads from fc.plan.dvInjection, alongside fc.plan.c3 and the asymptote fc.plan.vInfDep; for the Moon, whose transfer is solved round the Earth, the Lambert solution gives the post-injection velocity at the parking orbit directly. Missions and destinations describes the windows and the injection in full.

The example programs fly these ideas. The Mars flyby program matches its injection to the planned $v_{\infty}$ as a vector, from the parking orbit, in one burn. The Moon landing program sizes its capture from the speed the vehicle will have at perilune — energy gives $v^{2} + 2 μ (1/ r_{p} - 1/ r)$ — rather than from the speed at the edge of the Moon's sphere of influence, which it notes would under-read the burn by a kilometre a second, and it starts the burn half of itself early so that it straddles perilune. For orbit raising round the Earth the built-in guide is the Hohmann recipe: burn to raise the apoapsis, coast to it on fc.orbit.timeToApoapsis, and circularise prograde.

Try it

Set up a Mars flyby — destination Mars, arrival flyby, in the Mission panel — and fly it with the Odyssey stack and the Mars flyby example program. Once the stack is in its parking orbit, the console announces the injection it has planned. In a calm flight from Cape Canaveral in the late-2026 window it reads about injection in 34.3 min · 3634 m/s · v∞ 3.044 km/s, from a 186 × 216 km parking orbit — the 3.637 km/s derived above, give or take the parking orbit's own eccentricity. After the burn it reports what it achieved: 3,630 m/s spent for a $v_{\infty}$ of 3.027 km/s.

Then do the sum the other way. Escaping first and burning 3.03 km/s far from the Earth would have cost about 6.3 km/s; the stack does not carry it. The launch window, the parking orbit and the single deep burn are what make the mission possible at all.

What carries forward

Everything here assumed the start and target were circles in one plane and that the timing would take care of itself. Real transfers have to arrive where the target will be, not where it was, and the question becomes: given two positions and a time of flight, which conic joins them? That is Lambert's problem, and it is how the simulator finds the windows used above. Before that, the next two lessons stay in orbit: latitude, azimuth and the orbits you can reach, which is where the parking orbit's plane comes from, and spheres of influence and patched conics, which stitches the departure hyperbola above to the transfer beyond it.