The Kepler Problem

In 1705 Edmond Halley noticed that the comets of 1531, 1607 and 1682 traced suspiciously similar paths, applied Newton's brand-new gravitational mechanics, and announced that the same object would return around 1758. It did — sixteen years after his death. The machinery behind that prediction is the Kepler problem: a body moving under the attractive inverse-square force $\mathbf{F}=-(k/r^2)\hat{\mathbf{r}}$ , with $k=Gm_1m_2$ for gravity and $\mu$ the reduced mass of the two-body reduction. It is one of the very few problems in physics that can be solved exactly, and its answer — every orbit is a conic section with the force centre at one focus — is the reason planets ride ellipses, comets swing hyperbolas, and exoplanet hunters can read a planet's mass out of a star's wobble. The solution runs entirely on conservation laws. Angular momentum confines the motion to a plane and makes the radius sweep equal areas in equal times; energy fixes how far in and out the body can roam. One clever substitution, $u=1/r$ , then turns the messy radial equation into the linear Binet equation — an equation you already know how to solve — and the conic pops out. There is a bonus. The inverse-square law hides an extra conserved quantity, the Laplace–Runge–Lenz vector, which pins the ellipse's orientation in space. That is why Kepler orbits never precess — and why the tiny 43″-per-century drift of Mercury's perihelion became one of the decisive tests of general relativity.

Key takeaways

The Kepler problem is motion under the central attractive inverse-square force $F(r)=-k/r^2$ — gravity ( $k=Gm_1m_2$ ) or the attractive Coulomb force.
Conserved $E$ and $L$ collapse the 2-D motion to one radial equation in the effective potential $V_\mathrm{eff}(r)=-k/r+L^2/(2\mu r^2)$ .
The exact orbit is a conic section $r(\phi)=p/(1+e\cos\phi)$ with semi-latus rectum $p=L^2/(\mu k)$ and eccentricity $e=\sqrt{1+2EL^2/(\mu k^2)}$ .
The sign of $E$ classifies the orbit: $E<0$ ellipse, $E=0$ parabola, $E>0$ hyperbola; $e=0$ is a circle.
Bound orbits have $E=-k/(2a)$ and period $T=2\pi\sqrt{\mu a^3/k}$ — both set by the semi-major axis $a$ alone, not by $L$ .
Key idea: two conservation laws reduce the problem to a radial equation; the substitution $u=1/r$ linearizes it into Binet's form, whose sinusoidal solution is the conic $r=p/(1+e\cos\phi)$ .

The core insight

Two conservation laws (energy $E$ and angular momentum $L$ ) reduce the problem to a radial equation; the substitution $u=1/r$ then linearizes the orbit equation into Binet's form, whose sinusoidal solution is a conic section $r=p/(1+e\cos\phi)$ with $p=L^2/(\mu k)$ and $e=\sqrt{1+2EL^2/(\mu k^2)}$ .

Where this lives today

Every exoplanet count you have ever read is a census of solved Kepler problems. The radial-velocity method fits a star's periodic Doppler wobble with exactly the two-body machinery of this article — the period gives $a$ through the third law, the wobble amplitude and shape give the planet's minimum mass and eccentricity — and NASA's planet-hunting telescope was, fittingly, named Kepler. At the galaxy's heart, the star S2, tracked on a 16-year, $e=0.88$ ellipse around Sagittarius A*, weighed the central black hole at four million solar masses via equation (8); its measured orbital precession is now a live test of general relativity — the modern sequel to Mercury's 43 arcseconds. Spaceflight is applied Kepler. Mission designers stitch trajectories from conic arcs — "patched conics": a Hohmann transfer is half of one Kepler ellipse tangent to two others, and every gravity-assist flyby is a hyperbolic Kepler orbit in a planet's frame. When a launch window to Mars is announced, someone has evaluated equations (7) and (8) for you.

The physical picture

Imagine releasing a planet at some distance with some sideways speed. Energy conservation sets how far in and out it can roam; angular momentum sets how fast it sweeps angle. The radial motion is exactly like a ball rolling in a one-dimensional valley $V_\mathrm{eff}(r)$ : it oscillates between an inner turning point (perihelion) and an outer one (aphelion), while the angle keeps winding forward. Every possible fate of the body belongs to a single family of curves, sorted by one number — the energy. Launch at just the right speed and you circle forever; add energy and the circle stretches into an ellipse, snaps open into a parabola exactly at $E=0$ , then flattens toward a straight-line flyby as $E$ grows:

Read the figure from the inside out: all four trajectories share the same focus — the force centre — and pass through a perihelion on the right; only the energy differs. The eccentricity $e$ is a dimensionless energy dial at fixed $L$ : $e^2=1+2EL^2/(\mu k^2)$ . Push $E$ up from the circular minimum and $e$ grows from $0$ ; reach $E=0$ and $e=1$ (the orbit just barely unbinds); go positive and $e>1$ . The second picture to carry in your head is the radial valley itself. Gravity pulls the potential down as $-k/r$ ; the centrifugal term $L^2/(2\mu r^2)$ walls off the centre. Their sum has a genuine minimum, and where your energy line sits in this valley decides everything:

Figure

The effective potential — one radial valley sorts every orbit by its energy line

The magic of the inverse-square law is invisible in this picture but decisive: the radial oscillation period exactly matches the angular winding period — the radius returns to its minimum precisely when $\phi$ has advanced by $2\pi$ . That perfect 1:1 commensurability is why the orbit closes into a static ellipse instead of a rosette, and it is the geometric face of the conserved Laplace–Runge–Lenz vector you will meet at the end. A circular orbit sits exactly at the bottom of the valley; perturbed slightly, it oscillates about the circle — a near-circular ellipse, stable because the minimum is genuine.

From force law to one radial equation

Everything starts from the interaction itself. Write the inverse-square attraction together with the potential energy it comes from, choosing the zero of energy at infinite separation — the convention that makes "bound" and "unbound" read directly off the sign of $E$ :

\mathbf{F}=-\dfrac{k}{r^2}\hat{\mathbf{r}},\qquad U(r)=-\dfrac{k}{r} \qquad (1)

where: $k$ — force constant, $k=Gm_1m_2$ for gravity, $>0$ for attraction; $r$ — separation between the two bodies (the relative coordinate); $\hat{\mathbf r}$ — unit vector along the separation; $U(r)$ — potential energy, zero at infinity. Note that this is already the reduced two-body problem: a single fictitious particle of reduced mass $\mu=m_1m_2/(m_1+m_2)$ moving in the fixed potential $U(r)$ . Because the force always points along $\hat{\mathbf r}$ , it exerts no torque about the centre, and the angular momentum of the relative motion is locked. That single fact flattens the problem into a plane and hands us Kepler's second law before any orbit is computed:

L=\mu r^2\dot\phi=\text{const},\qquad \frac{\mathrm{d}A}{\mathrm{d}t}=\frac{L}{2\mu} \qquad (2)

where: $L$ — magnitude of the orbital angular momentum; $\dot\phi$ — angular velocity in the orbital plane; $\mathrm{d}A/\mathrm{d}t$ — areal velocity, the area swept per unit time. Constant areal velocity is the equal-areas law: the body must move fastest where $r$ is smallest. Nothing here used the inverse-square form — every central force does this. The force is also conservative, so the total energy of the relative motion is conserved. Using $L$ to eliminate the angular velocity turns that statement into a purely radial problem — the workhorse equation of the whole subject:

E=\tfrac12\mu\dot r^2+\underbrace{\frac{L^2}{2\mu r^2}-\frac{k}{r}}_{V_\mathrm{eff}(r)} \qquad (3)

where: $E$ — conserved total energy of relative motion; $\dot r$ — radial velocity; $V_\mathrm{eff}$ — the effective potential combining the gravitational well and the centrifugal barrier. The radius now behaves exactly like a particle in the 1-D valley drawn above; the angular motion rides along via (2). The elimination is two lines: write the planar kinetic energy as $\tfrac12\mu(\dot r^2+r^2\dot\phi^2)$ , then use (2) to replace $\dot\phi=L/(\mu r^2)$ , so the angular kinetic energy $\tfrac12\mu r^2\dot\phi^2$ reappears as the repulsive term $L^2/(2\mu r^2)$ — the barrier that prevents collapse to the centre for any $L\neq0$ .

The orbit is a conic section

Equation (3) can be integrated for $r(t)$ , but that road leads to the transcendental Kepler equation. The shape of the orbit, $r(\phi)$ , is far cleaner. The classic trick is to trade $t$ for $\phi$ and $r$ for its reciprocal $u=1/r$ ; the nonlinear radial dynamics then collapses into a linear equation you have solved a hundred times:

\frac{\mathrm{d}^2u}{\mathrm{d}\phi^2}+u=\frac{\mu k}{L^2},\qquad u\equiv\frac1r \qquad (4)

where: $u$ — reciprocal radius; $\phi$ — polar angle, now the independent variable. This is Binet's orbit equation: a simple harmonic oscillator in the variable $u(\phi)$ , driven by a constant. Its "oscillation" is the breathing of the radius around the circular value $u_c=\mu k/L^2$ , and the fact that its angular frequency is exactly $1$ (one radial cycle per $2\pi$ of angle) is precisely why Kepler orbits close. Constant-plus-sinusoid solves (4), and inverting it back to $r$ delivers the headline result of the whole problem — the polar equation of a conic section with the force centre at one focus:

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

where: $p$ — the semi-latus rectum, the radius where $\phi=\pi/2$ ; $e$ — the eccentricity, an integration constant fixed by the energy below; $\phi$ — the true anomaly, measured from perihelion. At $\phi=0$ the denominator is largest and $r$ smallest (perihelion, $r_\mathrm{min}=p/(1+e)$ ); for $e<1$ the denominator never vanishes and the orbit is a closed ellipse; at $e\geq1$ there are angles where $r\to\infty$ and the body escapes. For the bound case, the dictionary between $(E,L)$ and the ellipse elements looks like this:

Figure

Anatomy of a Kepler ellipse — the Sun at a focus, not the centre

Derivation AFrom Binet's equation to the conic section

The goal is the headline result of the whole problem: the orbit shape $r(\phi)$ , which will turn out to be the focal conic (5). Rather than integrating equation (3) for $r(t)$ — a road that dead-ends in the transcendental Kepler equation — we ask only for geometry, trading the time variable for the polar angle $\phi$ and the radius for its reciprocal $u=1/r$ . That double substitution is chosen because it collapses the nonlinear radial Newton equation into Binet's linear driven oscillator (4), which solves by inspection. The two things to watch: the time-for-angle trade is legitimate only because $L$ is conserved and $\dot\phi$ never changes sign, and the miraculous cancellation of every stray power of $r$ happens for the inverse-square law essentially alone — for other force laws the same substitution leaves a nonlinear mess.

Step 1. Trade time for angle. Using $L=\mu r^2\dot\phi$ , any time derivative becomes $\dfrac{\mathrm{d}}{\mathrm{d}t}=\dfrac{L}{\mu r^2}\dfrac{\mathrm{d}}{\mathrm{d}\phi}$ . Why: $L$ is conserved and $\dot\phi$ never changes sign, so $\phi$ is a legitimate clock — and it is the natural variable for a question about shape.

Step 2. Substitute $u=1/r$ . Since $\mathrm{d}r/\mathrm{d}\phi=-u^{-2}\,\mathrm{d}u/\mathrm{d}\phi$ , the radial velocity becomes $\dot r=-\dfrac{L}{\mu}\dfrac{\mathrm{d}u}{\mathrm{d}\phi}$ . Why: the awkward factors of $r^2$ from the time-to-angle conversion cancel exactly against the $u^{-2}$ from the reciprocal — this is the whole reason $u=1/r$ is the right substitution.

Step 3. Write the radial Newton equation. In the radial direction, $\mu\ddot r-\dfrac{L^2}{\mu r^3}=-\dfrac{k}{r^2}$ . Why: this is $F=ma$ along $\hat{\mathbf r}$ , with the centrifugal term appearing because $\hat{\mathbf r}$ rotates.

Step 4. Linearize. Applying Steps 1–2 to $\ddot r$ and dividing by common factors turns the nonlinear radial equation into $u''+u=\mu k/L^2$ — equation (4). Why: for the inverse-square law, and essentially only for it, every power of $r$ conspires to cancel, leaving a linear oscillator with a constant drive.

Step 5. Solve. Particular (constant) plus homogeneous (sinusoid): $u(\phi)=\dfrac{\mu k}{L^2}\big[1+e\cos(\phi-\phi_0)\big]$ ; choose $\phi_0=0$ so perihelion sits at $\phi=0$ . Why: a driven SHO oscillates about its displaced equilibrium; amplitude $e$ and phase $\phi_0$ are the two integration constants the second-order equation owes us.

Step 6. Invert and identify. Taking $r=1/u$ gives $r=\dfrac{L^2/(\mu k)}{1+e\cos\phi}$ — the polar equation of a conic with focus at the origin, equation (5). Why: comparison with the standard focal conic identifies $p=L^2/(\mu k)$ and the integration constant $e$ as the eccentricity.

Go deeper · Derivation B · 2 minEccentricity from the constants of motion

Derivation BEccentricity from the constants of motion

The conic solution (5) still carries one undetermined constant, the eccentricity $e$ , and the goal here is to pin it to the conserved quantities — the result will be equation (6), $e=\sqrt{1+2EL^2/\mu k^2}$ , which makes the ellipse/parabola/hyperbola classification automatic. Energy is what fixes it, and the cleanest way in is to stand at a turning point of the radial motion, where $\dot r=0$ and the energy equation (3) degenerates into pure algebra — a quadratic whose two roots are the perihelion and aphelion. We then compute the same two turning points from the geometric side, the conic (5), and force the two descriptions to agree. The care point is keeping the two roots straight ( $u_+$ is the reciprocal perihelion, so the largest $u$ is the smallest $r$ ) and remembering that for bound orbits $E<0$ , so the square root sits safely below 1.

Step 1. Evaluate the energy at a turning point. Where $\dot r=0$ , equation (3) reads $E=\dfrac{L^2}{2\mu r^2}-\dfrac{k}{r}$ . Why: at perihelion and aphelion the radial kinetic term vanishes by definition — all the energy is centrifugal plus potential.

Step 2. Recast as a quadratic in $u=1/r$ . Multiplying out: $\dfrac{L^2}{2\mu}u^2-ku-E=0$ . Why: the turning radii are now the two roots of a quadratic — no calculus left.

Step 3. Solve. The standard formula gives $u_\pm=\dfrac{\mu k}{L^2}\left[1\pm\sqrt{1+\dfrac{2EL^2}{\mu k^2}}\right]$ . Why: the two roots are the reciprocal perihelion ( $u_+$ , closest) and reciprocal aphelion ( $u_-$ , farthest).

Step 4. Match against the orbit. The conic (5) has its extremes at $\phi=0$ and $\phi=\pi$ : $u_\pm=\dfrac{\mu k}{L^2}(1\pm e)$ . Why: the same turning points must come out of the geometric description — matching them ties dynamics to geometry.

Step 5. Identify the eccentricity.

e=\sqrt{1+\dfrac{2EL^2}{\mu k^2}} \qquad (6)

Why: comparing Steps 3 and 4 forces this identification. Now the classification is automatic: $E<0\Rightarrow e<1$ (ellipse); $E=0\Rightarrow e=1$ (parabola); $E>0\Rightarrow e>1$ (hyperbola).

where: $e$ — orbit eccentricity, dimensionless; $E$ — total energy of relative motion. At fixed $L$ , $e$ is a monotonic dial on $E$ ; at fixed $E<0$ , larger $L$ means rounder orbit, with the circle $e=0$ at the maximum $L$ the energy allows.

Size, energy, and the year

Two more relations complete the toolbox, and both carry a surprise: the angular momentum drops out. Write the semi-major axis of the bound ellipse in terms of the energy alone. Averaging perihelion and aphelion distances, $a=\tfrac12(r_++r_-)=p/(1-e^2)$ , and substituting (6) collapses beautifully:

a=-\dfrac{k}{2E},\qquad E=-\dfrac{k}{2a}\quad(E<0) \qquad (7)

where: $a$ — semi-major axis; $E$ — bound-state energy, necessarily negative. All ellipses of the same energy have the same width $2a$ , however eccentric — a fat circle and a thin cigar with equal $a$ cost exactly the same energy. This degeneracy in $L$ is another fingerprint of the hidden symmetry. The period follows from pure geometry plus the equal-areas law, and inherits the same $L$ -independence — Kepler's third law:

T=2\pi\sqrt{\dfrac{\mu a^3}{k}}\;\Rightarrow\; T^2=\dfrac{4\pi^2}{G(m_1+m_2)}a^3 \qquad (8)

where: $T$ — orbital period; $a$ — semi-major axis; $G(m_1+m_2)$ — the gravitational parameter, obtained by inserting $k=Gm_1m_2$ and $\mu=m_1m_2/(m_1+m_2)$ . Note it is the total mass that enters — for a planet around the Sun, $m_1+m_2\approx M_\odot$ and all planets share one $T^2/a^3$ ratio; for a comparable-mass binary star the same law weighs the pair.

Go deeper · Derivation C · 1 minKepler's third law from the area integral

Derivation CKepler's third law from the area integral

We are after the period of a bound orbit — Kepler's third law, equation (8), $T^2=4\pi^2 a^3/G(m_1+m_2)$ . The route is deliberately lazy: instead of integrating the equations of motion in time, we exploit the equal-areas law (2), which says the orbit sweeps area at the constant rate $L/2\mu$ , so the period is simply the total area of the ellipse divided by that rate — geometry divided by a conserved constant, no dynamics integrated at all. The one piece of finesse is expressing the semi-minor axis through the semi-latus rectum, $b=\sqrt{ap}$ , because $p=L^2/(\mu k)$ is the hook through which the dynamics enters. Watch the angular momentum: it appears in both numerator and denominator and cancels identically, which is the content of the law — the clock of the orbit reads only its size $a$ , never its shape $e$ .

Step 1. Start from constant areal velocity. From (2), $\dfrac{\mathrm{d}A}{\mathrm{d}t}=\dfrac{L}{2\mu}$ . Why: over one full period the swept area is the entire ellipse, so the period is just total area over sweep rate — no integration of the motion needed.

Step 2. Write the ellipse area. $A_\text{ellipse}=\pi ab$ with $b=a\sqrt{1-e^2}$ , and since $p=a(1-e^2)$ , the semi-minor axis is $b=\sqrt{ap}$ . Why: expressing $b$ through $p$ lets the dynamical constant $L$ enter through $p=L^2/(\mu k)$ in the next step.

Step 3. Divide area by rate. $T=\dfrac{2\mu}{L}\pi ab=\dfrac{2\pi\mu}{L}a\sqrt{ap}$ . Why: time equals area over areal velocity, by Step 1.

Step 4. Insert $p=L^2/(\mu k)$ . $T=\dfrac{2\pi\mu}{L}a\sqrt{a\dfrac{L^2}{\mu k}}=2\pi\sqrt{\dfrac{\mu a^3}{k}}$ . Why: the angular momentum cancels identically — the period depends only on the size of the orbit, not its shape. Two orbits of equal $a$ , however different in eccentricity, take equal times.

Step 5. Specialize to gravity. With $k=Gm_1m_2$ and $\mu=m_1m_2/(m_1+m_2)$ , the masses combine into $T^2=\dfrac{4\pi^2}{G(m_1+m_2)}a^3$ — equation (8). Why: the product $\mu/k=1/[G(m_1+m_2)]$ involves only the total mass; the individual masses are invisible to the clock of the orbit. *Why: when $m_2\ll m_1$ , the result reduces to $T^2=4\pi^2 a^3/(GM_\odot)$ , the familiar form that holds for all planets orbiting the Sun.*

The hidden symmetry

Energy and angular momentum are conserved for every central force. The inverse-square law conserves one thing more — and it is the deep answer to "why do the ellipses not precess?". Form the following combination of momentum, angular momentum and direction:

\mathbf{A}=\mathbf{p}\times\mathbf{L}-\mu k\,\hat{\mathbf{r}}=\text{const} \qquad (9)

where: $\mathbf{A}$ — the Laplace–Runge–Lenz vector, magnitude $A=\mu ke$ , pointing from the focus toward perihelion; $\mathbf{p}=\mu\dot{\mathbf r}$ — momentum of the relative motion; $\mathbf{L}$ — the angular-momentum vector. A conserved vector in the orbital plane nails the orientation of the major axis for all time: the perihelion cannot drift. Any deviation from a pure $1/r$ potential — general-relativistic corrections, the Sun's oblateness, other planets — breaks the conservation and sets the axis slowly rotating: perihelion precession.

Which equation answers which question?

you are asked for	reach for
shape of the orbit	conic in polar form (5)
bound or unbound, how elongated	eccentricity from $E$ and $L$ (6)
size from energy, or vice versa	$a=-k/2E$ (7)
period, or a mass from a period	Kepler's third law (8)
speeds at two points of one orbit	angular momentum (2) at apsides, energy (3) elsewhere
why perihelion stays put	Laplace–Runge–Lenz (9)

Limiting cases

The single conic-section solution contains several physically distinct regimes selected by the value of the energy $E$ :

Case	Result	Meaning
Circular orbit	$E=E_\mathrm{min}=-\dfrac{\mu k^2}{2L^2},\qquad e=0,\qquad r=\dfrac{L^2}{\mu k}$	Minimum of $V_\mathrm{eff}$ ; radius constant, speed constant. The lowest-energy bound state for given $L$ .
Bound ellipse	$-\dfrac{\mu k^2}{2L^2}<E<0,\qquad 0<e<1,\qquad a=-\dfrac{k}{2E}$	Periodic motion between perihelion and aphelion; the generic planetary orbit.
Parabolic escape	$E=0,\qquad e=1,\qquad r=\dfrac{p}{1+\cos\phi}$	Marginally unbound: the particle reaches infinity with zero residual speed. The escape-velocity boundary.
Hyperbolic flyby	$E>0,\qquad e>1,\qquad r=\dfrac{p}{1+e\cos\phi}$	Unbound scattering trajectory; arrives from and departs to infinity with finite speed $v_\infty=\sqrt{2E/\mu}$ .
Radial fall, $L\to0$	$V_\mathrm{eff}\to -\dfrac{k}{r},\qquad e\to1\;\text{(degenerate)}$	No centrifugal barrier; the orbit degenerates to a straight line and the particle plunges into the centre.

Worked examples

1 · Weighing a year (confidence builder)

Given the Sun's gravitational parameter $GM_\odot=1.327\times10^{20}\ \mathrm{m^3/s^2}$ and the Earth–Sun distance $a=1.496\times10^{11}\ \mathrm{m}$ (1 AU), predict Earth's orbital period and speed.

Step 1. Since $m_\oplus\ll M_\odot$ , equation (8) becomes $T=2\pi\sqrt{a^3/(GM_\odot)}$ .

Step 2. Insert numbers: $T = 2\pi\sqrt{\dfrac{(1.496\times10^{11})^3}{1.327\times10^{20}}} = 2\pi\sqrt{2.523\times10^{13}\ \mathrm{s^2}} = 3.156\times10^{7}\ \mathrm{s}$ .

Step 3. Convert: $T = \dfrac{3.156\times10^{7}}{86400} = \boxed{365.3\ \mathrm{days}}$ — the year, recovered from a force law and a distance.

Step 4. Orbital speed (Earth's orbit is nearly circular, $e=0.0167$ ): $v=\sqrt{GM_\odot/a}=\sqrt{1.327\times10^{20}/1.496\times10^{11}} = 2.98\times10^{4}\ \mathrm{m/s} = \boxed{29.8\ \mathrm{km/s}}$ . The same one-liner runs everywhere in the Solar System: in units of years and AU, Kepler's third law for anything orbiting the Sun is simply $T^2=a^3$ .

2 · Halley's comet, end to end (exam level)

Halley's comet returns every $T=75.3$ years on an orbit of eccentricity $e=0.967$ . Find its semi-major axis, perihelion and aphelion distances, and the ratio of its fastest to slowest speeds.

Step 1. Semi-major axis from the third law in year–AU units: $a=T^{2/3}=(75.3)^{2/3}=\boxed{17.8\ \mathrm{AU}}$ — the comet's average scale sits out between Uranus (19.2 AU) and Saturn (9.6 AU).

Step 2. Perihelion: $r_\mathrm{min}=a(1-e)=17.8\times0.033=\boxed{0.59\ \mathrm{AU}}$ — it dives inside Venus's orbit.

Step 3. Aphelion: $r_\mathrm{max}=a(1+e)=17.8\times1.967=\boxed{35.1\ \mathrm{AU}}$ — out beyond Neptune. One orbit spans a factor of $60$ in distance.

Step 4. Speed ratio from angular momentum. At the apsides the velocity is purely tangential, so (2) gives $\mu r_\mathrm{min}v_\mathrm{max}=\mu r_\mathrm{max}v_\mathrm{min}$ , hence

\frac{v_\mathrm{max}}{v_\mathrm{min}}=\frac{r_\mathrm{max}}{r_\mathrm{min}}=\frac{1+e}{1-e}=\frac{1.967}{0.033}=\boxed{59.6}

Step 5. Absolute speeds from energy conservation: evaluating $\tfrac12v^2-GM_\odot/r=-GM_\odot/(2a)$ at each apsis gives $v_\mathrm{max}=54.5\ \mathrm{km/s}$ and $v_\mathrm{min}=0.91\ \mathrm{km/s}$ (ratio $59.6$ ✓). For most of its 75 years Halley drifts at strolling pace; the inner-system fireworks last months.

Origins · 1 minJohannes Kepler and how this was discovered

Johannes Kepler

1571–1630 · Prague

Kepler extracted the laws of planetary motion from Tycho Brahe's unmatched naked-eye observations of Mars — a decade of calculation he called his "war with Mars". His radical move was to trust data over two millennia of dogma: when the best circular model missed Tycho's positions by a mere eight arcminutes, he threw away the circles rather than the observations, and found the ellipse. He published the first and second laws in Astronomia Nova (1609) and the third in Harmonices Mundi (1619), all without knowing why they held — that answer waited seventy years for Newton. Kepler did it while casting horoscopes to pay the bills, defending his mother from a witchcraft trial, and founding modern optics on the side.

History

1601

— Kepler inherits Tycho Brahe's Mars data, the most precise astronomical record ever compiled without a telescope.

1609

— Astronomia Nova: orbits are ellipses with the Sun at a focus, sweeping equal areas in equal times.

1619

— Harmonices Mundi adds the third law: the square of the period scales as the cube of the mean distance.

1687

— Newton's Principia derives all three laws from universal inverse-square gravitation — the Kepler problem is solved.

1915

— Einstein explains Mercury's leftover 43″-per-century perihelion precession: the first post-Keplerian correction, and a founding test of general relativity.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 7 (Central Forces) · Goldstein, Poole & Safko, Classical Mechanics 3rd ed., Ch. 3 · Landau & Lifshitz, Mechanics Vol. 1, §15 (Kepler's problem) · Taylor, Classical Mechanics, Ch. 8 The key results of this article are: Binet's equation $u''+u=\mu k/L^2$ and its conic solution; eccentricity $e=\sqrt{1+2EL^2/\mu k^2}$ from the turning-point quadratic; size–energy relation $a=-k/(2E)$ and period $T=2\pi\sqrt{\mu a^3/k}$ ; and the Laplace–Runge–Lenz vector $\mathbf A=\mathbf p\times\mathbf L-\mu k\hat{\mathbf r}$ with magnitude $|\mathbf A|=\mu k e$ . Reference eccentricities: $e_\text{Earth}\approx0.0167$ , $e_\text{Mercury}\approx0.206$ ; Mercury's perihelion precession is $43''$ /century. Worked-example numbers follow from $GM_\odot=1.327\times10^{20}\ \mathrm{m^3/s^2}$ : Earth $T=365.3$ d, $v=29.8$ km/s; Halley $a=17.8$ AU, apsides $0.59/35.1$ AU, speeds $54.5/0.91$ km/s, ratio $59.6$ .

Builds on: Central Force Fundamentals · The Two-Body Problem and Reduced Mass · Angular Momentum Conservation in Central Fields · Reduction to the Radial Equation · The Effective Potential Leads to: Conic-Section Orbits · Kepler's First Law: Elliptical Orbits · Kepler's Third Law: Period-Semimajor Relation · Orbital Elements and Parameterization · Classification of Orbits by Energy

How can a real orbiting planet have negative total energy?

Only the kinetic term is forced to be positive. The potential $U(r)=-k/r$ was deliberately set to zero at infinite separation — equation (1) — so it is negative at every finite $r$ . A bound planet has too little kinetic energy to climb back out of that well, so the sum $E=\tfrac12\mu\dot r^2+V_\mathrm{eff}$ comes out negative: $E=-k/(2a)$ for an ellipse. The sign is bookkeeping relative to a chosen zero, but the physics is real: $|E|$ is exactly the energy you would have to supply to drag the pair apart to rest at infinity. That is why the sign of $E$ can classify orbits — negative means trapped in the valley, zero means barely escaping, positive means a flyby with speed $v_\infty=\sqrt{2E/\mu}$ left over.

Gravity only attracts — so where does the repulsive

L^2/2\mu r^2

term come from?

It is not a new force; it is the angular kinetic energy wearing a disguise. The planar kinetic energy is $\tfrac12\mu(\dot r^2+r^2\dot\phi^2)$ , and using $L=\mu r^2\dot\phi$ from (2) to eliminate $\dot\phi$ turns the tangential piece $\tfrac12\mu r^2\dot\phi^2$ into $L^2/(2\mu r^2)$ . Read it energetically: at fixed $L$ , moving inward forces $\dot\phi$ to spin up, so the tangential kinetic energy grows like $1/r^2$ — and that energy has to come from somewhere, which is exactly what a repulsive barrier does in the 1-D radial picture. Since gravity's pull only deepens like $-k/r$ , the $1/r^2$ barrier always wins at small $r$ : any body with $L\neq0$ can never reach the centre.

Why does the orbit close only for the inverse-square law?

Closure needs the radial breathing and the angular winding to stay in step. Binet's equation (4), $u''+u=\mu k/L^2$ , has angular frequency exactly $1$ : the radius completes one full in–out cycle in precisely $\Delta\phi=2\pi$ , so perihelion lands on the same spot every turn. That unit frequency is a conspiracy special to $1/r^2$ — in the derivation, every stray power of $r$ cancels only for this force law. A generic central force gives incommensurate radial and angular periods, and the "ellipse" precesses into a rosette. The deep statement is the Laplace–Runge–Lenz vector (9): a conserved arrow pointing at perihelion exists only for the pure Kepler potential. That is why Mercury's mere 43″-per-century drift was such a loud signal that something beyond $1/r^2$ was at work.

If energy alone fixes the size and the period, where did the angular momentum go?

It went into the shape, and only the shape. In $a=-k/(2E)$ and $T=2\pi\sqrt{\mu a^3/k}$ the angular momentum genuinely cancels — in the third-law derivation, $L$ appears in the swept-area rate $L/2\mu$ and in $p=L^2/(\mu k)$ and drops out identically. So at fixed $E<0$ , dialing $L$ changes nothing but the eccentricity $e=\sqrt{1+2EL^2/(\mu k^2)}$ : low $L$ gives a thin cigar, the maximum allowed $L$ gives the circle $e=0$ , all with the same $a$ , the same $E$ , and the same year. This $L$ -degeneracy is not an accident of algebra — it is a fingerprint of the same hidden symmetry that conserves the Laplace–Runge–Lenz vector.

Parabolic escape and radial fall both have

e=1

— are they the same orbit?

No — one number cannot specify an orbit that takes two constants, $E$ and $L$ , to pin down. The parabola is $E=0$ with finite $L$ : a genuine curved conic with semi-latus rectum $p=L^2/(\mu k)>0$ , grazing perihelion at $p/2$ and coasting to infinity. Radial fall is the opposite corner: $L\to0$ at any energy, so $p\to0$ and the conic degenerates into a straight line through the force centre — the limiting-cases table flags this $e\to1$ as "degenerate" for exactly this reason. Formally $e=\sqrt{1+2EL^2/\mu k^2}\to1$ whenever $L\to0$ , whatever $E$ is. So matching eccentricities mean nothing without matching $p$ : the parabola barely escapes, the radial orbit plunges into the centre.

Why is escape speed exactly

\sqrt{2}

times circular speed and not some messier number?

Both speeds are set by the same potential at the same radius, so the comparison is clean. A circular orbit requires $\mu v_c^2/r=k/r^2$ , giving $\tfrac12\mu v_c^2=k/(2r)$ — kinetic energy equal to half the well depth $k/r$ (this is the virial theorem in miniature). Escape means reaching the $E=0$ boundary: $\tfrac12\mu v_\mathrm{esc}^2=k/r$ , the full well depth. The ratio of the two kinetic energies is exactly 2, so $v_\mathrm{esc}=\sqrt2\,v_c$ at every radius. The factor traces directly to the exponent in $U=-k/r$ : a different power law would put a different fraction of the well depth into circular motion and change the $\sqrt2$ .

Which body actually moves on the ellipse — surely the Sun does not orbit the comet?

Strictly, neither body traces the conic of equation (5). The coordinate $r$ is the relative separation, and the thing moving on the ellipse is a fictitious particle of reduced mass $\mu=m_1m_2/(m_1+m_2)$ . Each real body orbits the common centre of mass on its own scaled-down copy of the same conic, with sizes in inverse proportion to the masses. For a planet around the Sun, $\mu\approx m_\mathrm{planet}$ and the Sun's ellipse is tiny — but not zero, and that residual wobble is precisely the periodic Doppler shift the radial-velocity method fits to discover exoplanets. For a comparable-mass binary star, both ellipses are large and the reduced-mass bookkeeping is not optional.

Why do the apsidal speeds follow from angular momentum alone, when speeds usually need energy conservation?

Because $L=\mu r v_\mathrm{tangential}$ only sees the tangential component of the velocity. At a generic point of the orbit the velocity has both radial and tangential parts, so knowing $L$ tells you only a piece of the speed and you must fall back on the full energy equation (3). But perihelion and aphelion are turning points: $\dot r=0$ there by definition, the velocity is purely tangential, and $L=\mu r v$ holds with the total speed. Equating the two apsides gives the ratio $v_\mathrm{max}/v_\mathrm{min}=(1+e)/(1-e)$ with no energy input at all — 59.6 for Halley. To get the absolute speeds (54.5 and 0.91 km/s) you still need energy, exactly as the worked example does in its final step.