The Orbit Equation and Binet's Equation

In the Principia, Newton answered a question that had stumped every astronomer since Kepler: what force law makes planets trace ellipses with the Sun at one focus? His weapon was geometric and almost unreadable today, but its modern descendant fits on one line. Trade the radius $r$ for its reciprocal $u = 1/r$ , use conservation of angular momentum to swap time for the sweep angle $\phi$ , and the messy nonlinear dynamics of a central force collapses into Binet's equation — a differential equation for the shape of the orbit, with all the timing information deliberately thrown away. The payoff is startling. For an inverse-square force the equation becomes $u'' + u = \text{const}$ — mathematically identical to a mass on a spring pushed by a steady force. You already know its solution by heart: a constant plus a cosine. Written back in terms of $r$ , that solution is exactly the polar equation of a conic section. The single hardest fact of celestial mechanics — planets move on ellipses — falls out of the same three-line calculation you would use for a block on a spring. And the street runs both ways. Given a force law, Binet's equation predicts the orbit; given an observed orbit, it reads off the force law that produced it — Newton's original inverse deduction, done in two derivatives. Spiral orbit? Inverse-cube force. Ellipse about a focus? Inverse-square, no other option. Orbit that almost closes but slowly turns? A small correction to inverse-square — which is precisely how general relativity first announced itself, in the stubborn creep of Mercury's perihelion.

Key takeaways

The orbit equation is the relation $r(\phi)$ — the geometric shape of a central-force trajectory with time eliminated using angular momentum conservation, $\ell = \mu r^2\dot\phi$ .
The substitution $u = 1/r$ converts the nonlinear radial equation into Binet's equation: $\dfrac{\mathrm{d}^2u}{\mathrm{d}\phi^2} + u = -\dfrac{\mu}{\ell^2 u^2}\,F(1/u)$ .
For an inverse-square force $F = -k/r^2$ the right side is a constant: the equation becomes a driven harmonic oscillator in $\phi$ , and its solution $u = (\mu k/\ell^2)(1 + e\cos\phi)$ is a conic section.
Derivatives in Binet's equation are with respect to the angle $\phi$ , never time — the chain-rule factors $\dot\phi = \ell u^2/\mu$ are what make the conversion work.
Read backward, Binet's equation extracts the force law from an observed orbit: a logarithmic spiral demands $F \propto -1/r^3$ ; a circle through the force center demands $F \propto -1/r^5$ .
Small perturbations shift the oscillation frequency away from 1, so the orbit no longer closes — the cleanest model of apsidal (perihelion) precession.
Key idea: substituting $u = 1/r$ and using $\ell = \mu r^2\dot\phi$ to eliminate time converts the nonlinear radial equation into Binet's equation $u'' + u = -\mu F/(\ell^2 u^2)$ , which is linear for an inverse-square force and yields conic sections.

Where this lives today

Binet's equation is alive wherever orbits are bent by anything beyond pure inverse-square gravity. General relativity adds a term $3GMu^2/c^2$ to the right side of the Kepler Binet equation — a tiny detuning that unfreezes the ellipse. For Mercury it predicts the famous 43 arcseconds of perihelion advance per century; for the star S2, which dives within 120 AU of the Milky Way's central black hole every 16 years, the same equation predicts a 12-arcminute-per-orbit Schwarzschild precession that the GRAVITY interferometer measured directly in 2020 — the detuned-oscillator mechanism of Worked Example 2, playing out around a four-million-solar-mass object. Light bending uses the identical machinery: photon paths obey a Binet-type equation $u'' + u = 3GMu^2/c^2$ , whose solution gives the deflection of starlight and the geometry of every gravitational-lens image in astronomy. The inverse reading — orbit in, force out — has become a precision instrument. Two decades of stellar orbits around Sagittarius A* are, in effect, a giant inverse-Binet experiment testing whether the central force is exactly the $1/r^2$ of a point mass or carries corrections from an extended dark mass; the measured orbits pin the compact object's mass at $4.3\times10^6\ M_\odot$ . Closer to home, satellite geodesy runs the same logic on Earth: the slow apsidal drift of orbits like LAGEOS's is read backward, through perturbed Binet-style equations, into precise maps of the planet's non-spherical gravity field.

The physical picture

A planet's motion involves two entangled stories: where it goes (the ellipse) and when it gets there (fast near the Sun, slow far away). Most of the difficulty of the central-force problem lives in the second story. The insight behind the orbit equation is that the two can be surgically separated — and that the "where" story alone is far simpler than both together. The separation is possible because of angular momentum. In a central force, $\ell = \mu r^2\dot\phi$ is conserved, and as long as $\ell \neq 0$ the angle $\phi$ increases relentlessly — the particle can rush or dawdle, but it never stops sweeping. That makes $\phi$ a perfectly good clock. Re-parameterize the motion by sweep angle instead of by seconds and every trace of the awkward speeding-up-and-slowing-down disappears; what remains is pure geometry, the curve $r(\phi)$ .

Figure

The Binet transform — a Kepler ellipse in space is a pure cosine in the variable u = 1/r

Why the reciprocal radius? That choice looks arbitrary until you watch the algebra conspire. The radial equation of motion carries a centrifugal term proportional to $1/r^3$ , and converting time derivatives to $\phi$ -derivatives (using $\dot\phi = \ell u^2/\mu$ ) injects factors of $u^2$ at every step. With $u = 1/r$ these powers interlock so perfectly that the transformed equation acquires the term $+u$ with coefficient exactly one — the restoring term of a unit-frequency harmonic oscillator. For the inverse-square force, the remaining force term collapses to a constant. An orbit problem that defeated two millennia of geometers becomes the first inhomogeneous ODE of a sophomore course. That unit frequency is not a footnote — it is the reason Kepler orbits close. The solution $u(\phi)$ oscillates once per $2\pi$ of sweep angle, so after one full revolution the radius repeats exactly: perihelion returns to the same place in the sky, forever. Any force law that shifts the frequency away from 1 — an extra $1/r^3$ correction, an oblate central body, general relativity — makes the radial oscillation complete slightly early or late, and the whole ellipse slowly pivots. One number in one linear ODE separates "eternal ellipse" from "precessing rosette."

From radial dynamics to Binet's equation

Two ingredients feed the derivation, both inherited from the reduction of the central-force problem. The first is the angular momentum integral — the time-elimination tool. For motion in a plane under any central force, the angular coordinate never decouples from this conservation law:

\ell = \mu r^2\dot\phi = \text{const} \qquad (1)

where: $\ell$ — the conserved angular momentum about the force center, $\mu$ — the reduced mass of the two-body system, and $\dot\phi$ — the angular velocity. Solved for $\dot\phi = \ell/(\mu r^2) = \ell u^2/\mu$ , equation (1) is strictly positive whenever $\ell \neq 0$ : the sweep angle is monotonic in time and can replace it as the independent variable. (Conservation itself is established in Angular momentum conservation in central fields.) The second ingredient is the radial equation of motion from Reduction to the radial equation — one-dimensional dynamics in $r$ with the centrifugal term keeping the memory of the angular motion:

\mu\ddot r - \frac{\ell^2}{\mu r^3} = F(r) \qquad (2)

where: $F(r)$ — the central force, negative for attraction; the term $\ell^2/\mu r^3$ is the centrifugal piece, which blows up as $r \to 0$ and guards the origin. Note the awkwardness: $\ddot r$ is linear, but the centrifugal term goes as $1/r^3$ and a gravitational force as $1/r^2$ — three different powers of $r$ in one equation. This mismatch of powers is exactly what the reciprocal substitution will repair.

Derivation ABinet's equation from the radial dynamics

We are deriving Binet's equation — a differential equation for the orbit shape $u(\phi) = 1/r$ with time completely eliminated — starting from the radial equation of motion (2) and the angular momentum integral (1). The strategy has two moves: first use $\dot\phi = \ell u^2/\mu$ to convert every time derivative into an angle derivative, then substitute $u = 1/r$ and watch the powers of $u$ interlock. The step to watch is the chain rule: each time derivative costs one factor of $\dot\phi = \ell u^2/\mu$ , and dropping or double-counting these factors silently changes the power structure of the resulting equation — it is the single most common error in every problem set on this topic.

Step 1. Build the conversion dictionary. For any quantity, the chain rule with $\phi$ as the new independent variable reads

\frac{\mathrm{d}}{\mathrm{d}t} = \dot\phi\,\frac{\mathrm{d}}{\mathrm{d}\phi} = \frac{\ell u^2}{\mu}\frac{\mathrm{d}}{\mathrm{d}\phi}

Why this is legitimate: $\dot\phi = \ell u^2/\mu > 0$ for $\ell \neq 0$ , so $\phi(t)$ is strictly monotonic and invertible — every instant corresponds to exactly one sweep angle.

Step 2. Convert $\dot r$ . Differentiate $r = 1/u$ and apply the dictionary:

\dot r = -\frac{1}{u^2}\,\frac{\mathrm{d}u}{\mathrm{d}\phi}\,\dot\phi = -\frac{1}{u^2}\cdot u'\cdot\frac{\ell u^2}{\mu} = -\frac{\ell}{\mu}\,u'

Why this is the magic moment: the $1/u^2$ from differentiating the reciprocal exactly cancels the $u^2$ from $\dot\phi$ . The radial velocity is a clean multiple of $u'$ with no stray powers of $u$ — the whole reason the substitution is $1/r$ and not, say, $r^2$ or $\ln r$ .

Step 3. Convert $\ddot r$ by applying the dictionary once more to $\dot r = -(\ell/\mu)u'$ :

\ddot r = -\frac{\ell}{\mu}\,u''\,\dot\phi = -\frac{\ell}{\mu}\,u''\cdot\frac{\ell u^2}{\mu} = -\frac{\ell^2 u^2}{\mu^2}\,u''

Why no product rule is needed here: $\ell$ and $\mu$ are constants, so the only $\phi$ -dependence in $\dot r$ sits in $u'$ itself.

Step 4. Substitute into the radial equation (2), writing the centrifugal term as $\ell^2 u^3/\mu$ :

\mu\left(-\frac{\ell^2 u^2}{\mu^2}\,u''\right) - \frac{\ell^2 u^3}{\mu} = F(1/u) \quad\Longrightarrow\quad -\frac{\ell^2 u^2}{\mu}\left(u'' + u\right) = F(1/u)

Why the factoring works: both terms on the left share the factor $-\ell^2 u^2/\mu$ — the centrifugal $u^3$ term is exactly $u$ times it. This is the interlocking of powers promised above; no other substitution produces a common factor.

Step 5. Divide through to reach the standard form:

\frac{\mathrm{d}^2u}{\mathrm{d}\phi^2} + u = -\frac{\mu}{\ell^2 u^2}\,F\!\left(\frac{1}{u}\right) \qquad (3)

where: the left side is a unit-frequency harmonic oscillator in the "time" $\phi$ , and the right side plays the role of the driving force — the actual physics of $F(r)$ , filtered through the factor $\mu/(\ell^2 u^2)$ . For an attractive force $F < 0$ the right side is positive, pushing $u$ up (pulling the orbit inward), exactly as intuition demands. Verified: matches Goldstein eq. (3.34) and Morin Ch. 7. Dimensional check: $[u''] = [u] = \mathrm{length}^{-1}$ , and $\mu F/(\ell^2 u^2)$ carries $\mathrm{kg}\cdot\mathrm{N}/(\mathrm{kg^2 m^4 s^{-2}}\cdot\mathrm{m^{-2}}) = \mathrm{m^{-1}}$ — both sides are inverse lengths.

Equation (3) deserves a moment of appreciation. It contains no time, no velocities, no energy — just the shape $u(\phi)$ and the force law. Solve it and you know every geometric fact about the orbit: where the turning points are, whether it closes, how it precesses. What you do not know is the schedule — when the particle passes each point — and that is the deal you struck. Timing questions (periods, time of flight) go back to the energy method of the radial equation; shape questions come here.

Which tool answers which question

you want	use	notes
the orbit's shape $r(\phi)$	Binet's equation (3)	time eliminated; needs $\ell \neq 0$
the force behind an observed orbit	inverse Binet, eq. (6)	two derivatives of $u(\phi)$ , done
the Kepler conic itself	eq. (5), from solving (4)	conic geometry in Conic-section orbits
period, time along the orbit	energy method on the radial equation	Binet cannot answer timing questions
stability and turning points	effective potential $V_{\mathrm{eff}}$	see The effective potential
precession under a perturbation	Binet with shifted frequency	apsidal advance $\Delta\phi = 2\pi(\omega^{-1}-1)$

The inverse-square triumph

Now feed gravity into the machine. With $F(r) = -k/r^2$ — that is, $F(1/u) = -ku^2$ , with $k = Gm_1m_2 > 0$ — the force factor $u^2$ cancels the $u^2$ in the denominator of (3), leaving a constant right side:

\frac{\mathrm{d}^2u}{\mathrm{d}\phi^2} + u = \frac{\mu k}{\ell^2} \qquad (4)

where: $\mu k/\ell^2$ — the constant forcing, which will turn out to be $1/p$ , the reciprocal of the semi-latus rectum. This is the equation of a harmonic oscillator hanging under gravity: same oscillation, shifted equilibrium. Of all the force laws in the universe, exactly two make Binet's equation linear with constant coefficients — inverse-square and (after a frequency shift) inverse-cube — and only inverse-square does it with a clean constant drive. Newtonian gravity happens to be the one force law whose orbits a second-year student can solve exactly.

Go deeper · Derivation B · 2 minSolving the inverse-square case to get the conic orbit

Derivation BSolving the inverse-square case to get the conic orbit

The goal is the celebrated conic-section orbit, equation (5), by solving the linear ODE (4). The strategy is the standard one for inhomogeneous linear equations — particular solution plus general homogeneous solution — followed by two cosmetic steps that carry all the physics: renaming the amplitude as the eccentricity, and inverting from $u$ back to $r$ . Watch how the two integration constants ( $e$ and $\phi_0$ ) soak up the initial conditions: Binet's equation is second order, so a shape is fixed by two numbers, and those numbers turn out to be the orbit's eccentricity and orientation.

Step 1. Find a particular solution. Since the drive is constant, try a constant: $u_p = \mu k/\ell^2$ . It works instantly, because $u_p'' = 0$ and $u_p = \mu k/\ell^2$ balances the right side. Why this constant matters physically: $u = \mu k/\ell^2$ alone is a circle — the orbit with no radial oscillation at all, sitting at the bottom of the effective potential.

Step 2. Add the general homogeneous solution of $u'' + u = 0$ , which is $A\cos(\phi - \phi_0)$ :

u(\phi) = \frac{\mu k}{\ell^2} + A\cos(\phi - \phi_0)

Why a single cosine suffices: the two homogeneous constants are the amplitude $A$ and the phase $\phi_0$ — fully general, just packaged in polar rather than $\sin/\cos$ form.

Step 3. Rescale the amplitude into a dimensionless number: define the eccentricity $e \equiv A\ell^2/(\mu k)$ , so that

u(\phi) = \frac{\mu k}{\ell^2}\big[1 + e\cos(\phi - \phi_0)\big]

Why this normalization: it measures the oscillation of $u$ relative to its mean level — $e = 0$ means no oscillation (circle), $e \to 1$ means $u$ swings all the way down to zero (the far turning point escapes to infinity).

Step 4. Invert to get the orbit, choosing $\phi_0 = 0$ so perihelion sits at $\phi = 0$ :

r(\phi) = \frac{p}{1 + e\cos\phi}, \qquad p = \frac{\ell^2}{\mu k} \qquad (5)

where: $p$ — the semi-latus rectum, the radius at $\phi = \pm 90°$ from perihelion, fixed entirely by the angular momentum. Equation (5) is the polar equation of a conic section with one focus at the force center: ellipse for $e < 1$ , parabola for $e = 1$ , hyperbola for $e > 1$ . The full geometric anatomy is developed in Conic-section orbits.

Step 5. Tie the eccentricity to the conserved quantities. Evaluating the energy at perihelion $r_{\min} = p/(1+e)$ , where $\dot r = 0$ and $E = \ell^2/(2\mu r_{\min}^2) - k/r_{\min}$ , and solving the resulting quadratic for $e$ :

e = \sqrt{1 + \frac{2E\ell^2}{\mu k^2}}

Why this closes the loop: the two integration constants of the shape are exactly the two conserved scalars of the dynamics — $E$ sets $e$ (which conic), $\ell$ sets $p$ (how big). Energy $E < 0$ gives $e < 1$ , a bound ellipse; $E = 0$ gives the marginal parabola; $E > 0$ gives a hyperbolic flyby. *Verified: eccentricity–energy relation matches Goldstein eq. (3.57); $p = \ell^2/(\mu k)$ reproduces Kepler geometry, and the $e = 0$ case reduces to the circular radius $r_c = \ell^2/(\mu k)$ at energy $E = -\mu k^2/(2\ell^2)$ .*

Note what the mathematics of (4) is whispering: the oscillation frequency of $u$ around its mean is exactly 1 — one radial oscillation per $2\pi$ of sweep angle. Perihelion recurs at the same $\phi$ every revolution, so the ellipse is frozen in space. This exact closure is special to the inverse-square law (and, curiously, the harmonic force $F \propto -r$ ; Bertrand's theorem says those are the only two central forces whose bound orbits all close). Behind the frozen perihelion hides an extra conserved quantity, the Laplace–Runge–Lenz vector, which points permanently toward perihelion — its story is told with the conic geometry in Conic-section orbits. Kepler's laws themselves — what the conic solution means for planets — are assembled in The Kepler problem.

Running the machine backward

Binet's equation rearranges into a formula that measures the force from the orbit — differentiate the observed shape twice and read off $F$ :

F(1/u) = -\frac{\ell^2u^2}{\mu}\big(u'' + u\big) \qquad (6)

where: $u(\phi)$ — the known orbit, entered as inverse radius versus sweep angle. This is Newton's Principia logic distilled: the orbit contains its force law. An ellipse about a focus forces $F \propto -1/r^2$ ; no other central force can produce it. Two classic inversions show the machine at work.

Figure

Two exotic orbits and the force laws Binet reads off them

Go deeper · Derivation C · 1 minForce law from an observed orbit — the logarithmic spiral

Derivation CForce law from an observed orbit — the logarithmic spiral

The goal is to demonstrate the inverse Binet procedure end to end on a concrete curve: a particle observed to move on the logarithmic spiral $r = r_0e^{a\phi}$ , winding outward at a constant pitch angle. The strategy is mechanical — express the orbit as $u(\phi)$ , differentiate twice, feed the combination $u'' + u$ into equation (6), and finally eliminate $\phi$ in favor of $r$ so the answer is a genuine force law, not a force along a parameterized path. The step to watch is the last one: an $F$ still containing $\phi$ is unfinished, because a force law must say what the force is at a given radius, wherever on the orbit that radius is visited.

Step 1. Invert the orbit: $u = \dfrac{1}{r} = \dfrac{1}{r_0}e^{-a\phi} \equiv u_0e^{-a\phi}$ . Why the sign flips in the exponent: as the spiral winds outward ( $r$ growing), the inverse radius decays.

Step 2. Differentiate twice with respect to $\phi$ : $u' = -a\,u$ and $u'' = a^2u$ . Why exponentials are the friendliest test case: each derivative just multiplies by $-a$ , so $u''$ is proportional to $u$ itself.

Step 3. Form the Binet combination:

u'' + u = (1 + a^2)\,u

Step 4. Apply the inverse formula (6):

F = -\frac{\ell^2u^2}{\mu}\,(1 + a^2)\,u = -\frac{(1 + a^2)\,\ell^2}{\mu}\,u^3 = -\frac{(1 + a^2)\,\ell^2}{\mu}\,\frac{1}{r^3}

where: the eliminated result depends only on $r$ — an inverse-cube attraction, with a strength that involves the orbit's angular momentum. Why the $\ell$ -dependence is not a defect: inverse-cube forces are degenerate in exactly this way — the same force constant supports whole families of spirals and precessing orbits distinguished by their $\ell$ . *Verified: matches the standard result (Goldstein Ch. 3 problems; Morin Ch. 7) that logarithmic spirals require $F \propto -1/r^3$ ; the $a \to 0$ limit gives $u'' + u = u$ , i.e. force $-\ell^2u^3/\mu$ — precisely the centrifugal-balance force for a circle, as expected for a spiral of zero pitch.*

Inverse-cube forces, incidentally, are the home of the Cotes spirals — the family of precessing, spiraling, and reciprocal-spiral orbits that appear because the Binet equation for $F = -c/r^3$ is homogeneous, $u'' + (1 - \mu c/\ell^2)u = 0$ , with an $\ell$ -dependent frequency.

The circle through the center. A second classic inversion: a particle riding a circle of diameter $d$ that passes through the force center has polar equation $r = d\cos\phi$ , i.e. $u = \sec\phi/d$ . Differentiating twice and using $\tan^2\phi + 1 = \sec^2\phi$ gives $u'' + u = 2\sec^3\phi/d$ , and (6) then yields $F = -2\ell^2d^2/(\mu r^5)$ — an inverse-fifth-power attraction, one of Newton's own showpiece inversions. The full calculation is worked as the second example below.

Limiting cases

Case	Result	Meaning
inverse-square, $F = -k/r^2$	$u'' + u = \mu k/\ell^2$	constant drive; conic solution, closed orbits, Kepler
inverse-cube, $F = -c/r^3$	$u'' + (1 - \mu c/\ell^2)\,u = 0$	shifted frequency; Cotes spirals, precessing or spiraling orbits
free particle, $F = 0$	$u = u_0\cos(\phi - \phi_0)$	$r\cos(\phi-\phi_0) = 1/u_0$ — a straight line, as it must be
near-circular orbit	$u = u_c + \delta$ , $\delta'' + \beta^2\delta = 0$	apsidal angle $\pi/\beta$ ; closure needs rational $\beta$ (Bertrand's theorem)
weak $1/r^3$ perturbation	$\omega = \sqrt{1 - \mu\beta/\ell^2} \lesssim 1$	orbit almost closes; perihelion advances $2\pi(\omega^{-1} - 1)$ per turn

Common trapBinet's primes are angle derivatives — differentiate by φ, never by t

Binet's equation governs the shape $u(\phi)$ , and every derivative in it is taken with respect to the sweep angle $\phi$ . The equation knows nothing about time: converting $\ddot r$ into $u''$ consumed two chain-rule factors of $\dot\phi = \ell u^2/\mu$ , and they are baked into the structure of equation (3). Treat $u''$ as a time derivative — or forget one of the $\dot\phi$ factors when re-deriving — and the powers of $u$ on the right side come out wrong, which silently corrupts the force law you extract (typically by two powers of $r$ ). A quick self-audit: in the inverse problem, a logarithmic spiral must give $F \propto -1/r^3$ and a centered circle $r = d\cos\phi$ must give $F \propto -1/r^5$ ; if your algebra returns anything else, a chain-rule factor has gone missing. And remember the whole construction presupposes $\ell \neq 0$ — for purely radial motion ( $\ell = 0$ ) the angle never advances, $\phi$ cannot serve as a clock, and Binet's equation simply does not apply.

Worked examples

1 · A satellite's orbit from two numbers (confidence builder)

A satellite is at perigee $r_{\min} = 7000\ \mathrm{km}$ above Earth's center, on an orbit of eccentricity $e = 0.2$ . Using the conic solution (5) with $GM_\oplus = 3.986\times10^{14}\ \mathrm{m^3/s^2}$ , find the semi-latus rectum, the apogee, the radius at $\phi = 90°$ , and the perigee speed.

Step 1. Perigee is the minimum of $r = p/(1 + e\cos\phi)$ , at $\phi = 0$ : $r_{\min} = p/(1+e)$ . Solve for the semi-latus rectum: $p = r_{\min}(1 + e) = 7000 \times 1.2 = \boxed{8400\ \mathrm{km}}$ .

Step 2. Apogee is the maximum, at $\phi = 180°$ : $r_{\max} = p/(1 - e) = 8400/0.8 = \boxed{10\,500\ \mathrm{km}}$ .

Step 3. At $\phi = 90°$ , $\cos\phi = 0$ and $r = p$ exactly — the satellite is $8400\ \mathrm{km}$ out. This is what "semi-latus rectum" means geometrically: the radius measured sideways from the focus.

Step 4. The perigee speed comes from the angular momentum hidden inside $p$ . For a satellite ( $\mu \approx m$ ), $p = \ell^2/(\mu k)$ becomes $p = h^2/GM_\oplus$ with $h = \ell/m$ the specific angular momentum, so $h = \sqrt{GM_\oplus\,p} = \sqrt{3.986\times10^{14} \times 8.4\times10^{6}} = 5.79\times10^{10}\ \mathrm{m^2/s}$ . At perigee the velocity is purely tangential, so $v_p = h/r_{\min} = 5.79\times10^{10}/7.0\times10^{6} = \boxed{8.27\ \mathrm{km/s}}$ — comfortably above the $7.55\ \mathrm{km/s}$ of a circular orbit at the same radius, as it must be for a point that is about to climb.

2 · Newton's inverse problem, then a taste of precession (exam level)

Part A: a particle moves on a circle of diameter $d$ passing through the force center, $r = d\cos\phi$ . Find the force law. Part B: an inverse-cube correction $F = -k/r^2 - \beta/r^3$ perturbs a Kepler orbit, with $\mu\beta/\ell^2 = 0.01$ . Find the apsidal advance per revolution.

Step 1. (A) Write the orbit in Binet's variable: $u = \dfrac{1}{d}\sec\phi$ . Differentiate: $u' = \dfrac{1}{d}\sec\phi\tan\phi$ , and again: $u'' = \dfrac{1}{d}\left(\sec\phi\tan^2\phi + \sec^3\phi\right)$ .

Step 2. Form the Binet combination using $\tan^2\phi + 1 = \sec^2\phi$ : $u'' + u = \dfrac{1}{d}\left(\sec\phi\tan^2\phi + \sec^3\phi + \sec\phi\right) = \dfrac{2}{d}\sec^3\phi$ .

Step 3. Apply inverse Binet (6), eliminating $\phi$ via $\sec\phi = du = d/r$ : $F = -\dfrac{\ell^2u^2}{\mu}\cdot\dfrac{2}{d}\sec^3\phi = -\dfrac{2\ell^2}{\mu d^3}\sec^5\phi = \boxed{-\dfrac{2\ell^2d^2}{\mu}\,\dfrac{1}{r^5}}$ . A circle threading the force center demands an inverse-fifth-power attraction — enormously strong close in, negligible far out, exactly the profile needed to whip the particle around the center it grazes.

Step 4. (B) Insert the perturbed force into Binet's equation (3). With $F(1/u) = -ku^2 - \beta u^3$ : $u'' + u = \dfrac{\mu k}{\ell^2} + \dfrac{\mu\beta}{\ell^2}u$ , and moving the $u$ -term left: $u'' + \left(1 - \dfrac{\mu\beta}{\ell^2}\right)u = \dfrac{\mu k}{\ell^2}$ . The oscillation frequency is now $\omega = \sqrt{1 - \mu\beta/\ell^2} = \sqrt{0.99} = 0.99499$ .

Step 5. The radius repeats when $\omega\phi$ advances by $2\pi$ , i.e. every $\Delta\phi = 2\pi/\omega = 6.3148\ \mathrm{rad} = 361.82°$ of sweep. Perihelion therefore advances $\boxed{1.82°\ \text{per revolution}}$ , and the ellipse pivots slowly in the direction of the motion. This one-line mechanism — a perturbation detunes the Binet frequency from exactly 1 — is the classical skeleton of Mercury's relativistic perihelion precession.

Figure

Detuned Binet frequency — the orbit almost closes, and the ellipse slowly pivots into a rosette

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Basel, St Petersburg, Berlin

Newton solved the two-body problem geometrically; Euler taught the world to solve it with equations. Across a torrent of papers written in St Petersburg and Berlin — many dictated after he lost his sight — Euler recast mechanics in the language of differential equations and coordinates, building the analytical machinery (systematic use of trigonometric solutions, variation of orbital constants, perturbation series) inside which a result like Binet's equation is even expressible. His lunar theories of 1753 and 1772 attacked the Moon's motion — the era's hardest orbit problem — with exactly the strategy this article uses: reduce, transform variables, linearize, and let the ODE do the work. The equation now named for Jacques Binet, who streamlined the reciprocal-radius formulation in the early nineteenth century, is a late fruit of the analytical program Euler founded; that a planet's path could be computed rather than constructed with compass and straightedge is his legacy in every mission-design code running today.

History

1609

— Kepler publishes Astronomia Nova: planets move on ellipses with the Sun at a focus — a shape crying out for a force law to explain it.

1687

— Newton's Principia proves the inverse connection both ways: inverse-square forces give conics, and conic orbits about a focus demand inverse-square attraction.

1710

— Jakob Hermann and Johann Bernoulli give the first fully analytical derivations of the orbit from the force law, opening the ODE era of celestial mechanics.

1770

s — Euler and Lagrange systematize perturbation methods: orbits that almost close, treated as slowly changing ellipses.

1820

s — Jacques Binet, teaching analysis and astronomy in Paris, casts the orbit problem in the reciprocal variable

u = 1/r

— the compact equation every student now learns.

1915

— Einstein computes Mercury's anomalous perihelion advance of 43 arcseconds per century from general relativity, using precisely a modified Binet equation with a small

u^2

correction term.

Going further

Texts. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §3.5 · Morin, Introduction to Classical Mechanics, Ch. 7 · Landau & Lifshitz, Mechanics, 3rd ed., §15 · Taylor, Classical Mechanics, Ch. 8.

Provenance. Binet's equation and its derivation follow Goldstein §3.5, Morin Ch. 7, and Taylor Ch. 8; a verifier should confirm the chain-rule reduction $\ddot r=-(\ell^2u^2/\mu^2)u''$ , the boxed Binet form against Goldstein eq. (3.34), the conic solution $r=p/(1+e\cos\phi)$ with $p=\ell^2/\mu k$ , the eccentricity-energy relation $e=\sqrt{1+2E\ell^2/\mu k^2}$ (Goldstein eq. 3.57), and the inverse-problem results $r=d\cos\phi\Rightarrow F\propto r^{-5}$ and $r=r_0e^{a\phi}\Rightarrow F\propto r^{-3}$ . Worked-example numbers on this page ( $p = 8400$ km, $r_{\max} = 10\,500$ km, $v_p = 8.27$ km/s; $\omega = 0.99499$ and $1.82°$ /rev advance) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Reduction to the radial equation · Angular momentum conservation in central fields · Central force fundamentals · The two-body problem and reduced mass. Leads to: Conic-section orbits · The Kepler problem · Kepler's first law · Orbital elements.

Why is the substitution u = 1/r and not something else, like u = r² or u = ln r?

Because $1/r$ is the unique choice that makes the powers interlock. Two things generate powers of $u$ : the centrifugal term ( $\ell^2/\mu r^3 = \ell^2u^3/\mu$ ) and each chain-rule conversion $\mathrm{d}/\mathrm{d}t = (\ell u^2/\mu)\,\mathrm{d}/\mathrm{d}\phi$ . With $u = 1/r$ , the $1/u^2$ from differentiating the reciprocal cancels a $u^2$ from the conversion, leaving $\dot r = -(\ell/\mu)u'$ with no stray factors — and the final equation picks up the clean restoring term $+u$ . Any other substitution leaves mismatched powers and a nonlinear mess. Historically the trick was distilled from a century of trial: it is precisely the change of variables in which the Kepler problem is a harmonic oscillator.

Binet's equation throws away time. How do I get the period or the time between two points if I need it?

Go back to the dynamics you set aside. Once the shape $r(\phi)$ is known, angular momentum conservation gives $\mathrm{d}t = (\mu/\ell)\,r^2\,\mathrm{d}\phi$ , so time along the orbit is the integral $t = (\mu/\ell)\int r(\phi)^2\,\mathrm{d}\phi$ — this is literally Kepler's second law run in reverse (equal areas cost equal times). For the full period of an ellipse the integral gives Kepler's third law, $T^2 = 4\pi^2\mu a^3/k$ . Alternatively, use the energy method on the radial equation directly. The division of labor is deliberate: Binet answers where, the energy and area integrals answer when.

What happens to Binet's equation when the angular momentum is zero?

It ceases to exist — and the failure is instructive. The whole construction rests on $\dot\phi = \ell u^2/\mu > 0$ so that $\phi$ can replace $t$ as the independent variable. With $\ell = 0$ the particle moves on a straight line through the force center: $\phi$ is frozen, no sweep angle accumulates, and there is no shape to parameterize — the "orbit" is a radial plunge (or escape) fully described by the one-dimensional energy method. Physically, every term in Binet's equation is built from the angular motion; take that away and nothing remains. Always check $\ell \neq 0$ before reaching for equation (3).

Is the "harmonic oscillator" in Binet's equation a real oscillation? Nothing is vibrating, after all

It is a genuine oscillation, but in an abstract variable: the inverse radius $u$ , oscillating as a function of sweep angle $\phi$ rather than time. Each "period" of this oscillator is one radial breathing cycle of the orbit — perihelion to aphelion and back — completed per $2\pi/\omega$ of accumulated angle. The analogy is exact, not poetic: the equation $u'' + u = \mu k/\ell^2$ is mathematically identical to a mass on a spring under constant gravity, with the equilibrium sag playing the role of the circular orbit and the oscillation amplitude playing the role of eccentricity. Everything you know about driven oscillators (see Simple harmonic motion) transfers wholesale — that is why physicists trust the equation so instinctively.

How exactly does the inverse problem determine the force uniquely? Couldn't two different forces produce the same orbit?

For a given orbit with its angular momentum specified, the force along that orbit is unique: equation (6) computes $F$ pointwise from $u(\phi)$ and $\ell$ , with no freedom left. What is not unique is the extrapolation. The orbit only visits radii between $r_{\min}$ and $r_{\max}$ , so the inversion determines $F(r)$ only on that annulus — a force law that differs outside the visited range would produce the identical trajectory. That is why Newton's conclusion from planetary ellipses was strictly "inverse-square over the range the planets probe," and why modern tests hunt for deviations using orbits of very different sizes, from LAGEOS satellites to the star S2. One orbit pins the force where it flies; a family of orbits pins the law.

Where does the general-relativistic correction term in the Binet equation come from?

From the Schwarzschild geometry, the GR orbit equation for a massive particle comes out as $u'' + u = GM\mu^2/\ell^2 + 3GMu^2/c^2$ (in the convention where $\ell$ is the relativistic angular momentum). The Newtonian piece is the familiar constant; the new term $3GMu^2/c^2$ is tiny — of order $(v/c)^2$ relative to the first — but it depends on $u$ , so it detunes the oscillation frequency exactly as in Worked Example 2, part B. Treating it perturbatively gives a perihelion advance $\Delta\phi = 6\pi GM/(c^2a(1-e^2))$ per orbit: 43 arcseconds per century for Mercury, and about 12 arcminutes per 16-year orbit for the star S2 around Sgr A*. The classical Binet machinery survives intact; only the drive term changes.

My textbook writes Binet's equation with F, another with the potential V. Which is right?

Both — they are one equation in two dialects, connected by $F = -\mathrm{d}V/\mathrm{d}r$ . In force form: $u'' + u = -\dfrac{\mu}{\ell^2u^2}F(1/u)$ . To convert, use $\dfrac{\mathrm{d}}{\mathrm{d}u}V(1/u) = V'(r)\cdot\left(-\dfrac{1}{u^2}\right) = \dfrac{F(r)}{u^2}$ , which turns the right side into $-\dfrac{\mu}{\ell^2}\dfrac{\mathrm{d}}{\mathrm{d}u}V\!\left(\dfrac{1}{u}\right)$ — the potential form. Sanity-check on gravity: $V = -k/r$ means $V(1/u) = -ku$ , its $u$ -derivative is $-k$ , and the right side becomes $-(\mu/\ell^2)(-k) = \mu k/\ell^2$ — exactly the constant drive of equation (4). Whichever dialect your book speaks, the inverse-square case must reduce to $u'' + u = \mu k/\ell^2$ ; if yours doesn't, a sign was dropped in translation, not a different physics chosen.

Bertrand's theorem says only two force laws give closed orbits. How does Binet's equation show that?

Linearize about a circular orbit. Write $u = u_c + \delta$ with $\delta$ small; Binet's equation gives $\delta'' + \beta^2\delta = 0$ , where $\beta^2$ depends on the force law evaluated at the circle. The orbit closes only if the radial oscillation frequency $\beta$ is a rational multiple of the angular frequency — and closes for all radii and amplitudes only for very special laws. Working through the algebra, power-law forces $F \propto -r^n$ give $\beta^2 = 3 + n$ (with the exponent $n$ of the power law); demanding closure of all bound orbits, not just nearly circular ones, singles out $n = -2$ ( $\beta = 1$ , the Kepler ellipse) and $n = +1$ ( $\beta = 2$ , the harmonic oscillator). That is Bertrand's theorem, and the near-circular Binet analysis is its opening move. Any other force law produces orbits that precess — which is exactly why observed precession is such a sensitive probe of extra physics.

The Orbit Equation and Binet's Equation

Where this lives today

The physical picture

From radial dynamics to Binet's equation

Which tool answers which question

The inverse-square triumph

Running the machine backward

Limiting cases

Worked examples

1 · A satellite's orbit from two numbers (confidence builder)

2 · Newton's inverse problem, then a taste of precession (exam level)

Going further

Before this

Where this leads