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11.4 — Orbits in Full
An orbit needs six numbers to specify completely — three for position and three for velocity at any instant, or equivalently six geometric parameters. This chapter sets out those six, derives the equation that gives a spacecraft's speed anywhere in its orbit, and computes from scratch the altitude at which a satellite stays above one point on the Earth.
The six orbital elements
Size and shape:
Semi-major axis a — half the long axis, and it alone determines the period and the energy.
Eccentricity e — how elongated. e = 0 is a circle, e \to 1 is a parabola.
Orientation:
Inclination i — the tilt of the orbital plane relative to a reference plane (the equator for Earth orbits, the ecliptic for solar orbits).
Longitude of the ascending node \Omega — where the orbit crosses the reference plane going north.
Argument of periapsis \omega — where the closest point sits within the orbital plane.
Position:
True anomaly \nu — the angle from periapsis to the object right now. This is the only one that changes with time in an unperturbed orbit.
Useful derived quantities:
r_p = a(1-e), \qquad r_a = a(1+e), \qquad a = \frac{r_p+r_a}{2}
Periapsis, apoapsis, and the fact that the semi-major axis is their average. The prefixes change with the central body: perigee and apogee for Earth, perihelion and aphelion for the Sun, periapsis and apoapsis generally.
Energy and the vis-viva equation
The total energy of an orbiting body:
E = \frac{1}{2}mv^2-\frac{GMm}{r}
Kinetic plus potential, with the potential from Chapter 1.9 measured from zero at infinity.
The key result — and it is not obvious — is that this depends only on a:
\boxed{E = -\frac{GMm}{2a}}
Derive it at periapsis, where the velocity is purely tangential so L = mv_pr_p:
E = \frac{L^2}{2mr_p^2}-\frac{GMm}{r_p}
Substituting L^2 = GMm^2a(1-e^2) from Chapter 11.3 and r_p = a(1-e):
E = \frac{GMm\,a(1-e^2)}{2a^2(1-e)^2}-\frac{GMm}{a(1-e)}
Using 1-e^2 = (1-e)(1+e):
E = \frac{GMm(1+e)}{2a(1-e)}-\frac{GMm}{a(1-e)} = \frac{GMm}{a(1-e)}\left[\frac{1+e}{2}-1\right] = \frac{GMm}{a(1-e)}\cdot\frac{e-1}{2}
E = -\frac{GMm}{2a}
The eccentricity has cancelled completely. Two orbits with the same semi-major axis have the same energy regardless of shape, which is the same cancellation that made the period independent of eccentricity in Chapter 11.3.
Now set the two expressions for E equal:
\frac{1}{2}mv^2-\frac{GMm}{r} = -\frac{GMm}{2a}
\boxed{v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)}
The vis-viva equation — Latin for "living force", an old term for kinetic energy. It gives the speed at any point of any orbit from two numbers, and it is the single most-used equation in orbital mechanics.
Three special cases
Circular orbit, r = a:
v_c = \sqrt{\frac{GM}{r}}
Escape, a \to \infty:
v_{\text{esc}} = \sqrt{\frac{2GM}{r}} = \sqrt{2}\,v_c
Escape velocity is exactly \sqrt{2} times circular velocity, at any radius. A 41 % increase in speed converts a bound orbit into an escape.
Hyperbolic, a < 0, giving v^2 > 2GM/r and a leftover speed at infinity:
v_\infty^2 = -\frac{GM}{a}
Called the characteristic energy, C_3 = v_\infty^2, and it is how interplanetary launch requirements are specified.
Worked example: the International Space Station
Altitude 420 km, so r = 6371+420 = 6791 km. Earth's GM = 3.986\times10^{14} m³/s².
v = \sqrt{\frac{3.986\times10^{14}}{6.791\times10^{6}}} = \sqrt{5.870\times10^{7}} = 7662\ \text{m/s}
7.66 km/s, or 27,600 km/h.
T = \frac{2\pi r}{v} = \frac{2\pi(6.791\times10^{6})}{7662} = 5569\ \text{s} = 92.8\ \text{minutes}
And 15.5 orbits per day, which is why astronauts see 16 sunrises daily.
Escape velocity from that altitude:
v_{\text{esc}} = \sqrt{2}\times7662 = 10{,}836\ \text{m/s}
So a spacecraft in low Earth orbit is already more than 70 % of the way to escape, in speed terms — which is why the phrase "halfway to anywhere in the solar system" is used for low Earth orbit.
Geostationary orbit, derived
The requirement: the satellite's orbital period must equal the Earth's rotation period, so it stays above one point.
Use the sidereal day, not the solar day (Chapter 11.1), because the satellite must match the Earth's rotation relative to the stars:
T = 23\ \text{h}\ 56\ \text{m}\ 4.1\ \text{s} = 86{,}164.1\ \text{s}
From Kepler's third law (Chapter 11.3):
T^2 = \frac{4\pi^2a^3}{GM} \quad\Longrightarrow\quad a = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}
Substituting:
a = \left(\frac{(3.986\times10^{14})(86164.1)^2}{4\pi^2}\right)^{1/3}
T^2 = (8.61641\times10^{4})^2 = 7.4242\times10^{9}
GMT^2 = (3.986\times10^{14})(7.4242\times10^{9}) = 2.9593\times10^{24}
\frac{2.9593\times10^{24}}{39.478} = 7.4961\times10^{22}
Cube root. Take logs: \log_{10}(7.4961\times10^{22}) = 22.8748, divided by 3 is 7.6249:
a = 10^{7.6249} = 4.2164\times10^{7}\ \text{m} = 42{,}164\ \text{km}
Altitude above the surface:
h = 42164-6371 = 35{,}793\ \text{km}
\boxed{\text{Geostationary altitude} = 35{,}786\ \text{km}}
(The standard value uses the equatorial radius of 6378 km rather than the mean.)
Orbital speed there:
v = \frac{2\pi a}{T} = \frac{2\pi(4.2164\times10^{7})}{86164} = 3075\ \text{m/s}
Compare with 7.66 km/s in low Earth orbit. Higher orbits are slower, which is one of the counterintuitive facts of orbital mechanics.
The constraints on geostationary orbit:
It must be equatorial, i = 0. Any inclination makes the satellite trace a figure-of-eight in the sky rather than staying put. An inclined orbit at the same altitude is geosynchronous but not geostationary.
It must be circular, or the satellite would speed up and slow down and drift east and west.
There is only one such orbit, and it is a finite resource. Satellites are allocated slots by international agreement, typically 2° apart, giving about 180 usable positions.
And the round-trip signal delay:
t = \frac{2\times35786\times10^{3}}{3\times10^{8}} = 0.239\ \text{s}
A quarter of a second, which is the pause you hear on satellite phone calls and why geostationary internet is poor for anything interactive. Starlink's satellites orbit at 550 km, giving a delay of about 4 ms, which is the entire commercial argument for low-orbit constellations.
Arthur C. Clarke proposed the geostationary orbit in a 1945 article in Wireless World, correctly computing the altitude and noting that three satellites would cover the world. He did not patent it. The orbit is now sometimes called the Clarke Belt.
Useful orbits
| Orbit | Altitude | Period | Use |
|---|---|---|---|
| Low Earth (LEO) | 160–2000 km | 90–130 min | ISS, imaging, Starlink |
| Sun-synchronous | ~700 km, i \approx 98° | ~99 min | Earth observation |
| Medium Earth (MEO) | ~20,200 km | 12 h | GPS |
| Geostationary | 35,786 km | 24 h | Communications, weather |
| Molniya | 200\times40{,}000 km, i = 63.4° | 12 h | High-latitude coverage |
Sun-synchronous, and why it needs an oblate Earth
A Sun-synchronous orbit crosses the equator at the same local solar time every day, so images are always taken under the same lighting — essential for comparing them.
This requires the orbital plane to rotate by 360° per year, about 0.9856° per day, to keep up with the Earth's motion around the Sun.
And it happens for free, because the Earth is not a sphere. The equatorial bulge exerts a torque on the orbit that makes the ascending node drift:
\dot{\Omega} = -\frac{3}{2}J_2\left(\frac{R_E}{a}\right)^2\frac{\cos i}{(1-e^2)^2}n
where J_2 = 1.0826\times10^{-3} measures the oblateness and n = 2\pi/T is the mean motion.
The \cos i is what makes it work. For i < 90° the drift is westward; for i > 90° — a retrograde orbit — it is eastward, which is the direction needed.
Worked example. For a 700 km altitude orbit, a = 7.071\times10^{6} m, T = 5919 s, n = 1.0616\times10^{-3} rad/s. Setting \dot{\Omega} = 1.991\times10^{-7} rad/s (one revolution per year):
\cos i = -\frac{2\dot{\Omega}}{3J_2n}\left(\frac{a}{R_E}\right)^2 = -\frac{2(1.991\times10^{-7})}{3(1.0826\times10^{-3})(1.0616\times10^{-3})}\times(1.1099)^2
= -\frac{3.982\times10^{-7}}{3.448\times10^{-6}}\times1.232 = -0.1423
i = 98.2°
Slightly retrograde, and every Earth observation satellite — Landsat, Sentinel, weather satellites — flies at close to this inclination. A deviation from a perfect sphere, turned into a design feature.
Molniya
The problem: a geostationary satellite sits over the equator, so from high latitudes it appears very low in the sky or below the horizon. Useless for Russia.
The solution: a highly elliptical orbit with perigee low over the southern hemisphere and apogee at 40,000 km over the northern. By Kepler's second law the satellite crawls through apogee, spending about 8 of its 12 hours over the target region.
And the inclination of 63.4° is not arbitrary. The same oblateness that drives nodal regression also rotates the argument of perigee:
\dot{\omega} \propto \left(5\cos^2 i-1\right)
Setting the bracket to zero:
\cos^2 i = \frac{1}{5} \quad\Longrightarrow\quad i = 63.435°
At that inclination the perigee does not drift, so the apogee stays over the northern hemisphere indefinitely without correction. A number that falls out of the Earth's shape and determines the design of an entire communications system.
Ground tracks
As the satellite orbits, the Earth turns beneath it, so the path traced on the surface is not a simple circle.
In one orbital period the Earth rotates:
\Delta\lambda = 360°\times\frac{T_{\text{orbit}}}{86164\ \text{s}}
For the ISS, T = 5569 s:
\Delta\lambda = 360\times\frac{5569}{86164} = 23.3°
Each pass is 23.3° of longitude west of the last. The ISS covers the whole planet between 51.6°N and 51.6°S — its inclination — and repeats roughly every three days.
And this is why launch inclination cannot be less than the launch site's latitude without an expensive plane change. Kennedy Space Center at 28.5°N cannot directly reach an orbit of lower inclination, which is why equatorial launch sites like Kourou at 5.2°N are valuable.
Launching eastward also gains the Earth's rotation speed for free:
v_{\text{rot}} = \frac{2\pi R_E\cos\phi}{86164}
At the equator: 465 m/s. At Kennedy: 409 m/s. At Baikonur (45.6°N): 326 m/s. Kourou's 463 m/s is why the Ariane rockets launch from French Guiana.
Perturbations
Real orbits are not conic sections, and the departures matter enormously in practice.
| Perturbation | Where it matters |
|---|---|
| Earth's oblateness (J_2) | All Earth orbits; drives Sun-synchronous and Molniya |
| Atmospheric drag | Below ~600 km |
| Sun and Moon's gravity | Above ~20,000 km |
| Solar radiation pressure | Large light structures |
| Higher gravity harmonics | Geostationary station-keeping |
Atmospheric drag is the dominant effect in low orbit, and it produces a genuinely paradoxical result.
Drag removes energy, so a decreases, so from the vis-viva equation the satellite speeds up.
v = \sqrt{\frac{GM}{r}} \quad\Longrightarrow\quad \text{smaller }r\text{ means larger }v
A drag force that opposes motion makes the satellite go faster. This is the orbital paradox, and the resolution is that the potential energy released as it falls exceeds the energy drag takes away.
Practical consequence: the ISS loses about 2 km of altitude per month and needs reboosting several times a year.
And drag depends on solar activity. During solar maximum the upper atmosphere heats and expands, increasing density at a given altitude by up to an order of magnitude. Skylab reentered in 1979 years earlier than planned because of an unexpectedly active solar cycle.
In February 2022, SpaceX lost 38 of 49 newly launched Starlink satellites to a geomagnetic storm that raised drag by about 50 % while they were still in their low insertion orbit.
Geostationary satellites need north–south station-keeping against the Sun and Moon, which would otherwise pull the inclination to 15° over 50 years. This costs about 50 m/s of delta-v per year, and it is the main consumable that determines a satellite's operational life.
Where this shows up in your life
Satellite television comes from 35,786 km, and your dish points at a fixed spot in the sky because of the calculation above.
GPS uses 20,200 km orbits, chosen so that each satellite completes exactly two orbits per sidereal day and repeats its ground track daily.
Weather forecasting uses both geostationary satellites for continuous coverage of one hemisphere and Sun-synchronous ones for high-resolution polar-orbiting passes.
Satellite internet is moving to low orbit for the latency reason above.
Earth observation — crop monitoring, deforestation tracking, disaster response, ice sheet measurement — is almost all Sun-synchronous.
And space debris is now a serious problem. Over 30,000 tracked objects and millions of untracked fragments occupy low Earth orbit, and the Kessler syndrome — a cascade in which collisions generate debris that causes more collisions — is a genuine risk for the most used altitudes.
What the next chapter fixes
Orbits have been treated as fixed. Getting from one to another is the actual business of spaceflight, and it costs fuel in a way that is completely unlike moving on the ground. Chapter 11.5 derives the Hohmann transfer and its cost, explains why plane changes are so expensive, works out when a bi-elliptic transfer beats a Hohmann one, and computes a full mission delta-v budget from launch pad to geostationary orbit.