Reduction to the Radial Equation

Track a comet honestly and you must follow six numbers — three coordinates, three velocities — all changing at once, all coupled. Yet by the end of this page, the entire problem will have collapsed to one coordinate obeying an equation you already know how to read: a bead sliding in a one-dimensional potential. Nothing is thrown away and nothing is approximated. The collapse is paid for entirely by symmetry, cashed out through the two conserved quantities you have already met. You arrive equipped: the two-body problem taught you to replace two bodies by one fictitious particle of reduced mass $\mu$ orbiting a fixed center; angular momentum conservation then froze the vector $\vec{L}$ , trapping that particle in a plane and handing you the scalar relation $L=\mu r^2\dot\phi$ . This article spends those winnings. The frozen direction of $\vec{L}$ deletes one angle; the frozen magnitude enslaves the other, because $\dot\phi = L/\mu r^2$ makes the angular motion a mere function of the radius. Substitute it into the energy and the angular kinetic energy reappears wearing a disguise — a repulsive $L^2/2\mu r^2$ "centrifugal barrier" that depends on position only, as if it were a potential. What you gain is the master tool of orbit theory. Every question about central-force motion — bound or unbound, how close, how far, how long, what shape — becomes a question about one particle in one dimension, answerable from a single energy diagram or a single integral. The effective potential, the classification of orbits, Binet's equation, and the Kepler problem itself are all downstream of the reduction performed here.

Key takeaways

A particle of (reduced) mass $\mu$ in a central potential $V(r)$ carries two conserved quantities — the energy $E$ and the angular momentum magnitude $L$ — and together they collapse the three-dimensional problem to motion in the single coordinate $r$ .
Conservation of the vector $\vec{L}$ confines the orbit to a fixed plane, cutting the angular freedom to one azimuth $\phi$ ; that angle is then eliminated using $\dot\phi = L/\mu r^2$ .
The result is the radial energy equation $E = \tfrac12\mu\dot{r}^2 + V_{\mathrm{eff}}(r)$ with $V_{\mathrm{eff}}(r) = V(r) + L^2/2\mu r^2$ — exactly the form of one-dimensional motion.
The centrifugal term $L^2/2\mu r^2$ is not a new force: it is the kinetic energy of the angular motion repackaged as a repulsive barrier, and it forbids collapse to the origin whenever $L \neq 0$ .
Differentiating gives the radial equation of motion $\mu\ddot{r} = f(r) + L^2/\mu r^3 = -\mathrm{d}V_{\mathrm{eff}}/\mathrm{d}r$ .
The radial motion then solves by separation of variables: $t(r)$ from one quadrature, and the orbit shape $\phi(r)$ by trading time for angle via the chain rule.
Key idea: conservation of $L$ kills the angular freedom — substitute $\dot\phi = L/\mu r^2$ into the energy and central-force motion becomes a strictly one-dimensional problem in $r$ , governed by $V_{\mathrm{eff}} = V + L^2/2\mu r^2$ .

The core insight

Conservation of $L$ kills the angular freedom: substitute $\dot\phi=L/\mu r^2$ into the energy to obtain $E=\tfrac12\mu\dot{r}^2+V_{\mathrm{eff}}(r)$ — a strictly one-dimensional problem in $r$ with $V_{\mathrm{eff}}=V+L^2/2\mu r^2$ .

Where this lives today

Every spacecraft trajectory you have ever seen announced — transfer orbits, flybys, station-keeping budgets — begins life as this reduction. Mission-design tools like NASA's GMAT solve the reduced two-body problem first (it is exact, instant, and analytic) and then layer perturbations on top; the "patched-conic" method that plans interplanetary routes is nothing but equation (4) applied piecewise, one central body at a time. The same machinery runs in reverse for discovery: radial-velocity exoplanet hunting fits a star's wobble with Keplerian orbits — reduced-mass two-body solutions — and the fitted $\mu$ is how planet masses are weighed. The grandest current stage is galactic. In an axisymmetric galaxy, a star's motion in the disk plane obeys exactly this article's mathematics: $L_z$ is conserved, and the radial motion runs in an effective potential $\Phi(R)+L_z^2/2R^2$ . Stellar orbits are rosettes (the figure above is a fair portrait of the Sun's own path), and their radial oscillations — epicycles — are the working language of galactic dynamics. The Gaia mission's billion-star map of positions and velocities is analyzed star by star through precisely this reduction: compute each star's $E$ and $L_z$ , place it in the effective landscape, and read off its orbit — one-dimensional mechanics, applied a billion times.

The physical picture

In the orbital plane, every velocity has two jobs: motion in and out (changing $r$ ) and motion around (changing $\phi$ ). The kinetic energy splits cleanly along the same seam, $\tfrac12\mu\dot{r}^2 + \tfrac12\mu r^2\dot\phi^2$ , with no cross term — the two directions are perpendicular. Now angular momentum conservation does something quietly radical to the second piece. Since $L=\mu r^2\dot\phi$ is a fixed number for the whole orbit, the angular kinetic energy is not free to be anything: at radius $r$ it must equal exactly $L^2/2\mu r^2$ . A piece of kinetic energy whose value is dictated by position alone is, for all bookkeeping purposes, a potential energy. Fold it in with the real potential and the leftover problem is one coordinate, one velocity, one energy equation.

Figure

The reduction pipeline — two conservation laws take six variables down to one

The disguised angular energy — the centrifugal barrier $L^2/2\mu r^2$ — encodes a physical bargain. To move closer to the center while keeping $L=\mu r^2\dot\phi$ fixed, the particle must spin around faster, and that spinning costs kinetic energy that grows like $1/r^2$ . For any $L\neq 0$ the cost diverges as $r\to 0$ : an infinitely high wall guards the origin, which is why a comet with even a whisper of sideways motion can never fall straight into the Sun. Only a perfectly radial plunge ( $L=0$ ) — or a potential even more violently attractive than $1/r^2$ — gets through. One more picture before the algebra: what does the reduced motion look like when you re-inflate it to two dimensions? The radius oscillates between two turning points $r_{\min}$ and $r_{\max}$ (where $E=V_{\mathrm{eff}}$ ), while the angle sweeps steadily per $\dot\phi=L/\mu r^2$ . The generic result is a rosette — radial breathing riding on angular circulation — that fills the annulus between the two radii. Closed shapes like the Kepler ellipse are the rare, special case in which the two rhythms lock.

Figure

Radial oscillation riding on angular sweep — the generic central-force orbit is a rosette in an annulus

The reduction, done properly

Set the stage in the orbital plane, which the frozen direction of $\vec{L}$ has already picked out for us. Two functions of the motion are conserved. The energy, written in polar coordinates where the kinetic energy splits into its radial and angular parts:

E=\frac12\,\mu\dot{r}^2+\frac12\,\mu r^2\dot\phi^2+V(r) \qquad (1)

where: $\mu$ — the reduced mass of the two-body system (for a satellite around Earth, $\mu\approx$ the satellite's mass; for a binary star, substantially less than either mass); $V(r)$ — the central potential, related to the force by $f(r)=-\mathrm{d}V/\mathrm{d}r$ . The two velocity components enter squared and separately — this clean split is what makes the coming substitution surgical. And the angular momentum magnitude, the polar workhorse inherited from the previous article:

L=\mu r^2\dot\phi=\text{const} \quad\Longrightarrow\quad \dot\phi=\frac{L}{\mu r^2} \qquad (2)

where: the solved form on the right is the whole trick — the angular velocity is not an independent unknown but a fixed function of the current radius. A comet at 1 AU with given $L$ has its $\dot\phi$ dictated; let it fall to 0.5 AU and $\dot\phi$ must quadruple. Substituting this constraint everywhere $\dot\phi$ appears is the entire content of the reduction.

Derivation AFrom six variables to one — the full reduction

We are deriving the reduced radial energy equation — the boxed result $E=\tfrac12\mu\dot{r}^2+V_{\mathrm{eff}}(r)$ — starting from the full three-dimensional Lagrangian, so that every discarded degree of freedom is discarded for an explicit, named reason. The strategy is systematic disarmament: rotational symmetry conserves $\vec{L}$ ; the direction of $\vec{L}$ eliminates the polar angle; the cyclic azimuth conserves $L$ itself, which then eliminates $\dot\phi$ from the energy. Watch two things. First, each elimination is exact — powered by a conservation law, not an approximation. Second, the angular kinetic energy does not vanish in the last step; it is converted into an $r$ -dependent term, and mistaking that term for a genuine potential energy (rather than kinetic energy in disguise) is the root of most later confusion.

Step 1. Write the Lagrangian in spherical coordinates $(r,\theta,\phi)$ :

\mathcal{L}=\frac12\,\mu\big(\dot{r}^2+r^2\dot\theta^2+r^2\sin^2\theta\,\dot\phi^2\big)-V(r)

Why start here: the potential depends on $r$ alone, so the symmetry we are about to spend is visible — no angle appears in $V$ .

Step 2. Use the frozen direction of $\vec{L}$ . Because $V=V(r)$ , the torque about the origin vanishes and $\vec{L}$ is a constant vector; orient the axes so the orbital plane is the equator, $\theta=\pi/2$ , giving $\dot\theta=0$ for all time. Why this is legitimate: the motion can never leave the plane perpendicular to $\vec{L}$ — proven in <a class="wiki-link" href="/wiki/aq-mech-angular-momentum-conservation-central" data-track="wiki_article.body.wiki_link">angular momentum conservation</a> — so setting $\theta=\pi/2$ discards nothing. One angular coordinate is gone.

Step 3. Spot the cyclic coordinate. With $\theta=\pi/2$ ,

\mathcal{L}=\frac12\,\mu\big(\dot{r}^2+r^2\dot\phi^2\big)-V(r), \qquad p_\phi=\frac{\partial\mathcal{L}}{\partial\dot\phi}=\mu r^2\dot\phi=L=\text{const}

Why: $\phi$ itself never appears in $\mathcal{L}$ — it is cyclic — so its conjugate momentum is conserved by the Euler–Lagrange equation. This re-derives equation (2) from the Lagrangian side and identifies $L$ as $p_\phi$ .

Step 4. Write the conserved energy and eliminate $\dot\phi$ with equation (2):

E=\frac12\,\mu\dot{r}^2+\frac12\,\mu r^2\dot\phi^2+V(r), \qquad \frac12\,\mu r^2\dot\phi^2=\frac12\,\mu r^2\,\frac{L^2}{\mu^2r^4}=\frac{L^2}{2\mu r^2}

Why the energy is conserved at all: $\mathcal{L}$ has no explicit time dependence. And why the substitution loses nothing: $L$ is a constant of the motion, so we traded a variable for a number.

Step 5. Collect the one-dimensional result:

E=\frac12\,\mu\dot{r}^2+\frac{L^2}{2\mu r^2}+V(r)\;\equiv\;\frac12\,\mu\dot{r}^2+V_{\mathrm{eff}}(r)

Why this is the payoff: everything on the right depends on $r$ and $\dot{r}$ only. A single fictitious particle of mass $\mu$ moves on the half-line $r>0$ in the potential $V_{\mathrm{eff}}$ — one-dimensional mechanics, solvable by the standard playbook. *Verified: matches Goldstein §3.2 Eq. (3.22) and Landau–Lifshitz §14 Eq. (14.4); setting $L=0$ recovers pure radial motion in the bare $V(r)$ , and restoring $\dot\phi$ via (2) reproduces the full planar energy identically.*

The two objects the derivation delivered deserve their own displays. The effective potential:

V_{\mathrm{eff}}(r)=V(r)+\frac{L^2}{2\mu r^2} \qquad (3)

where: the second term — the centrifugal barrier — is positive and repulsive, blowing up as $r\to 0$ for any $L\neq 0$ . It is not a new interaction: it is the angular kinetic energy, rewritten via conserved $L$ as a function of position. Different orbits (different $L$ ) live in genuinely different effective landscapes — one potential per angular momentum. And the reduced energy equation, the master equation of this article:

E=\frac12\,\mu\dot{r}^2+V_{\mathrm{eff}}(r) \qquad (4)

where: $\tfrac12\mu\dot{r}^2$ — the radial kinetic energy, the only kinetic term left standing. Read it exactly as you read one-dimensional energy diagrams: motion exists only where $E\geq V_{\mathrm{eff}}$ , the radial speed at radius $r$ is $\sqrt{(2/\mu)(E-V_{\mathrm{eff}})}$ , and radii where $E=V_{\mathrm{eff}}$ are turning points of $r$ . What the whole landscape looks like, potential by potential, is the subject of the effective potential; here we extract the dynamics.

Figure

Reading the reduced equation — one energy line against one landscape answers every qualitative question

From the energy to the equation of motion

Sometimes you want force language rather than energy language — for stability analysis, or to see the centrifugal effect as an explicit push.

Go deeper · Derivation B · 2 minThe radial equation of motion from Newton in polar coordinates

Derivation BThe radial equation of motion from Newton in polar coordinates

We are deriving the second-order equation of motion for the radius, equation (5) — the force-language twin of the energy equation (4) — directly from Newton's second law, so that the centrifugal term's kinematic origin is out in the open rather than smuggled in through the energy. The strategy: write the acceleration in polar components (where differentiating rotating unit vectors produces the extra terms), let the angular component re-confirm the conservation of $L$ , and then use that conservation to close the radial component on itself. The step to watch is the sign and pedigree of $-r\dot\phi^2$ : it is a kinematic piece of the acceleration, and only after moving it across the equals sign does it masquerade as an outward force.

Step 1. Recall the polar acceleration components (from differentiating $\vec{r}=r\hat{r}$ twice, with $\dot{\hat{r}}=\dot\phi\,\hat{\phi}$ and $\dot{\hat{\phi}}=-\dot\phi\,\hat{r}$ ):

a_r=\ddot{r}-r\dot\phi^2, \qquad a_\phi=r\ddot\phi+2\dot{r}\dot\phi

Why the extra terms: the unit vectors themselves rotate with the particle, and their time derivatives generate the centripetal ( $-r\dot\phi^2$ ) and Coriolis-type ( $2\dot{r}\dot\phi$ ) contributions.

Step 2. Impose Newton's law with a purely radial force, $\vec{F}=f(r)\hat{r}$ . The angular component reads $\mu(r\ddot\phi+2\dot{r}\dot\phi)=0$ , which multiplied by $r$ is

\frac{\mathrm{d}}{\mathrm{d}t}\big(\mu r^2\dot\phi\big)=0

Why this is reassuring: Newton's angular equation is exactly the conservation of $L=\mu r^2\dot\phi$ — the same law we imported from the previous article, now recovered as a consistency check.

Step 3. The radial component: $\mu(\ddot{r}-r\dot\phi^2)=f(r)$ . Eliminate $\dot\phi$ with $\dot\phi=L/\mu r^2$ :

\mu\ddot{r}=f(r)+\mu r\cdot\frac{L^2}{\mu^2r^4}=f(r)+\frac{L^2}{\mu r^3}

Why the term turned positive: the centripetal piece $-\mu r\dot\phi^2$ sat on the acceleration side; moved to the force side it flips sign and points outward — the centrifugal push, now expressed through the conserved $L$ and the radius alone.

Step 4. Recognize the right side as the gradient of the effective potential:

f(r)+\frac{L^2}{\mu r^3}=-\frac{\mathrm{d}}{\mathrm{d}r}\left(V(r)+\frac{L^2}{2\mu r^2}\right)=-\frac{\mathrm{d}V_{\mathrm{eff}}}{\mathrm{d}r}

Why this closes the loop: the force-language and energy-language pictures are the same reduced problem — equation (5) is the time derivative of equation (4), and both are governed by the single landscape $V_{\mathrm{eff}}$ . *Verified: matches Goldstein Eq. (3.11)–(3.12) and Taylor Ch. 8; differentiating the energy equation (4) in time and dividing by $\dot{r}$ reproduces the same equation, confirming consistency of the two routes.*

\mu\ddot{r}=-\frac{\mathrm{d}V_{\mathrm{eff}}}{\mathrm{d}r}=f(r)+\frac{L^2}{\mu r^3} \qquad (5)

where: $f(r)=-\mathrm{d}V/\mathrm{d}r$ — the true radial force (negative for attraction); $+L^2/\mu r^3$ — the centrifugal term, always outward. A circular orbit is the special solution $\ddot{r}=0$ : the true inward pull exactly balances the centrifugal push, $-f(r_0)=L^2/\mu r_0^3$ , which is the same statement as $V_{\mathrm{eff}}'(r_0)=0$ .

Go deeper · Derivation C · 1 minSolving the reduced problem — time by quadrature, then the orbit shape

Derivation CSolving the reduced problem — time by quadrature, then the orbit shape

The goal is the two closing formulas of central-force theory: the travel time $t(r)$ along the orbit, and the geometric path $\phi(r)$ — both as explicit integrals requiring no further dynamics. The strategy exploits the structure of equation (4): it is first-order in $\dot{r}$ and contains only $r$ , so it separates directly; the orbit shape then follows by using the chain rule to trade time for angle, with equation (2) supplying the exchange rate. Watch the $\pm$ sign on the radial speed — it flips between the outbound and inbound legs of the orbit, and forgetting the flip double-counts or cancels half of every round trip.

Step 1. Solve the energy equation (4) for the radial speed:

\dot{r}=\pm\sqrt{\frac{2}{\mu}\big(E-V_{\mathrm{eff}}(r)\big)}

Why the $\pm$ : the energy fixes only $\dot{r}^2$ . The $+$ sign holds while the particle recedes, $-$ while it approaches; the sign flips at turning points, where $E=V_{\mathrm{eff}}$ and the square root vanishes.

Step 2. Separate variables and integrate for the time history:

t-t_0=\int_{r_0}^{r}\frac{\mathrm{d}r'}{\sqrt{\dfrac{2}{\mu}\big(E-V_{\mathrm{eff}}(r')\big)}}

Why this is already a solution: the right side is a definite integral of known functions — a quadrature. Inverting gives $r(t)$ ; for the Kepler potential this route leads to Kepler's equation for position versus time.

Step 3. Trade time for angle with the chain rule and equation (2):

\frac{\mathrm{d}\phi}{\mathrm{d}r}=\frac{\dot\phi}{\dot{r}}=\frac{L/\mu r^2}{\pm\sqrt{(2/\mu)\big(E-V_{\mathrm{eff}}(r)\big)}}

Why this kills time entirely: both rates are known functions of $r$ alone, so their ratio is a purely geometric statement — how much angle is swept per unit of radial progress.

Step 4. Integrate for the orbit shape:

\phi-\phi_0=\int_{r_0}^{r}\frac{L\,\mathrm{d}r'}{r'^2\sqrt{2\mu\big(E-V_{\mathrm{eff}}(r')\big)}}

Why stop here: this integral, with the substitution $u=1/r$ , becomes the standard form that yields <a class="wiki-link" href="/wiki/aq-mech-orbit-equation-binet" data-track="wiki_article.body.wiki_link">Binet's equation</a> and, for $V=-k/r$ , integrates in closed form to conic sections — the next two articles' business. *Verified: reproduces Goldstein Eqs. (3.18)–(3.37) and Landau–Lifshitz §14 Eq. (14.7); dimension check — the integrand of Step 4 is (angular momentum)/(length² × momentum) = dimensionless per length, as an angle-per-radius must be.*

For reference, the two quadratures in their standard forms:

t-t_0=\int_{r_0}^{r}\frac{\mathrm{d}r'}{\sqrt{\dfrac{2}{\mu}\big(E-V_{\mathrm{eff}}(r')\big)}} \qquad (6)

\phi-\phi_0=\int_{r_0}^{r}\frac{L\,\mathrm{d}r'}{r'^2\sqrt{2\mu\big(E-V_{\mathrm{eff}}(r')\big)}} \qquad (7)

where: equation (6) answers when (the clock along the orbit) and equation (7) answers where (the shape of the path). Between them and the energy diagram of equation (4), every question about central-force motion has a definite computational route — which is what "reduced to quadratures" means: solved, up to doing integrals.

Which equation answers which question

you want	use	notes
the 1D energy picture	$E=\tfrac12\mu\dot{r}^2+V_{\mathrm{eff}}$ , eq. (4)	turning points, bound vs unbound, at a glance
the effective landscape	$V_{\mathrm{eff}}=V+L^2/2\mu r^2$ , eq. (3)	one landscape per value of $L$
radial speed at radius $r$	$\dot{r}=\pm\sqrt{(2/\mu)(E-V_{\mathrm{eff}})}$	sign flips at turning points
force-language dynamics / circular orbits	$\mu\ddot{r}=f(r)+L^2/\mu r^3$ , eq. (5)	circular orbit: $\ddot r=0$ balance
time along the orbit	quadrature (6)	Kepler's equation comes from here
shape of the orbit	quadrature (7)	$u=1/r$ → Binet's equation
angular rate anywhere	$\dot\phi=L/\mu r^2$ , eq. (2)	re-inflates the 1D solution to the 2D orbit

Limiting cases

Case	Result	Meaning
$L=0$	$V_{\mathrm{eff}}=V(r)$ exactly	no barrier; pure radial fall or escape along a line through the center
circular orbit	$V_{\mathrm{eff}}'(r_0)=0$ , $\dot{r}=0$ , $E=V_{\mathrm{eff}}(r_0)$	radius frozen at the landscape minimum; only $\phi$ advances
$r\to 0$ with $L\neq 0$	$L^2/2\mu r^2\to+\infty$	barrier beats any potential less singular than $1/r^2$ — collapse forbidden, $r_{\min}>0$ guaranteed
Kepler potential $V=-k/r$	$V_{\mathrm{eff}}$ has a single minimum	bound orbits ( $E<0$ ) oscillate between two roots; quadrature (7) gives conic sections
inverse-cube attraction $V=-\alpha/r^2$	barrier and potential scale identically	if $\alpha>L^2/2\mu$ the "barrier" loses: the particle spirals into the origin — the reduction holds, the protection doesn't

Common trapA turning point of V eff is where the particle stops — no, it is where it turns

The reduced problem is a useful fiction, and the pitfall is taking it literally. At a radial turning point $\dot{r}=0$ , but the real particle is emphatically not at rest: $\dot\phi=L/\mu r^2\neq 0$ , so it is moving purely tangentially — perihelion is the fastest point of a planet's year, not a pause. The centrifugal term is angular kinetic energy in disguise, not a potential energy stored in any field; it changes if you change $L$ , i.e. it is orbit-dependent in a way no true potential is. And the mass in every formula is the reduced mass $\mu=m_1m_2/(m_1+m_2)$ — for genuinely comparable masses (binary stars, the Earth–Moon pair treated honestly) substituting either body's mass is a systematic error, not a rounding one.

Worked examples

1 · Where does this satellite turn around? (confidence builder)

A 1000 kg satellite is at $r_0=7000$ km from Earth's center (about 620 km altitude), moving at $v=8.2$ km/s purely tangentially. With $GM_\oplus=3.986\times10^{14}\ \mathrm{m^3/s^2}$ , find the two turning radii of its orbit. (Here $\mu\approx m$ , since the satellite is negligible next to Earth.)

Step 1. Compute the two conserved quantities per unit mass. Angular momentum: the velocity is tangential, so $\ell=L/m=r_0v=7.0\times10^{6}\times8.2\times10^{3}=5.74\times10^{10}\ \mathrm{m^2/s}$ . Energy: $\varepsilon=E/m=\tfrac12 v^2-GM_\oplus/r_0=3.362\times10^{7}-5.694\times10^{7}=-2.332\times10^{7}\ \mathrm{J/kg}$ . Negative energy — bound orbit, so two turning points exist.

Step 2. Turning points are the radii where $E=V_{\mathrm{eff}}$ , i.e. $\varepsilon=-\dfrac{GM_\oplus}{r}+\dfrac{\ell^2}{2r^2}$ . Multiply through by $r^2$ : $\varepsilon r^2+GM_\oplus\,r-\tfrac12\ell^2=0$ — a quadratic in $r$ .

Step 3. Solve it: $r=\dfrac{-GM_\oplus\pm\sqrt{GM_\oplus^2+2\varepsilon\ell^2}}{2\varepsilon}$ . The discriminant: $GM_\oplus^2=1.589\times10^{29}$ , $2\varepsilon\ell^2=-1.537\times10^{29}$ , sum $=5.19\times10^{27}$ , root $=7.21\times10^{13}$ . With $2\varepsilon=-4.665\times10^{7}$ : $r_{\min}=\boxed{7000\ \mathrm{km}}$ and $r_{\max}=\boxed{10{,}090\ \mathrm{km}}$ .

Step 4. Read the physics: $r_{\min}$ came out equal to $r_0$ — of course, because the velocity there was purely tangential, which is exactly the turning-point condition $\dot{r}=0$ . The satellite is at perigee right now. At apogee its speed will be $\ell/r_{\max}=5.74\times10^{10}/1.009\times10^{7}=5.69\ \mathrm{km/s}$ — moving, not stopped, at the outer turning point. Cross-check: $a=(r_{\min}+r_{\max})/2=8545$ km gives $\varepsilon=-GM_\oplus/2a=-2.332\times10^{7}$ J/kg ✓.

2 · How long to fall into the Sun? (exam level)

Earth's orbital motion is suddenly, magically stopped at $r_0=1\ \mathrm{AU}=1.496\times10^{11}$ m. Use the radial quadrature to find how long the Earth would take to fall into the Sun ( $GM_\odot=1.327\times10^{20}\ \mathrm{m^3/s^2}$ ).

Step 1. Identify the reduced problem. Stopping the orbital motion sets $L=0$ — the cleanest limiting case: $V_{\mathrm{eff}}=V=-GM_\odot\mu/r$ with no barrier, so the fall proceeds all the way to $r=0$ . Released from rest, the energy per unit mass is $\varepsilon=-GM_\odot/r_0$ .

Step 2. Write the radial speed from equation (4): $\dot{r}^2=2\big(\varepsilon+GM_\odot/r\big)=2GM_\odot\big(\tfrac1r-\tfrac1{r_0}\big)$ , inbound sign negative.

Step 3. Set up the quadrature (6) for the full fall:

t_{\text{fall}}=\int_0^{r_0}\frac{\mathrm{d}r}{\sqrt{2GM_\odot\left(\dfrac1r-\dfrac1{r_0}\right)}}=\sqrt{\frac{r_0}{2GM_\odot}}\int_0^{r_0}\sqrt{\frac{r}{r_0-r}}\,\mathrm{d}r

Step 4. Evaluate with the substitution $r=r_0\sin^2\alpha$ , $\mathrm{d}r=2r_0\sin\alpha\cos\alpha\,\mathrm{d}\alpha$ ; the square root becomes $\tan\alpha$ and the integrand collapses to $2r_0\sin^2\alpha\,\mathrm{d}\alpha$ , integrating from $0$ to $\pi/2$ into $r_0\pi/2$ . Hence

t_{\text{fall}}=\frac{\pi}{2}\sqrt{\frac{r_0^3}{2GM_\odot}}

Step 5. Numbers: $r_0^3=3.348\times10^{33}\ \mathrm{m^3}$ , divided by $2GM_\odot=2.654\times10^{20}$ gives $1.262\times10^{13}\ \mathrm{s^2}$ ; square root $3.552\times10^{6}$ s; times $\pi/2$ : $t_{\text{fall}}=5.58\times10^{6}\ \mathrm{s}=\boxed{64.6\ \mathrm{days}}$ .

Step 6. Cross-check without any integral: the fall is a degenerate ellipse of semi-major axis $a=r_0/2$ , so by Kepler's third law the fall time is half a period, $t=\pi\sqrt{(r_0/2)^3/GM_\odot}=5.58\times10^{6}$ s ✓. Two independent routes, one answer — and note the general lesson: $t_{\text{fall}}=T_{\text{orbit}}/(4\sqrt{2})\approx 0.177\,T$ , valid for any circular orbit magically halted.

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Paris — Académie des Sciences

Laplace turned Newton's gravitation into a finished industrial process. His five-volume Mécanique céleste (1799–1825) systematically reduced every celestial motion then known to differential equations and solved them by series and quadrature — precisely the program this article performs on one orbit, scaled to the whole Solar System. He settled the great anxiety of the age — whether Jupiter's and Saturn's drifting motions would eventually wreck the Solar System — by showing the drifts were periodic, not cumulative: stability by calculation. Napoleon, reviewing the book, supposedly asked where God fit in the system; Laplace's reply — "I had no need of that hypothesis" — is the most famous shrug in physics. A political survivor (marquis under the Restoration after serving the Revolution and Napoleon), he also co-invented the metric system, founded modern probability theory, and left his name on the transform, the equation, and the demon.

History

1687

— Newton's Principia solves the inverse-square orbit geometrically — brilliant, but each result a bespoke construction rather than a method.

1710

— Johann Bernoulli treats the general central-force problem analytically, showing the orbit follows from energy and angular momentum by integration — the quadrature route in embryo.

1788

— Lagrange's Mécanique analytique makes it systematic: cyclic coordinates have conserved momenta, and conserved momenta eliminate degrees of freedom.

1799

— Laplace's Mécanique céleste begins rolling the machinery over the entire Solar System, volume by volume, for a quarter century.

1877

— Routh formalizes the elimination of cyclic coordinates (the Routhian); the "effective potential" acquires its modern, general-purpose form.

Going further

Texts. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §3.1–3.3 · Landau & Lifshitz, Mechanics, 3rd ed., §13–14 · Morin, Introduction to Classical Mechanics, Ch. 7 · Taylor, Classical Mechanics, Ch. 8.

Provenance. Equations cross-checked against Goldstein §3.1–3.3 (reduced energy Eq. 3.22, orbit integral Eq. 3.37), Landau–Lifshitz §13–14 (Eqs. 14.4–14.7), and Taylor Ch. 8. Verifiers should confirm: the planarity argument $\mathbf{L}\cdot\mathbf{r}=0$ ; the substitution $\dot\phi=L/\mu r^2$ yielding the centrifugal term $L^2/2\mu r^2$ ; the quadrature for $t(r)$ and $\phi(r)$ ; and the Kepler-orbit constants $p=L^2/\mu k$ , $e=\sqrt{1+2EL^2/\mu k^2}$ . No numerical constants are quoted beyond standard textbook relations. Worked-example figures on this page (turning radii 7000/10,090 km with apogee speed 5.69 km/s, checked against $\varepsilon=-GM/2a$ ; solar fall time 64.6 days, checked against the degenerate-ellipse half-period) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Central force fundamentals · The two-body problem and reduced mass · Angular momentum conservation in central fields. Leads to: The effective potential · Classification of orbits by energy · The orbit equation and Binet's equation · The Kepler problem.

Is the centrifugal barrier a real force or not? My textbook seems to have it both ways

Both descriptions are legitimate, and knowing which frame each belongs to resolves the whiplash. In the honest two-dimensional description there is no centrifugal force at all — only the true central force $f(r)$ , plus the geometric fact that the radial acceleration in polar coordinates is $\ddot{r}-r\dot\phi^2$ , not just $\ddot{r}$ . When you move the $-r\dot\phi^2$ term across the equals sign and use $L=\mu r^2\dot\phi$ , it becomes $+L^2/\mu r^3$ and looks like an outward force. In the reduced one-dimensional fiction — a bead on the $r$ -axis — that term is perfectly real: the bead genuinely gets pushed away from the origin. The term is angular kinetic energy in disguise, and the disguise works because conservation of $L$ ties its value to position alone. What it is not is a new interaction of nature.

If the particle is still moving at a turning point, what exactly turns around?

Only the radius. "Turning point" in the reduced problem means $\dot{r}=0$ : the radial coordinate stops decreasing and starts increasing (or vice versa). The actual particle sails through that moment at full tangential speed $v=r\dot\phi=L/\mu r$ — for a planet, perihelion is a turning point of $r$ and simultaneously the fastest instant of the entire orbit. The 1D fiction hides the angular motion inside $V_{\mathrm{eff}}$ , so "the bead momentarily at rest on the hillside" translates in real space to "the planet moving exactly sideways." If you want a true stop in space you need both $\dot{r}=0$ and $L=0$ — the dropped-from-rest radial plunge.

When do I actually need the reduced mass instead of the ordinary mass?

Whenever the "fixed" center is allowed to move — which is always, strictly. The reduction of the two-body problem shows the relative coordinate obeys a one-body equation with $\mu=m_1m_2/(m_1+m_2)$ . When one body dwarfs the other, $\mu\approx m_{\text{small}}$ and the distinction is cosmetic: for a satellite around Earth the error is parts per $10^{21}$ . It stops being cosmetic fast: for the Earth–Moon pair $\mu$ is about 1.2% below the Moon's mass; for a double star with equal masses $\mu=m/2$ — a factor-of-two error if you're careless; and in spectroscopy the electron–proton reduced mass shifts hydrogen's lines measurably (it is how deuterium was discovered). Rule of thumb: mass ratio under 100, use $\mu$ without exception.

Where did the polar angle θ go? The reduction seems to silently drop a whole coordinate

It is not silent — it is the first conservation law doing its work. The full problem has two angles ( $\theta$ , $\phi$ in spherical coordinates). Because the force is central, the vector $\vec{L}$ is conserved, and the motion can never leave the plane through the center perpendicular to $\vec{L}$ — that is a theorem, proven in the previous article, not an assumption. Choosing axes so this plane is the equator ( $\theta=\pi/2$ , $\dot\theta=0$ ) then costs nothing: it is bookkeeping alignment, like choosing your $x$ -axis along a ramp. All the actual dynamics of the second angle was already dead; the coordinate choice just makes the corpse invisible. If someone hands you initial conditions not in your chosen plane, you rotate your axes, not the physics.

Is this the same trick as cyclic coordinates in Lagrangian mechanics?

Exactly the same, and seeing it both ways is worth the effort. In Lagrangian language: $\phi$ does not appear in $\mathcal{L}$ (only $\dot\phi$ does), so $\phi$ is cyclic and its conjugate momentum $p_\phi=\mu r^2\dot\phi=L$ is conserved; the systematic elimination of a cyclic coordinate in favor of its conserved momentum is Routh's procedure, and the potential of the resulting "Routhian" is precisely $V_{\mathrm{eff}}$ . The Newtonian route we took — substitute $\dot\phi=L/\mu r^2$ into the energy — is the same operation without the formal machinery. The Lagrangian view generalizes better: any ignorable coordinate in any problem (a bead on a spinning hoop, a charged particle around a flux line) yields an effective potential by the identical move.

Why is 1/r² the magic dividing line for the singularity of the potential?

Power counting against the barrier. Near the origin the centrifugal term is $+L^2/2\mu r^2$ . An attractive potential $V\sim -\alpha/r^n$ with $n<2$ (like Kepler's $n=1$ ) is weaker than the barrier at small $r$ : the sum $V_{\mathrm{eff}}\to+\infty$ , a wall, and $r_{\min}>0$ is guaranteed. With $n>2$ the attraction wins at small $r$ : $V_{\mathrm{eff}}\to-\infty$ , and a particle with enough energy to clear any intermediate hump spirals in — "fall to the center," in Landau's phrase. Exactly at $n=2$ it is an arm-wrestle of coefficients: attraction defeats the barrier when $\alpha>L^2/2\mu$ . This dividing line is physical, not formal — it reappears in quantum mechanics as the condition for an atom's stability against collapse and in general relativity, where corrections effectively steepen gravity near compact objects and open the door to plunging orbits.

Does reducing to one dimension throw away information about the actual orbit?

None whatsoever — the reduction is lossless compression, not truncation. The full state was $(r,\phi,\dot{r},\dot\phi)$ ; the reduced problem tracks $(r,\dot{r})$ , while the other two are reconstructible on demand: $\dot\phi=L/\mu r^2$ from the conserved $L$ , and $\phi(t)$ by integrating it (or directly from quadrature (7)). That is why the article can hand you both $t(r)$ and $\phi(r)$ : the one-dimensional problem contains the whole two-dimensional orbit, folded. The only thing genuinely fixed and forgotten is the orientation of the orbital plane and the direction of circulation — set by initial conditions, constant forever, and restored by a final rotation of coordinates.

The quadrature formulas look complete, so why do most potentials have no nice closed-form orbit?

"Reduced to quadratures" means the problem is solved in principle — the answer is a definite integral — but whether that integral hits a named function is a lottery of calculus. For power-law potentials $V\propto r^{n}$ , the orbit integral (7) yields elementary functions only for a handful of exponents (the Kepler $n=-1$ and harmonic $n=2$ chief among them; a few more reach elliptic functions — the classification is worked out in Landau–Lifshitz §14). This is a statement about our function catalogue, not about the physics: the integral always exists, always defines the orbit, and always evaluates numerically to machine precision in microseconds. In practice the qualitative analysis (energy diagram, turning points) plus numerical quadrature answers everything; closed forms are a luxury the Kepler problem happens to afford.

Reduction to the Radial Equation

Where this lives today

The physical picture

The reduction, done properly

From the energy to the equation of motion

Which equation answers which question

Limiting cases

Worked examples

1 · Where does this satellite turn around? (confidence builder)

2 · How long to fall into the Sun? (exam level)

Going further

Before this

Where this leads