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Unit 1 · Lesson 1-3
Angular Momentum
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Angular Momentum

Specific angular momentum, planarity of orbits and Kepler's second law

Why are orbits planar?
Here is something odd about every orbit you have ever seen. They are flat. The ISS traces a closed loop that lies in a single plane. So does the Moon, near enough. So does every planet, every comet, every satellite ever launched. Not one of them wanders off into a three-dimensional tangle. Why should that be? Gravity pulls in all directions. Space has three dimensions. Nothing obviously forbids a spiralling, wobbling, corkscrewing path. And yet: flat. Always flat. Newton called momentum the quantity of motion. There is a second kind of quantity of motion, one his contemporaries very nearly missed — and it is the reason orbits are flat, the reason they sweep out equal areas, and ultimately the reason they close at all.
Watch first · presented by Shoaib Aziz
The equal-area wedges and the three same-h conics at the end are exact, not sketched, and every number in the lab segment is printed by this lesson's own lab solution.
Start with the familiar kind. Linear momentum is p⃗=mv⃗\vec{p} = m\vec{v} — mass times velocity, a vector. Newton's second law is really a statement about it: force is the rate at which momentum changes. No force, no change: momentum is conserved.
A satellite in orbit clearly does not conserve linear momentum. Gravity is pulling on it constantly, and its velocity vector swings through a full 360° every orbit. So linear momentum is not the quantity we want.
The other kind is angular momentum: how much rotational motion something has about a chosen point. For a point mass it is the cross product of position and velocity. In astrodynamics we almost always divide out the satellite's mass and work with the specific angular momentum:
h⃗=r⃗×v⃗\vec{h} = \vec{r} \times \vec{v}

Specific angular momentum — units of km²/s. "Specific" throughout this course means "per unit mass", which is why the satellite's mass never appears in orbital mechanics.

Figure 1.6
rvh= r × vh is perpendicular to the plane — so the plane cannot move without h moving
h = r × v points perpendicular to the plane containing both r and v. Since r and v are the only vectors describing the orbit, that plane *is* the orbital plane — and h is its unique normal direction.
💡Read the cross product as a question
A cross product a⃗×b⃗\vec{a}\times\vec{b} asks: how much of b is perpendicular to a, and which way does the resulting rotation point? So h⃗=r⃗×v⃗\vec{h} = \vec{r}\times\vec{v} measures how much of the velocity is sideways rather than straight in or out. A satellite falling directly toward Earth has v⃗\vec{v} parallel to r⃗\vec{r}, so h=0h = 0 — and it will hit. Any orbit at all requires h≠0h \neq 0. The magnitude is h=rvsin⁡γh = rv\sin\gamma, where γ\gamma is the angle between the two. At perigee and apogee the velocity is exactly perpendicular to the radius, so there h=rvh = rv — a fact we will use constantly.
Now the payoff. Angular momentum changes only when a torque acts, and torque is τ⃗=r⃗×F⃗\vec{\tau} = \vec{r}\times\vec{F}.
What is the gravitational force? It points from the satellite directly at the centre of the Earth — that is, exactly antiparallel to r⃗\vec{r}. And the cross product of two parallel vectors is zero.
τ⃗=r⃗×F⃗g=r⃗×(−μmr3r⃗)=0⃗\vec{\tau} = \vec{r} \times \vec{F}_g = \vec{r} \times \left(-\frac{\mu m}{r^3}\vec{r}\right) = \vec{0}

A central force can never exert a torque about its own centre. Therefore h is constant — in magnitude and direction — forever.

Figure 1.7
F is always parallel to r ⟹ τ = r × F = 0 ⟹ h never changesA force that always points at the centre can never twist the orbit.
Wherever the satellite is, gravity points straight back at the centre. A force with no lever arm cannot twist anything. This one line of vector algebra is the reason orbits are flat.
💡Why every orbit is flat
h⃗\vec{h} is constant, and h⃗\vec{h} is perpendicular to the orbital plane. A fixed perpendicular means a fixed plane. So the flatness of orbits is not a coincidence, or a special case, or an approximation. It is a direct consequence of gravity pointing at the centre. Any central force whatsoever — gravity, electrostatics, a spring — produces planar motion. And this also tells you exactly how to break it: apply a force that is not central. That is what a thruster does when you change planes (Unit 6), and what Earth's equatorial bulge does slowly and for free (Unit 7).
Figure 1.8
h straight uph tiltedTilt h and the entire orbital plane tilts with it. One vector, three degrees of freedom, the plane fixed for good.
Tilt h and the entire orbital plane tilts with it. Three numbers specify the direction and magnitude of h; two of them will become the orbital elements i and Ω in Unit 3.
There is one more thing hiding in h⃗\vec{h}, and it is a result that took Kepler years of observation to extract from data — yet falls out of conservation in about three lines.
Consider the tiny sliver of area the radius vector sweeps out in time dtdt. That sliver is very nearly a triangle with base rr and height vsin⁡γ dtv\sin\gamma\,dt, so its area is dA=12rvsin⁡γ dt=12h dtdA = \tfrac{1}{2} r v \sin\gamma\, dt = \tfrac{1}{2}h\,dt. Therefore:
dAdt=h2=constant\frac{dA}{dt} = \frac{h}{2} = \text{constant}

Kepler's Second Law. The rate at which area is swept out is literally half the angular momentum — so it cannot change.

🕰Kepler's second law before Newton
Kepler published the equal-areas law in 1609, seventy-eight years before the Principia. He had no concept of angular momentum, no vector calculus, and a physically wrong model — he believed the Sun swept the planets along with a kind of rotating magnetic broom. He found the law by grinding through Tycho Brahe's observations of Mars for years, in an era before logarithms, doing every division by hand. The law is exactly right. The reasoning behind it was entirely wrong. This is worth sitting with. A correct empirical law can survive a completely mistaken explanation — and it took Newton's mechanics to reveal that Kepler's second law was never really about areas at all. It was conservation of angular momentum, wearing a disguise.
I used to measure the heavens, now I measure the shadows of Earth. Although my mind was heaven-bound, the shadow of my body lies here.
– Johannes Kepler, the epitaph he wrote for himself, 1630
Figure 1.9
fastslowEqual areas in equal timesdA/dt = h/2, and h is constant — so this is conservation, not coincidence.
Two shaded wedges of equal area, swept in equal times. Near perigee the satellite is close and fast, covering a wide angle over a short radius; near apogee it is far and slow, creeping through a narrow angle over a long radius. The product balances exactly, because h is constant.
📐Worked Example: the Molniya speed ratio
A Molniya orbit has perigee 500 km altitude (rp=6871r_p = 6871 km) and apogee 39,873 km altitude (ra=46,244r_a = 46{,}244 km). At both apsides the velocity is perpendicular to the radius, so h=rpvp=ravah = r_p v_p = r_a v_a. Rearranging: vpva=rarp=46,2446871=6.73\dfrac{v_p}{v_a} = \dfrac{r_a}{r_p} = \dfrac{46{,}244}{6871} = \mathbf{6.73} The satellite moves nearly seven times faster at perigee than at apogee — and we obtained that from conservation alone, without knowing μ\mu, the period, or the shape of the orbit. This is exactly why Molniya orbits exist: the satellite loiters for hours near apogee over the northern hemisphere, then whips through perigee in minutes.
↓But there is a problem
Angular momentum has told us a great deal. The orbit lies in a fixed plane. It sweeps equal areas in equal times. Speed and radius trade off against each other in a precise ratio. But notice what it has not told us: the shape. A circle and a wildly elongated ellipse can have the same angular momentum. h⃗\vec{h} cannot distinguish between them. To pin down the shape we need a second conserved quantity — and it is the one you already met in school, wearing a sign that confuses almost everybody.