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Unit 2 · Lesson 2-2
The Orbit Equation
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The Orbit Equation

The eccentricity vector and the closed-form solution of the two-body problem

Crossing the equation of motion with h
There is a moment in this derivation where you will feel cheated. We have a nonlinear second-order vector differential equation. The honest approaches — separation of variables, series solutions, integrating factors — all fail on it. What works instead is a move that looks like sleight of hand: cross the entire equation with a vector we already know is constant, and then notice that both sides are perfect time derivatives of something. Integrate, and a new constant of integration falls out. It is a vector. Nobody put it there; it emerges. It points at perigee, its length is the eccentricity, and its existence is the reason orbits close at all instead of slowly rotating. Four lines. That is the whole cost of the two-body problem.
Watch first · presented by Shoaib Aziz
The worked example in this film is the lesson's own: r = [8000, 2000, 0] km and v = [−1.5, 7.2, 0] km/s give p = 9,213 km, e = 0.125 and ν = 20.7°, and the orbit equation hands back 8,246.2 km. The 56.3° turn of e over six laps is printed by the second lab's solution.
Line one. Take the equation of motion and cross it, on the right, with the angular momentum h⃗=r⃗×v⃗\vec{h} = \vec{r} \times \vec{v}:
r⃗¨×h⃗=−μr3 r⃗×h⃗\ddot{\vec{r}} \times \vec{h} = -\frac{\mu}{r^3}\,\vec{r} \times \vec{h}

Legal because h is a fixed vector — Unit 1 proved ḣ = 0, and that is what makes the next step work.

Line two. The left-hand side is already a derivative in disguise. Because h⃗\vec{h} is constant,
ddt(r⃗˙×h⃗)=r⃗¨×h⃗+r⃗˙×h⃗˙⏟= 0=r⃗¨×h⃗\frac{d}{dt}\left(\dot{\vec{r}} \times \vec{h}\right) = \ddot{\vec{r}} \times \vec{h} + \dot{\vec{r}} \times \underbrace{\dot{\vec{h}}}_{=\,0} = \ddot{\vec{r}} \times \vec{h}
So the left side is ddt(v⃗×h⃗)\dfrac{d}{dt}(\vec{v} \times \vec{h}). One side is done.
Line three is where the work is. Expand r⃗×h⃗=r⃗×(r⃗×v⃗)\vec{r} \times \vec{h} = \vec{r} \times (\vec{r} \times \vec{v}) with the BAC-CAB identity A⃗×(B⃗×C⃗)=B⃗(A⃗⋅C⃗)−C⃗(A⃗⋅B⃗)\vec{A} \times (\vec{B} \times \vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B}):
r⃗×(r⃗×v⃗)=r⃗ (r⃗⋅v⃗)−v⃗ (r⃗⋅r⃗)=rr˙ r⃗−r2v⃗\vec{r} \times (\vec{r} \times \vec{v}) = \vec{r}\,(\vec{r}\cdot\vec{v}) - \vec{v}\,(\vec{r}\cdot\vec{r}) = r\dot{r}\,\vec{r} - r^2\vec{v}

Using r·v = r ṙ — the dot product of position and velocity is radius times the rate of change of radius.

Substitute and divide through. The right-hand side becomes
−μr3(rr˙ r⃗−r2v⃗)=μ(v⃗r−r˙ r⃗r2)-\frac{\mu}{r^3}\left(r\dot{r}\,\vec{r} - r^2\vec{v}\right) = \mu\left(\frac{\vec{v}}{r} - \frac{\dot{r}\,\vec{r}}{r^2}\right)
and now stare at the bracket. That is exactly the quotient rule applied to r⃗/r\vec{r}/r:
ddt(r⃗r)=v⃗r−r˙ r⃗r2\frac{d}{dt}\left(\frac{\vec{r}}{r}\right) = \frac{\vec{v}}{r} - \frac{\dot{r}\,\vec{r}}{r^2}
Both sides are now derivatives of things.
ddt(v⃗×h⃗)=μ ddt(r⃗r)=μ dr^dt\frac{d}{dt}\left(\vec{v} \times \vec{h}\right) = \mu\,\frac{d}{dt}\left(\frac{\vec{r}}{r}\right) = \mu\,\frac{d\hat{r}}{dt}

The equation of motion, rearranged until it is nothing but a statement about two rates of change being equal.

Figure 2.3
1Cross with h(d²r/dt²) × h = −(μ/r³) r × hlegal because dh/dt = 02Left side is a derivative(d²r/dt²) × h = d/dt (v × h)product rule — the second term dies3Expand the triple product−(μ/r³)(r × h) = μ d/dt (r/r)BAC-CAB, then the quotient rule4Integrate oncev × h = μ (r/|r| + e)e is the constant of integrationr = p / (1 + e cos ν)
The four lines as a flow: cross with h, recognise the left side as a derivative, expand the triple product, recognise the right side as a derivative. Integrate once and the problem is over.
Line four. Integrate both sides with respect to time. The constant of integration is a vector — call it μe⃗\mu\vec{e}, for reasons that are about to become obvious:
v⃗×h⃗=μ(r^+e⃗)\vec{v} \times \vec{h} = \mu\left(\hat{r} + \vec{e}\right)

The eccentricity vector e appears as nothing more than a constant of integration. It is conserved because it is a constant of integration — that is the entire proof.

💡A third conserved quantity: the eccentricity vector
Unit 1 found two conserved quantities by physical argument: h⃗\vec{h} and ε\varepsilon. This derivation produces a third, for free, and it is the one that carries the information the other two were missing. e⃗\vec{e} is known as the Laplace–Runge–Lenz vector. It is conserved only for an exact inverse-square force. Change the exponent even slightly — say by making the central body oblate — and e⃗\vec{e} starts to rotate. That rotation has a name you already know: apsidal precession. Mercury's perihelion advances 43 arcseconds per century more than Newton predicts, and explaining that was the first observational triumph of general relativity. In Unit 7 you will compute the far larger precession that Earth's equatorial bulge inflicts on every satellite. So: orbits close because e⃗\vec{e} is fixed. Whenever a real orbit fails to close, something has broken the inverse-square law, and e⃗\vec{e} is the thing keeping score.
One rearrangement gives e⃗\vec{e} directly from a state vector, which is how you will actually compute it:
e⃗=1μ[(v2−μr)r⃗−(r⃗⋅v⃗) v⃗]\vec{e} = \frac{1}{\mu}\left[\left(v^2 - \frac{\mu}{r}\right)\vec{r} - (\vec{r}\cdot\vec{v})\,\vec{v}\right]

Everything on the right is measurable right now, at a single instant. The result does not change from one instant to the next.

Figure 2.4
rνefocusperigeeOver one orbitr — swings all the way roundν — advances 0° to 360°e — does not move at all|e| is the shape;its direction is perigee.
The eccentricity vector lies in the orbital plane, points from the focus toward perigee, and has magnitude e. It is the orbit's built-in compass needle: it remembers which way the ellipse is aimed.
Now finish. Dot both sides of v⃗×h⃗=μ(r^+e⃗)\vec{v} \times \vec{h} = \mu(\hat{r} + \vec{e}) with r⃗\vec{r}.
The left side yields to a scalar triple product identity: r⃗⋅(v⃗×h⃗)=(r⃗×v⃗)⋅h⃗=h⃗⋅h⃗=h2\vec{r} \cdot (\vec{v} \times \vec{h}) = (\vec{r} \times \vec{v}) \cdot \vec{h} = \vec{h} \cdot \vec{h} = h^2.
The right side gives μ(r+r⃗⋅e⃗)=μ(r+recos⁡ν)\mu(r + \vec{r}\cdot\vec{e}) = \mu(r + re\cos\nu), where ν\nu is defined as the angle between e⃗\vec{e} and r⃗\vec{r}.
Set them equal and solve for rr:
  r=h2/μ1+ecos⁡ν=p1+ecos⁡ν  \boxed{\;r = \frac{h^2/\mu}{1 + e\cos\nu} = \frac{p}{1 + e\cos\nu}\;}

The orbit equation. p = h²/μ is the semi-latus rectum. This is the closed-form solution to the two-body problem, and it took four lines.

Figure 2.5
pν = 90°rνfocus — not the centreperigeeapogeer = p / (1 + e cos ν)p sets the scale · e sets the shape · ν is where you are now
Every symbol, located on the picture. p sets the scale, e sets the shape, ν is where you are right now, measured from the direction e points. The focus — not the centre — is the origin.
💡True anomaly: definition and a common error
The true anomaly ν\nu is the angle at the focus, measured from perigee, in the direction of motion. It is not measured from the centre of the ellipse. It is not measured from any axis you chose. It is measured from e⃗\vec{e}, a direction the orbit itself defines — which is why the orbit equation contains no arbitrary constants at all. Notice also what ν\nu is not doing: it is not increasing uniformly in time. The satellite races through perigee and crawls through apogee, exactly as Kepler's second law demands. The orbit equation tells you the shape completely and tells you nothing whatsoever about when. That omission is the subject of Unit 5, and it is the hardest problem in the book.
📐Worked Example: state vector to shape, in one pass
A satellite is tracked at r⃗=[8000, 2000, 0]\vec{r} = [8000,\ 2000,\ 0] km, v⃗=[−1.5, 7.2, 0]\vec{v} = [-1.5,\ 7.2,\ 0] km/s. Magnitudes. r=80002+20002=8246.2r = \sqrt{8000^2 + 2000^2} = 8246.2 km, v2=1.52+7.22=54.09v^2 = 1.5^2 + 7.2^2 = 54.09 km²/s². Angular momentum. h⃗=r⃗×v⃗=[0, 0, 8000(7.2)−2000(−1.5)]=[0,0,60,600]\vec{h} = \vec{r}\times\vec{v} = [0,\ 0,\ 8000(7.2) - 2000(-1.5)] = [0, 0, 60{,}600] km²/s. Semi-latus rectum. p=h2/μ=606002/398600=9212.9p = h^2/\mu = 60600^2/398600 = 9212.9 km. Eccentricity vector. r⃗⋅v⃗=8000(−1.5)+2000(7.2)=2400\vec{r}\cdot\vec{v} = 8000(-1.5) + 2000(7.2) = 2400. v2−μ/r=54.09−48.337=5.7528v^2 - \mu/r = 54.09 - 48.337 = 5.7528. e⃗=1398600(5.7528[8000,2000,0]−2400[−1.5,7.2,0])=1398600[49,622, −5774, 0]\vec{e} = \tfrac{1}{398600}\left(5.7528[8000, 2000, 0] - 2400[-1.5, 7.2, 0]\right) = \tfrac{1}{398600}[49{,}622,\ -5774,\ 0] e⃗=[0.12449, −0.01449, 0]\vec{e} = [0.12449,\ -0.01449,\ 0], so e=0.1253e = \mathbf{0.1253}. Where is perigee? e⃗\vec{e} points 6.6° below the x-axis, so the satellite passed perigee a little while ago. Current true anomaly. cos⁡ν=e⃗⋅r⃗er=0.12449(8000)+(−0.01449)(2000)0.12533×8246.2=0.9356\cos\nu = \dfrac{\vec{e}\cdot\vec{r}}{er} = \dfrac{0.12449(8000) + (-0.01449)(2000)}{0.12533 \times 8246.2} = 0.9356, and since r⃗⋅v⃗>0\vec{r}\cdot\vec{v} > 0 the satellite is climbing, so ν=+20.67°\nu = +\mathbf{20.67°}. Check. r=p/(1+ecos⁡ν)=9213.1/(1+0.12533×0.9356)=8246.2r = p/(1 + e\cos\nu) = 9213.1/(1 + 0.12533 \times 0.9356) = 8246.2 km. Exactly the measured rr — which is the point: the shape was recovered from a single instant of tracking data.
🕰History of the [Laplace–Runge–Lenz vector](https://en.wikipedia.org/wiki/Laplace%E2%80%93Runge%E2%80%93Lenz_vector)
The vector e⃗\vec{e} is called the Laplace–Runge–Lenz vector. Jakob Hermann found it in 1710. Johann Bernoulli generalised it the same year. Laplace rediscovered it in 1799, Hamilton again in 1847, Gibbs again in 1901, and Runge put it in a textbook that Lenz happened to cite in a 1924 paper about the hydrogen atom — which is how a piece of eighteenth-century celestial mechanics ended up named after two twentieth-century physicists. The hydrogen connection is not a coincidence. The Coulomb force is also inverse-square, so the LRL vector is conserved there too, and its conservation is exactly why the energy levels of hydrogen are degenerate in a way that no other atom's are. The same algebra you just did explains a line in an emission spectrum.
Nature uses only the longest threads to weave her patterns, so each small piece of her fabric reveals the organization of the entire tapestry.
– Richard Feynman, The Character of Physical Law, 1965
↓But there is a problem
Look at what came out of four lines of algebra. r=p1+ecos⁡νr = \frac{p}{1 + e\cos\nu} Two parameters. pp sets the size. ee sets the shape. Everything else is bookkeeping. And here is the part that should stop you: the equation does not restrict ee. Nothing in the derivation said ee had to be small, or less than one, or positive. So what happens at e=1e = 1, where the denominator can reach zero and rr runs off to infinity? What happens beyond? Slice a cone with a plane, and you get a circle. Tilt the plane, and you get an ellipse. Tilt further, a parabola. Further still, a hyperbola. Four shapes, one continuous family, one parameter. Every orbit that has ever existed is somewhere on that cone.