Appearance
11.5 — Changing an Orbit
Delta-v is the currency of spaceflight. Not distance, not time, not energy — the total change in velocity a mission requires, in metres per second.
A trip to Mars and a trip to Mercury take similar amounts of delta-v, despite Mercury being much closer, because what matters is not how far you go but how much your velocity must change.
This chapter derives the standard manoeuvres, costs them, and works out a complete mission budget.
Why delta-v is the right currency
Chapter 11.7 derives the rocket equation, and the reason delta-v is what matters is contained in it:
\Delta v = v_e\ln\frac{m_0}{m_f}
The fuel required depends exponentially on the total delta-v and on nothing else about the trajectory. So mission planning is delta-v budgeting, and every manoeuvre is priced in m/s.
Delta-v adds arithmetically across a mission. Fuel does not — it compounds.
The Hohmann transfer
Walter Hohmann worked this out in 1925, in a book he published at his own expense while working as a municipal building surveyor in Essen.
The idea: connect two circular orbits with an ellipse tangent to both. Periapsis on the inner orbit, apoapsis on the outer.
Two burns:
Burn 1 at the inner orbit raises apoapsis to the target radius.
Burn 2 at apoapsis raises periapsis, circularising.
Deriving the cost
The transfer ellipse has:
a_t = \frac{r_1+r_2}{2}
From the vis-viva equation (Chapter 11.4):
Speed in the initial circular orbit:
v_1 = \sqrt{\frac{\mu}{r_1}}
writing \mu = GM for brevity.
Speed at periapsis of the transfer:
v_{p} = \sqrt{\mu\left(\frac{2}{r_1}-\frac{2}{r_1+r_2}\right)} = \sqrt{\frac{2\mu}{r_1}\cdot\frac{r_2}{r_1+r_2}} = \sqrt{\frac{\mu}{r_1}}\sqrt{\frac{2r_2}{r_1+r_2}}
\boxed{\Delta v_1 = \sqrt{\frac{\mu}{r_1}}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)}
Similarly at the far end:
\boxed{\Delta v_2 = \sqrt{\frac{\mu}{r_2}}\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right)}
And the transfer time is half the ellipse's period:
t = \pi\sqrt{\frac{a_t^3}{\mu}}
Worked example: LEO to geostationary
r_1 = 6771 km (400 km altitude), r_2 = 42{,}164 km, \mu = 3.986\times10^{14}.
Initial circular speed:
v_1 = \sqrt{\frac{3.986\times10^{14}}{6.771\times10^{6}}} = \sqrt{5.887\times10^{7}} = 7673\ \text{m/s}
First burn:
\frac{2r_2}{r_1+r_2} = \frac{2(42164)}{48935} = 1.7233, \qquad \sqrt{1.7233} = 1.3128
\Delta v_1 = 7673(1.3128-1) = 7673\times0.3128 = 2400\ \text{m/s}
Second burn:
v_2 = \sqrt{\frac{3.986\times10^{14}}{4.2164\times10^{7}}} = \sqrt{9.454\times10^{6}} = 3075\ \text{m/s}
\frac{2r_1}{r_1+r_2} = \frac{2(6771)}{48935} = 0.27670, \qquad \sqrt{0.27670} = 0.52602
\Delta v_2 = 3075(1-0.52602) = 3075\times0.47398 = 1458\ \text{m/s}
Total:
\Delta v = 2400+1458 = 3858\ \text{m/s}
Transfer time:
a_t = \frac{6771+42164}{2} = 24{,}468\ \text{km}
t = \pi\sqrt{\frac{(2.4468\times10^{7})^3}{3.986\times10^{14}}} = \pi\sqrt{\frac{1.4649\times10^{22}}{3.986\times10^{14}}} = \pi\sqrt{3.675\times10^{7}}
= \pi\times6063 = 19{,}047\ \text{s} = 5.29\ \text{hours}
This is the standard geostationary transfer, and every communications satellite launched has flown it.
Note the asymmetry. The first burn is 2400 m/s and the second only 1458, despite the second being at a much greater distance. Burns at high speed are more effective, which is the next section.
The Oberth effect
A burn deep in a gravity well gives more energy gain than the same burn far out.
The reason is that energy is quadratic in speed. Adding \Delta v to a vehicle already moving at v changes the kinetic energy by:
\Delta E = \frac{1}{2}m(v+\Delta v)^2-\frac{1}{2}mv^2 = mv\Delta v+\frac{1}{2}m\Delta v^2
The cross term mv\Delta v is proportional to the existing speed. Burn where you are already moving fast and you get more energy for the same fuel.
Worked example. A 100 m/s burn at 7673 m/s (LEO) versus at 3075 m/s (GEO), per kilogram:
\Delta E_{\text{LEO}} = (7673)(100)+\frac{1}{2}(100)^2 = 767{,}300+5000 = 772{,}300\ \text{J}
\Delta E_{\text{GEO}} = (3075)(100)+5000 = 312{,}500\ \text{J}
Two and a half times more energy from the same fuel, purely from being in a faster orbit.
Practical consequences:
Escape burns are done at perigee, as low and fast as possible.
Interplanetary missions use gravity assists to dive close to a planet before burning (Chapter 11.6).
And a counterintuitive result: to escape the solar system it is more efficient to first fall towards the Sun and burn at perihelion than to burn directly outwards. The Parker Solar Probe uses this, though for the opposite goal.
Hermann Oberth published this in 1929, and it was one of the results that made him one of the three founders of astronautics alongside Tsiolkovsky and Goddard.
Bi-elliptic transfers
For very large ratios, three burns beat two.
The scheme: burn to a very high apoapsis r_b, far beyond the target; at that apoapsis, raise periapsis to r_2; then at r_2, lower apoapsis to circularise.
Why it can win. The middle burn happens where the vehicle is moving very slowly, so changing the orbit shape costs almost nothing. You pay for the extra distance and save on the shape change.
The crossover:
\frac{r_2}{r_1} > 15.58
Below 11.94 the Hohmann always wins; between 11.94 and 15.58 it depends on r_b.
Worked example: r_2/r_1 = 20. Take r_1 = 6771 km, r_2 = 135{,}420 km, r_b = 400{,}000 km.
Hohmann total: working through the formulas gives \Delta v = 4177 m/s.
Bi-elliptic total: \Delta v = 4149 m/s.
A saving of 28 m/s — under 1 %. And the transfer time goes from 3.1 days to about 79 days.
Which is why bi-elliptic transfers are rarely used. The saving is tiny and the time cost is enormous. They matter for one thing: plane changes.
Plane changes
The most expensive manoeuvre in orbital mechanics.
To change the orbital plane by angle \theta at speed v:
\Delta v = 2v\sin\frac{\theta}{2}
Derive it. The velocity vector must be rotated by \theta without changing magnitude, so the required change is the third side of an isosceles triangle with two sides of length v and included angle \theta.
Worked example: 28.5° plane change in LEO, at v = 7673 m/s:
\Delta v = 2(7673)\sin(14.25°) = 2(7673)(0.24615) = 3777\ \text{m/s}
Almost as much as the entire LEO-to-GEO transfer. For a 28.5° turn.
And a 90° plane change:
\Delta v = 2(7673)\sin(45°) = 10{,}851\ \text{m/s}
More than it took to reach orbit. This is why launch inclination is chosen at the pad and essentially never changed.
How it is made affordable:
Do it at apoapsis. At the transfer orbit's apoapsis the speed is only about 1600 m/s, so:
\Delta v = 2(1600)\sin(14.25°) = 788\ \text{m/s}
Nearly five times cheaper, simply for being done where the vehicle is slow.
Combine it with the circularisation burn. Instead of two separate manoeuvres, do one burn at an angle. Vector addition, not scalar:
\Delta v = \sqrt{v_1^2+v_2^2-2v_1v_2\cos\theta}
Worked example: geostationary insertion from a 28.5° transfer orbit. At apoapsis, v_1 = 1597 m/s in the transfer orbit and v_2 = 3075 m/s circular, with a 28.5° turn needed:
\Delta v = \sqrt{(1597)^2+(3075)^2-2(1597)(3075)\cos(28.5°)}
= \sqrt{2.550\times10^{6}+9.456\times10^{6}-2(1597)(3075)(0.87882)}
= \sqrt{1.2006\times10^{7}-8.631\times10^{6}} = \sqrt{3.375\times10^{6}} = 1837\ \text{m/s}
Against 1458+788 = 2246 m/s for doing them separately. A saving of 409 m/s, purely from doing one combined burn instead of two.
This is why geostationary insertion is always a combined manoeuvre, and it is one of the clearest cases where vector arithmetic saves real money.
Rendezvous
Getting to the same orbit is not the same as getting to the same place.
The phasing problem. If you are 90° behind the target in the same orbit, you cannot simply accelerate towards it — thrusting forward raises your orbit, which makes your period longer, which makes you fall further behind.
\boxed{\text{To catch up, slow down.}}
Slowing lowers the orbit, shortens the period, and you gain angle. Then speed up again to match.
Worked example. You are in a 400 km circular orbit, 10° behind the target. Drop to a phasing orbit with apogee at 400 km and perigee at 350 km.
a_{\text{phasing}} = \frac{6771+6721}{2} = 6746\ \text{km}
T_{\text{target}} = 2\pi\sqrt{\frac{(6.771\times10^{6})^3}{3.986\times10^{14}}} = 5545\ \text{s}
T_{\text{phasing}} = 2\pi\sqrt{\frac{(6.746\times10^{6})^3}{3.986\times10^{14}}} = 5514\ \text{s}
Gain per orbit:
\Delta t = 31\ \text{s} \quad\Longrightarrow\quad \Delta\theta = \frac{31}{5545}\times360° = 2.01°
Five orbits to close 10°, taking about 7.7 hours.
Cost: two small burns of about 14 m/s each.
This is exactly what every visit to the ISS does. The launch is timed so the phasing required is small, and then a sequence of small burns closes the gap over hours or days. Modern fast rendezvous reaches the station in about three hours — two orbits — which requires launching within a few seconds of the ideal moment.
And the final approach uses Clohessy–Wiltshire relative motion, in which the target is treated as the origin of a rotating frame. In that frame the dynamics are genuinely strange: thrusting towards the target moves you away, and a free-floating object traces a small ellipse around the station once per orbit.
This is why docking is done with a slow, controlled approach along a specific corridor, at about 0.1 m/s in the final metres.
A full mission budget
Launch to geostationary, everything included.
| Stage | Delta-v (m/s) | Note |
|---|---|---|
| Ideal orbital speed | 7673 | |
| Gravity losses | ~1200 | Thrusting against gravity |
| Drag losses | ~100 | Lower atmosphere |
| Steering losses | ~50 | Non-optimal thrust angle |
| Earth rotation credit | -409 | Eastward from 28.5°N |
| Total to LEO | ~9400 | |
| Transfer injection | 2400 | Hohmann first burn |
| Combined circularise + plane change | 1837 | At apogee |
| Total to GEO | ~13{,}600 | |
| Station-keeping, 15 years | ~750 | 50 m/s per year |
| Disposal to graveyard orbit | ~11 | End of life |
About 13.6 km/s to place a satellite in geostationary orbit, of which only 7.7 is the orbital speed itself. The rest is losses and manoeuvring.
Gravity losses deserve explanation. During ascent the rocket must support its own weight, and every second spent doing that is g of acceleration wasted. A vertical hover costs 9.81 m/s of delta-v per second and achieves nothing. This is why rockets pitch over into a gravity turn as early as the aerodynamics allow, and why high thrust-to-weight ratios matter at liftoff.
Selected delta-v costs, all from low Earth orbit:
| Destination | Delta-v (km/s) |
|---|---|
| Geostationary transfer | 2.4 |
| Escape Earth | 3.2 |
| Moon transfer | 3.1 |
| Lunar surface | 5.9 |
| Mars transfer | 3.6 |
| Mars surface (with aerobraking) | ~4.8 |
| Mercury orbit | ~13 |
| Jupiter transfer | 6.3 |
| Solar escape | 8.8 |
Mercury costs more than Jupiter. Getting to the inner solar system means shedding the Earth's orbital velocity, and there is no atmosphere at Mercury to brake against. BepiColombo uses nine gravity assists over seven years to arrive with a manageable budget.
Launch windows
Transfers only work if the target is where you are going when you arrive.
For a Hohmann transfer to Mars, the transfer takes 259 days, during which Mars moves:
\theta_{\text{Mars}} = \frac{259}{687}\times360° = 135.7°
So Mars must be 180°-135.7° = 44.3° ahead of Earth at departure.
How often does that happen? The synodic period:
\frac{1}{T_{\text{syn}}} = \frac{1}{T_E}-\frac{1}{T_M} = \frac{1}{365.25}-\frac{1}{687} = 2.7379\times10^{-3}-1.4556\times10^{-3}
= 1.2823\times10^{-3} \quad\Longrightarrow\quad T_{\text{syn}} = 780\ \text{days} = 2.14\ \text{years}
A Mars launch window every 26 months, lasting a few weeks. Miss it and you wait two years, which is why Mars missions cluster: 2018, 2020, 2022, 2024.
And the windows are not equal, because Mars's orbit is eccentric (e = 0.093). The delta-v required varies by about 20 % between the best and worst windows on a 15-year cycle.
Electric propulsion
Chapter 11.7 shows the rocket equation favours high exhaust velocity. Ion engines achieve 30,000 m/s against chemical rockets' 4,500 — but with thrust measured in millinewtons.
The trade: a chemical burn is impulsive, taking minutes. An ion engine thrusts for months or years, spiralling slowly outward.
And a low-thrust spiral costs more delta-v than a Hohmann transfer — for LEO to GEO it is about 5900 m/s against 3858 — because you are continuously fighting gravity rather than making two efficient impulsive burns.
But the fuel saving is enormous. From the rocket equation, the mass ratio for 5900 m/s at v_e = 30{,}000 is e^{0.197} = 1.22, against e^{0.857} = 2.36 for 3858 m/s at v_e = 4500.
\boxed{\text{22 \% fuel instead of 136 \%.}}
Which is why most modern geostationary satellites use electric propulsion for orbit raising and station-keeping, accepting several months of transfer time for a large increase in payload. Dawn, which orbited both Vesta and Ceres, would have been impossible chemically.
Where this shows up in your life
Every satellite you use flew a transfer computed exactly this way.
Launch scheduling is entirely driven by windows and phasing.
The ISS reboosts several times a year, each a small delta-v against drag.
Interplanetary missions are planned years ahead around window arithmetic, which is why the launch of a Mars mission slipping by a month can mean a two-year delay.
And the reason space is expensive is the exponential in the rocket equation acting on a 13.6 km/s budget. Chapter 11.7 makes that exponential explicit.
What the next chapter fixes
Everything so far has assumed two bodies. Add a third and the problem becomes unsolvable in closed form, and it also becomes far richer: there are points where a spacecraft can sit indefinitely, trajectories that steal energy from planets, and regions where the motion is genuinely chaotic. Chapter 11.6 derives the five Lagrange points, explains gravity assists with the energy bookkeeping done properly, and shows what chaos means for a solar system that has been stable for four billion years.