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Unit 2 · Lesson 2-1
The Two-Body Equation of Motion
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The Two-Body Equation of Motion

Reduction to relative motion, the gravitational parameter μ and the assumptions of the model

From physical argument to a differential equation
Everything in Unit 1 was earned by physical reasoning and good instinct. Balance gravity against the centripetal requirement and out drops circular speed. Notice a central force exerts no torque and out drops a fixed plane. Add up two energies and out drops vis-viva. Good instinct is how the subject was discovered. It is not how the subject is known. So here is the promise from the end of the last lesson, stated as bluntly as it deserves: give me hh and ε\varepsilon and I will tell you the exact shape of the path — and prove that no other shape is possible. Not a good approximation. Not for special cases. Every object, every orbit, every central-gravity problem in the universe. A claim that strong needs a proof, and a proof needs something to start from. This lesson builds the starting line: one vector differential equation from which every result in this course — Kepler's three laws, the conic sections, Hohmann transfers, Lambert targeting, the lot — follows as a consequence. It also builds something less comfortable: an explicit list of the things we are about to pretend are true, each one signed and dated for demolition in Unit 7.
Watch first · presented by Shoaib Aziz
The barycentre in this film is drawn to scale — 4,700 km from the Earth's centre, inside the planet — and the lab segment's numbers (a ≈ 7990 km, T ≈ 7108 s, a closure error of 0.0000 km, the Moon's 1.21%) are printed by this lesson's own lab solutions.
Start honestly. There are two bodies, and Newton's third law says they pull on each other. Write both, in some inertial frame, with position vectors R⃗1\vec{R}_1 and R⃗2\vec{R}_2 measured from an arbitrary origin:
MR⃗¨1=GMmr2r^,mR⃗¨2=−GMmr2r^,r⃗=R⃗2−R⃗1M\ddot{\vec{R}}_1 = \frac{GMm}{r^2}\hat{r}, \qquad m\ddot{\vec{R}}_2 = -\frac{GMm}{r^2}\hat{r}, \qquad \vec{r} = \vec{R}_2 - \vec{R}_1

Two coupled second-order vector equations — twelve scalar unknowns. Nobody wants to solve this.

Twelve unknowns, coupled. But notice what we actually care about. Nobody has ever asked where the Earth is in some absolute cosmic sense. The question is always where the satellite is relative to the Earth. So change variables: stop tracking two positions and start tracking the vector between them.
Divide each equation by its own mass and subtract:
r⃗¨=R⃗¨2−R⃗¨1=−G(M+m)r3 r⃗\ddot{\vec{r}} = \ddot{\vec{R}}_2 - \ddot{\vec{R}}_1 = -\frac{G(M+m)}{r^3}\,\vec{r}

The relative equation of motion. Twelve unknowns collapsed to three — and the two masses appear only as their sum.

Figure 2.1
What is actually happeningWhat we solve insteadrMmbarycentreboth bodies move — twelve unknownssubtractrμone relative motion — three unknownscentre fixed, μ = G(M + m)
Two bodies orbiting their common barycentre (left) become one fictitious body orbiting a fixed centre (right). The reduction is exact — no approximation has been made yet. Earth wobbles about the barycentre too; for a satellite the wobble is smaller than an atom.
💡Why μ, and not G and M separately
For an Earth satellite, mm is perhaps 10310^3 kg and MM is 6×10246 \times 10^{24} kg. Their sum is MM to twenty-one decimal places, so we write μ=GM\mu = GM and call it the gravitational parameter. This is not laziness — it is better data. Nobody can measure GG to more than about four significant figures; it is the worst-measured constant in physics. But the product GMGM is measured superbly, because it is exactly what orbits respond to. Track a spacecraft for a few weeks and you get μ⊕=398,600.4418 km3/s2\mu_{\oplus} = 398{,}600.4418\ \text{km}^3/\text{s}^2 — eleven digits. So when you see μ\mu instead of GMGM, read it as: this is a number we actually know.
  r⃗¨=−μr3 r⃗  \boxed{\;\ddot{\vec{r}} = -\frac{\mu}{r^3}\,\vec{r}\;}

The two-body equation of motion. Everything else in this course is a consequence of this line.

Figure 2.2
rμ/r²2rμ/4r²twice as far out, a quarter the pull — and the arrow is drawn to scaled²r/dt² = −μ r / r³direction — always at the focusmagnitude — falls as 1/r²no mass — the satellite's cancels
Reading the equation as a picture: r̂ = r/r sets the direction, μ/r² sets the magnitude, and the minus sign points the acceleration back down the line to the focus. Written as μr/r³ rather than μr̂/r², the two jobs are done by one expression.
💡Reading the equation before solving it
Three things are already visible in r⃗¨=−μr⃗/r3\ddot{\vec{r}} = -\mu\vec{r}/r^3, before any algebra: It is a central force. The acceleration is parallel to r⃗\vec{r}. So r⃗×r⃗¨=0\vec{r} \times \ddot{\vec{r}} = 0, no torque, h⃗\vec{h} conserved — the entire content of lesson 1-3 sits in the fact that r⃗\vec{r} appears on the right-hand side. The satellite's mass has vanished. A cubesat and a space station on identical initial conditions fly identical orbits. Gravity accelerates everything the same, which is Galileo's tower dressed for space. It is nonlinear. The 1/r31/r^3 makes it so, and nonlinearity is why three bodies destroy everything. Two bodies is the last problem in celestial mechanics that surrenders cleanly, and we are about to take full advantage.
Now the uncomfortable part. That equation is not the truth. It is a model, and it is built on four assumptions, each one false in a specific and interesting way.
AssumptionWhat is actually trueWhere it breaksRepaired in
Only two bodies existThe Sun and Moon pull on every Earth satelliteAbove ~15,000 km altitude the Moon's tug is no longer ignorable7-3
Both bodies are spherical point massesEarth is 21 km fatter at the equatorRotates every orbital plane on Earth, continuously7-1
No forces except gravityThere is measurable air up to ~1000 km, and sunlight carries momentumDrag dominates below ~500 km; light pressure dominates for balloons7-2, 7-3
Mass is constantEvery burn throws mass overboardAny time an engine is lit6-1
⚠How accurate the two-body model is
Having just called it a lie, here is the defence: propagate a typical LEO satellite with pure two-body motion and after a full day you are wrong by roughly a kilometre in position. Out of an orbit forty-two thousand kilometres around. That is a model that is 99.998% right while being 100% wrong in principle — and it is why the two-body problem is not merely a teaching exercise. Real operations run a two-body core with perturbations layered on top as corrections. You are not learning a toy version. You are learning the load-bearing part.
Essentially, all models are wrong, but some are useful.
– George E. P. Box, Empirical Model-Building and Response Surfaces, 1987
One more thing before we solve it. A computer cannot integrate a second-order equation directly; integrators eat first-order systems. The standard move is to stack position and velocity into a single six-element state vector and let the derivative do both jobs:
x⃗=[r⃗v⃗],x⃗˙=[v⃗−μr3r⃗]\vec{x} = \begin{bmatrix}\vec{r}\\ \vec{v}\end{bmatrix}, \qquad \dot{\vec{x}} = \begin{bmatrix}\vec{v}\\ -\dfrac{\mu}{r^3}\vec{r}\end{bmatrix}

The same physics, reshaped for a machine. The top three rows say "the derivative of position is velocity" — free. The bottom three carry all the gravity.

🕰1687: what the Principia left open
Newton published the Principia having proved that an inverse-square force produces conic-section orbits. What he could not do was write the answer as a formula in time — and neither can anyone else, as Unit 5 will show in painful detail. He also could not solve three bodies. Nor could Euler, Lagrange, Laplace, or Poincaré, who in 1890 finally explained why by discovering that the three-body problem is chaotic — an answer that won him a prize and founded a branch of mathematics. So the equation on this page is the last one in celestial mechanics that has a clean, exact, complete solution. That is worth knowing before we spend the next lesson extracting it.
One will be struck by the complexity of this figure, which I do not even try to draw.
– Henri Poincaré, on the tangle of trajectories in the three-body problem, Les Méthodes nouvelles de la mécanique céleste, vol. III, 1899
↓But there is a problem
Look at what we are holding. r⃗¨=−μr⃗/r3\ddot{\vec{r}} = -\mu\vec{r}/r^3 is a second-order, nonlinear, coupled vector differential equation. Equations of that description essentially never have closed-form solutions. Add one more body and this one doesn't either — the three-body version has been open for three hundred years and will stay open. And yet this one gives up completely. Not numerically. Not approximately. There is an exact, finite, closed-form answer, and it is reached by a piece of vector algebra so slick that the first time you see it, it looks like cheating. Next lesson, we cheat.