Orbits as conic sections

The engines stop. From that instant until they light again, nothing the flight computer does changes where the vehicle goes. Its path is fixed by two vectors — where it is and how fast it is moving — and by the gravity of the body it is falling round. This lesson is about that path: what shape it has, why it has that shape, and how to read everything that matters about it from the state at cutoff.

The answer is older than rocketry. Every unpowered trajectory around a single body is a conic section — a circle, an ellipse, a parabola or a hyperbola — with the centre of the body at one focus. A booster that has just shut down, a satellite in low orbit and a probe leaving for Mars are each following one of those four curves, and which one is decided by a single number.

That is why it matters when you fly. At cutoff you stop steering and start choosing a conic. Cut off a little fast and the far side of the orbit is higher than you wanted. Cut off with the nose a degree high and the near side is lower, possibly inside the atmosphere. Every burn in the rest of these courses is a move from one conic to another, so the curves are worth knowing well.

Falling and missing

Newton's picture is still the best one. Put a cannon on a mountain tall enough to be above the air and fire it horizontally. A slow shot falls to the ground nearby. A faster one lands further away, and because the ground curves away beneath it, it has further to fall before it gets there. At one particular speed the ground curves away exactly as fast as the ball falls: it never lands, and comes round behind the cannon. That is a circular orbit. For something skimming the surface of the Earth the speed is about 7.9 km/s.

Fire faster still and the ball rises as it goes round, climbing to a high point on the far side and falling back to the mountain: an ellipse, with the cannon at its lowest point. Faster again and the far point recedes, until at $\sqrt{2}$ times the circular speed it never comes back and the ellipse has opened into a parabola. Beyond that the path is a hyperbola, and the ball leaves with speed to spare.

Two things in that picture are worth holding on to. An orbit is not a balance between gravity and some outward force; it is a fall that keeps missing. And the shots slower than circular are ellipses too — ellipses whose highest point is the cannon and whose lowest point is somewhere inside the Earth. A sub-orbital trajectory is an orbit that happens to meet the ground before it completes.

Two things that do not change

The setting is the two-body problem. A vehicle moves under the gravity of a body much heavier than itself, which is treated as a point — or as a perfect sphere, which by Newton's shell theorem pulls the same way from outside. Nothing else acts: no thrust, no air, no third body. Put the origin at the centre of the heavy body and let $\mathbf{r}$ be the vehicle's position, $r=|\mathbf{r}|$ its distance. Newton's law of gravitation, divided through by the vehicle's mass, is

where the dots are time derivatives and $\mu =GM$ is the body's gravitational parameter — the gravitational constant times its mass, 398,600.4418 km³/s² for the Earth. The vehicle's own mass has dropped out, which is why a 30 kg satellite and the 100 t stage that released it fly the same path. (Strictly both bodies circle their common centre of mass; for a spacecraft and a planet the difference is far below anything measurable.)

Energy

Take the dot product of the equation with the velocity $\mathbf{v}=\dot{\mathbf{r}}$. The left side becomes $\mathbf{v}\cdot \dot{\mathbf{v}}$, which is the time derivative of $v^2/2$. On the right, $\mathbf{r}\cdot \dot{\mathbf{r}}=r\dot{r}$ — differentiate $\mathbf{r}\cdot \mathbf{r}=r^2$ to see it — so the right side is $-\mu \dot{r}/r^2$, the time derivative of $\mu /r$. Collecting both on one side:

$\epsilon$ is the specific mechanical energy: kinetic plus potential energy per kilogram, with the potential taken as zero infinitely far away, which is why it is negative anywhere near the body. One km²/s² is one MJ/kg. A coasting vehicle that climbs trades speed for height, and $\epsilon$ does not move.

Angular momentum

Now take the cross product of $\mathbf{r}$ with the equation. The right side vanishes, because $\mathbf{r}\times \mathbf{r}=0$. The left side, $\mathbf{r}\times \ddot{\mathbf{r}}$, is the time derivative of $\mathbf{r}\times \dot{\mathbf{r}}$, since the other term of that derivative is $\dot{\mathbf{r}}\times \dot{\mathbf{r}}=0$. So

$\mathbf{h}$ is the specific angular momentum, and it is constant as a vector, which says two things. Its direction is fixed and $\mathbf{r}$ is always perpendicular to it, so the motion stays in one plane through the centre of the body — the orbital plane. And its magnitude, $h=rv\cos \gamma$, where $\gamma$ is the flight-path angle of the velocity above the local horizontal, is twice the rate at which the line from the centre to the vehicle sweeps out area. That is Kepler's second law, equal areas in equal times. Close in, the radius is short and the vehicle has to move fast to sweep its share; far out it crawls.

The only thing this used was that gravity points at the centre. It is also why nothing short of thrust out of the plane can change the plane.

Why the shape is a conic

Energy and angular momentum nearly pin the orbit down, but they do not yet give its shape. That needs a third conserved vector. Take the cross product of the equation of motion with $\mathbf{h}$:

Because $\mathbf{h}$ is constant, the left side is the time derivative of $\mathbf{v}\times \mathbf{h}$. The triple product on the right expands as $\mathbf{r}\times (\mathbf{r}\times \mathbf{v})=\mathbf{r}(\mathbf{r}\cdot \mathbf{v})-\mathbf{v}{r}^2=\mathbf{r}r\dot{r}-\mathbf{v}{r}^2$, so

Two quantities with the same derivative differ by a constant, and it is conventional to write that constant as $\mu \mathbf{e}$:

$\mathbf{e}$ is the eccentricity vector. It lies in the orbital plane and never changes. Now dot both sides with $\mathbf{r}$. On the left, $\mathbf{r}\cdot (\mathbf{v}\times \mathbf{h})=(\mathbf{r}\times \mathbf{v})\cdot \mathbf{h}=h^2$. On the right, $\mu (r+re\cos \nu )$, where $e=|\mathbf{e}|$ and $\nu$ is the angle from $\mathbf{e}$ to $\mathbf{r}$. So $h^2=\mu r(1+e\cos \nu )$, or

This is the polar equation of a conic section with its focus at the origin. The semi-latus rectum $p$ sets its size and the eccentricity $e$ its shape. The vector $\mathbf{e}$ points at periapsis, the closest point, where $\nu =0$; the angle $\nu$, measured from periapsis in the direction of motion, is the true anomaly. With $e=0$ the radius never changes: a circle. With $e$ between 0 and 1 the denominator never reaches zero: a closed ellipse. With $e=1$ it reaches zero only as $\nu \to 180°$: a parabola. With $e>1$ it reaches zero at $\cos {\nu }_{\infty }=-1/e$: a hyperbola, whose two arms run out along those angles.

Nothing in this was about rockets. It holds for anything under an inverse-square central force, which is why the same four curves describe planets, comets and the path of a stage after cutoff.

Energy decides the class

What remains is to tie the shape to the two numbers you can read off at cutoff. Square both sides of $\mathbf{v}\times \mathbf{h}=\mu (\mathbf{r}/r+\mathbf{e})$. The velocity is perpendicular to $\mathbf{h}$, so the left side is $v^2{h}^2$; the right is ${\mu }^2(1+2e\cos \nu +e^2)$. The conic equation gives $e\cos \nu =h^2/(\mu r)-1$, and substituting and collecting terms leaves

So the sign of the energy decides the class of the curve. Negative energy gives $e<1$ and a closed orbit; zero gives the parabola; positive gives a hyperbola. The angular momentum then decides how far from round the curve is: for a given energy the largest possible $h$ makes $e=0$, the circle, and a smaller $h$ stretches it.

For a closed orbit it helps to name the semi-major axis $a$, half the longest diameter, so that periapsis is at $a(1-e)$, apoapsis at $a(1+e)$, and $p=a(1-e^2)$. Put $h^2=\mu a(1-e^2)$ into the last equation and the eccentricity cancels:

The energy depends only on the size of the orbit, not on its shape. Put that back into the definition of $\epsilon$ and you have the most useful single line in orbital mechanics, the vis-viva equation:

It gives the speed anywhere on an orbit from the distance and the size alone. Three cases fall straight out of it. On a circle $a=r$, so the circular speed is $v_c=\sqrt{\mu /r}$. As $a\to \infty$ the orbit opens into a parabola and the speed is the escape speed $v_{\text{esc}}=\sqrt{2\mu /r}=\sqrt{2}{v}_c$. For a hyperbola $a$ is negative, and far from the body, where $2/r\to 0$, the speed settles at the hyperbolic excess speed $v_{\infty }=\sqrt{-\mu /a}=\sqrt{2\epsilon }$.

The period comes from the second law. The orbit's area, $\pi ab$ with the semi-minor axis $b=a\sqrt{1-e^2}$, is swept at the constant rate $h/2$, so $T=2\pi ab/h$. With $h=\sqrt{\mu a(1-e^2)}$ everything but the size cancels:

That is Kepler's third law. For the Earth: 200 km above the equatorial radius ($r=6,578$ km) the circular speed is 7.784 km/s, the escape speed 11.009 km/s and the period 88.5 minutes; at 400 km they are 7.669 km/s, 10.845 km/s and 92.6 minutes. Escape from the surface itself is 11.2 km/s.

A worked example: cut off slightly fast

An upper stage shuts down 200 km above the equatorial radius, so $r=6,378.1+200=6,578.1$ km, moving horizontally ($\gamma =0$) at 7.900 km/s. Circular speed there is 7.784 km/s, so it is 116 m/s fast. What orbit is it in?

The energy first. $v^2/2=31.205$ MJ/kg and $\mu /r=60.595$ MJ/kg, so $\epsilon =-29.390$ MJ/kg. That is negative, so the orbit is closed, and its semi-major axis is $a=-\mu /2\epsilon =6,781.3$ km. The angular momentum is $h=rv=51,967$ km²/s, and the eccentricity $e=\sqrt{1+2\epsilon h^2/{\mu }^2}=0.0300$.

Because the velocity is horizontal and faster than circular, the cutoff point is periapsis, $a(1-e)=6,578.1$ km, which checks. Apoapsis is $a(1+e)=6,984.4$ km, 606 km up, on the far side. The period is 92.6 minutes. Near circular speed the far side climbs by about 3.6 km for every extra metre per second at cutoff, which is why an insertion burn is cut off on a precise number and not a round one.

Now cut off at the same speed with the velocity 2° above the horizontal. The energy has not changed, so neither has $a$, nor the period. But $h=rv\cos \gamma =51,936$ km²/s — smaller by six hundredths of a percent — and that is enough to raise the eccentricity to 0.0460. Periapsis falls to 91 km, apoapsis rises to 715 km, and the cutoff point is no longer periapsis at all: it sits 51° past it. A 91 km periapsis is in the upper atmosphere, and an orbit that dips there does not last. At 5° above the horizontal the periapsis is 222 km below the surface. The speed was right in both cases; the direction was not, and direction acts on a different element.

The "Transfer to GEO" setting is the ellipse the landing page flies, from 15,178 km out to the geostationary radius, 42,164 km: 6.215 km/s at the start point and 2.237 km/s at the far end, and the dots show the vehicle spending most of its 13.4-hour period out there. Push the speed towards 7.247 km/s and watch the period grow without limit as the ellipse opens; the "Escape speed" setting is exactly on the parabola. Then return to circular speed and move only the flight-path angle. The energy does not change, so the period stays at 5.17 hours while the orbit stretches and its apse line swings round; beyond about 35° either way the periapsis is inside the Earth and the trajectory meets the ground. (At circular speed the eccentricity is exactly $\sin \gamma$, so the periapsis is $r(1-\sin \gamma )$, which drops below 6,371 km at 35.5°.)

Six numbers

In a plane a conic needs three numbers to fix it — size, shape and which way it points — and one more to say where the vehicle is on it. In three dimensions the plane itself needs two. The classical set is:

• $a$, semi-major axis. The size. It fixes the energy, $\epsilon =-\mu /2a$, and the period.
• $e$, eccentricity. The shape, from circle to hyperbola.
• $i$, inclination. The tilt of the orbital plane to the equator: the angle between $\mathbf{h}$ and the body's spin axis. Below 90° the orbit runs eastward with the rotation, above 90° it is retrograde.
• $\Omega$, right ascension of the ascending node. Where, around the equator, the orbit crosses it going north, measured from a fixed direction in space.
• $\omega$, argument of periapsis. The angle within the orbital plane from the ascending node to periapsis: which way the ellipse points.
• $\nu$, true anomaly. Where the vehicle is now. The time since periapsis, or the mean anomaly, says the same thing.

Six numbers for six numbers: the state vector — three components of position, three of velocity — has exactly as much information in it, and the elements are another way of writing it down. Every position and velocity maps to one set of elements and back. That conversion is what a flight computer does when it reports an orbit, and the derivation above is the conversion: $\epsilon$ gives $a$, $\mathbf{h}$ gives $i$ and $\Omega$, $\mathbf{e}$ gives $e$ and $\omega$, and the angle from $\mathbf{e}$ to $\mathbf{r}$ is $\nu$.

Two of them go undefined in common cases. On a circle there is no periapsis, so $\omega$ and $\nu$ are replaced by their sum, the argument of latitude, measured from the node. On an equatorial orbit there is no node, so $\Omega$ is dropped and angles are measured from the reference direction. These are not physical problems; they are places where one of the numbers has nothing to describe.

What each one moves

A pilot needs the elements the other way round: not what they mean but which burn changes which of them.

• Along the velocity — prograde to speed up, retrograde to slow down. This changes the energy and so $a$, and with it $e$. Burn at periapsis and the apoapsis moves while the periapsis stays where it is; burn at apoapsis and the reverse. The worked example is a prograde burn at periapsis.
• Radially, towards or away from the centre. This does almost nothing to the energy, since it is at right angles to the velocity, but it tilts the velocity and so changes $h$. The eccentricity changes and the line of apsides swings round: $\omega$ moves. The 2° error above is, in effect, a small radial component at cutoff.
• Normal to the plane, along $\mathbf{h}$ or against it. This tilts the plane. Fired where the orbit crosses the equator it changes $i$ alone; fired anywhere else it moves $\Omega$ as well. It is expensive: turning a velocity $v$ through an angle $\Delta i$ costs $2v\sin (\Delta i/2)$, which at 7.78 km/s is 136 m/s for a single degree. This is why a launch goes out on the azimuth that puts it straight into the right plane — the subject of latitude, azimuth and the orbits you can reach.
• Nothing at all, to change $\nu$. Waiting is free.

Real planets are not points, and the Earth's equatorial bulge — the $J_2$ term of its gravity field — pulls slightly off-centre. The orbit is then not a fixed conic but a slowly turning one: for a 400 km orbit inclined at 51.6°, like the International Space Station's, the node regresses westward by 5.0° a day. The elements at any instant describe the osculating conic, the one the vehicle would follow if the perturbation stopped at that moment.

In Vivapse

The simulator integrates the vehicle's position and velocity directly and never flies a conic. Conics are how it reports and predicts. keplerElements(r, v) in src/sim/orbit.ts is the derivation above as code: it forms the eccentricity vector as $((v^2-\mu /r)\mathbf{r}-(\mathbf{r}\cdot \mathbf{v})\mathbf{v})/\mu$, which is $\mathbf{v}\times \mathbf{h}/\mu -\mathbf{r}/r$ with the triple product expanded, takes the node line from $\mathbf{h}$, measures the argument of latitude for circular orbits and uses the reference axis when the orbit is equatorial.

What a program sees is fc.orbit: apoapsis, periapsis, semiMajorAxis, eccentricity, period, timeToApoapsis, timeToPeriapsis, argPeriapsis, trueAnomaly, inclination, raan and closed, with the full list and units in the flight computer reference. Two details matter. The apsides are measured as heights above the equatorial radius, 6,378.137 km, so a sub-orbital arc has a periapsis of several thousand kilometres below zero — a real number, the closest approach of its conic to the centre of the Earth, less 6,378 km. And because the simulated Earth has its $J_2$ bulge, the orbit is not an exact conic: orbitOf reports the heights the coast will really reach, from the exact radial solution in the equatorial plane and by integrating one revolution for an inclined orbit. The physics model covers the gravity field; the same file's propagate and the rails behind fc.sleep coast the vehicle on Kepler plus $J_2$, and keplerStep in src/sim/lambert.ts propagates any of the four conics by universal variables for the interplanetary predictions.

Try it

Open the playground with the default vehicle and the full-mission program, and add a line that reports the conic every five seconds. At the top of the program, beside the other state:

let nextOrbitLog = 0;

and as the first lines inside update(fc):

if (fc.t >= nextOrbitLog) {
  nextOrbitLog += 5;
  fc.log("pe", (fc.orbit.periapsis / 1e3).toFixed(0), "km  ap",
    (fc.orbit.apoapsis / 1e3).toFixed(0), "km  e", fc.orbit.eccentricity.toFixed(4));
}

Launch and watch the console. On the pad the periapsis reads about −6,370 km and the eccentricity 0.997: a point resting on the spinning Earth is on a thin ellipse whose far end is the pad itself and whose near end is a few kilometres from the centre. Through the first-stage burn the periapsis hardly moves — it is still near −6,030 km at staging — because the booster's speed is mostly upward, and upward speed adds almost nothing to the angular momentum that holds the near side of the orbit up. The upper stage does the work, flying nearly horizontally. The periapsis climbs by hundreds and then thousands of kilometres a minute, crosses zero a little over seven minutes after liftoff, and the orbit event fires when it clears 140 km. By then the eccentricity has fallen to a few thousandths. The whole ascent, read this way, is one conic being bent into another.

What carries forward

Everything here was for a vehicle left alone. The next question is how to move between conics on purpose, and vis-viva is the tool: it turns every burn into a change of $a$ at a known $r$. Transfers and the Oberth effect uses it to find the cheapest way from one orbit to another, and to explain why a burn is worth more the faster the vehicle is going when it makes it.