Scattering in a Central Potential

How do you measure a force you cannot see? You throw things at it. Almost everything physics knows about interactions at inaccessible scales — the charge structure of the atom, the size of the nucleus, the shape of interatomic potentials, the gravity of a planet a spacecraft will never touch — was learned by firing projectiles past a center of force and reading the story written in their deflections. In 1911 that method cracked the atom open: the huge deflections of alpha particles skimming gold nuclei could only come from a pointlike concentration of charge. In 1974 the same mathematics, run in reverse as engineering, swung Mariner 10 around Venus to reach Mercury — the first gravity assist — and every outer-planet mission since has flown on hyperbolae designed by these equations. The classical theory is a jewel of economy. A particle arrives from infinity with energy $E > 0$ , aimed to pass the force center at perpendicular offset $b$ — the impact parameter. Angular momentum and energy conservation then determine everything: how close it gets, how hard it turns, and the angle $\theta$ between its incoming and outgoing straight-line asymptotes. The whole subject collapses into a single function, the deflection function $\theta(b, E)$ , delivered by one quadrature — the same effective-potential machinery that handles bound orbits, now run on the unbound branch. This article builds that function: the master integral from conservation laws, its exact evaluation for the Coulomb/Kepler force and the hard sphere, the small-angle impulse limit, and the geometry of the gravitational slingshot. What an experiment with a beam of particles measures — the differential cross-section — is the deflection function's companion story, told in its own article.

Key takeaways

Scattering is central-force mechanics on the unbound branch: $E > 0$ , a single turning point $r_{\min}$ , and a trajectory with two straight asymptotes whose angle of intersection is the scattering angle $\theta$ .
The impact parameter $b$ fixes the angular momentum from pure asymptotics: $\ell = \mu v_\infty b = b\sqrt{2\mu E}$ — so $(b, E)$ alone determine the entire orbit.
Conservation laws deliver the deflection function by quadrature: $\theta(b,E) = \pi - 2\int_{r_{\min}}^{\infty}\frac{(\ell/r^2)\,\mathrm{d}r}{\sqrt{2\mu(E-U) - \ell^2/r^2}}$ , with the factor 2 from the orbit's mirror symmetry about the apsidal line.
For the Coulomb/Kepler potential $U = k/r$ the integral closes exactly: $\tan(\theta/2) = |k|/(2Eb)$ — the formula behind both Rutherford's nucleus and every gravity assist.
A hard sphere gives pure reflection geometry, $b = R\cos(\theta/2)$ for $b \le R$ ; at large $b$ or high $E$ the impulse approximation $\theta \approx$ (transverse momentum kick)/(forward momentum) takes over.
Converting $\theta(b)$ into what a beam experiment measures is the job of the differential cross-section — the companion article to this one.
Key idea: all of central scattering collapses to one function, the deflection angle $\theta(b,E)$ , obtained by quadrature from energy and angular-momentum conservation; the measurable differential cross-section is its geometric Jacobian, $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega} = \frac{b}{\sin\theta}\big|\mathrm{d}b/\mathrm{d}\theta\big|$ .

The core insight

All of central scattering collapses to one function, the deflection angle $\theta(b,E)$ , obtained by quadrature from energy and angular-momentum conservation; the measurable differential cross section is just its geometric Jacobian $\frac{\mathrm{d}\sigma}{\mathrm{d}\Omega}=\frac{b}{\sin\theta}\big|\mathrm{d}b/\mathrm{d}\theta\big|$ .

Where this lives today

Hyperbolic-orbit navigation is a production technology in 2026. NASA's Europa Clipper, launched in 2024, swung past Mars in early 2025 and returns for its Earth gravity assist in December 2026 — one more solved instance of equation (5) — before its hyperbola is bent toward Jupiter arrival in 2030; ESA's JUICE is threading a parallel chain of Earth and Venus flybys. Each encounter is designed by choosing $(v_\infty, b)$ so the deflection function delivers the required turn, with perijove checked against radiation and atmosphere constraints exactly as in the worked example. The same mechanics runs in reverse for planetary defense: when an asteroid passes close to Earth, its future path is the outgoing asymptote of a hyperbolic encounter, and the uncertainty in $b$ — through keyholes a few hundred meters wide — decides whether a later return threatens an impact. At the opposite end of the size scale, Coulomb deflection functions are the workhorse of plasma physics and materials engineering. In a fusion plasma, transport coefficients are built from accumulated small-angle Coulomb scatterings — the $1/b$ tail of equation (5) integrated over aims from the Debye screening length down to the $90°$ impact parameter $b_{90} = d/2$ , producing the Coulomb logarithm $\ln\Lambda$ that appears in every collision rate in ITER's design documents. And in semiconductor fabs, ion implantation profiles are computed from millions of Monte Carlo scattering events using screened-Coulomb deflection functions — Rutherford's 1911 integral, run at industrial scale to place dopant atoms in silicon.

The physical picture

Fix the force center and fly a particle past it. Two numbers are set while the particle is still infinitely far away: its energy $E = \tfrac12\mu v_\infty^2$ , and its aim — the perpendicular distance $b$ between the force center and the straight line it would travel if the force were switched off. Everything that follows is determined. Aimed dead-on ( $b = 0$ ), a repulsive center stops the particle at $U(r_{\min}) = E$ and throws it straight back: $\theta = \pi$ . Aimed generously wide, the particle barely notices: $\theta \to 0$ . In between, the deflection function $\theta(b)$ interpolates — and for a repulsive potential it does so monotonically, one angle for each aim.

Figure

The anatomy of one scattering event — impact parameter in, deflection angle out, with the orbit mirror-symmetric about the line of closest approach

Two conservation laws do all the work, and it pays to see what each one carries. Angular momentum is fixed at infinity, where the geometry is trivial: the lever arm of the incoming straight line is exactly $b$ , so $\ell = \mu v_\infty b$ . This is the only door through which the aim enters the dynamics — after this, $b$ lives inside $\ell$ . Energy conservation then governs the radial motion in the effective potential $U_{\mathrm{eff}} = U + \ell^2/2\mu r^2$ , exactly as for bound orbits, except that with $E > 0$ there is a single turning point $r_{\min}$ instead of two: the particle falls in, turns once, and leaves forever. The last structural fact is a symmetry that halves every calculation in this subject. The orbit is mirror-symmetric about the apsidal line — the ray from the force center through the point of closest approach. Whatever polar angle $\varphi_0$ the particle sweeps coming in from infinity to $r_{\min}$ , it sweeps the same angle going back out. With no force at all the velocity direction would rotate by $\pi$ as the particle passes (in and out along the same line); each leg of the actual swept angle eats into that, so the net turning is $\theta = \pi - 2\varphi_0$ . Every deflection function in this article is this one identity plus an integral for $\varphi_0$ .

Figure

Two deflection functions — the hard sphere cuts off sharply at its radius, while the Coulomb tail deflects at every aim, however wide

The two curves above are the extremes that bracket everything. The hard sphere is the finite-range archetype: past $b = R$ the deflection is exactly zero, full stop. The Coulomb potential is the infinite-range archetype: $\theta$ falls only as $1/b$ , so there is no aim so wide that the particle escapes unturned — the source of the famous forward divergence that its companion article confronts. Real potentials — screened charges, van der Waals tails, nuclear wells — live between these extremes, and attractive ones can do richer things: pull orbits into rainbow extrema, or even capture them onto spirals. Those phenomena get their own article; here we build the machine.

The deflection machinery

Start where the particle starts: at infinity. There the potential is gone, all the energy is kinetic, and both conserved quantities can be read off the straight incoming line:

\ell = \mu v_\infty b = b\sqrt{2\mu E} \qquad (1)

where: $\mu$ — the reduced mass of projectile and target (the two-body problem reduced as usual); $v_\infty = \sqrt{2E/\mu}$ — the asymptotic speed; $b$ — the impact parameter. This one line is the interface between aim and dynamics: the geometry of the experiment enters the equations of motion only through $\ell$ . Larger $b$ at fixed energy means a taller centrifugal barrier, hence a gentler encounter. The radial motion is the standard central-force story with $E > 0$ . Energy conservation gives

\dot r^2 = \frac{2}{\mu}\big(E - U(r)\big) - \frac{\ell^2}{\mu^2r^2} \qquad (2)

where: the last term is the centrifugal barrier written per unit mass squared. The turning point — the distance of closest approach — is the radius where the radial speed dies:

E = U(r_{\min}) + \frac{\ell^2}{2\mu r_{\min}^2} \qquad (3)

where: $r_{\min}$ — the outermost root of this equation. Note what it is not: it is not the solution of $U(r) = E$ unless $b = 0$ . At any finite impact parameter the centrifugal term does part of the stopping, so a repulsive center turns the particle around farther out than a head-on collision would. Head-on Coulomb gives the cleanest special case: $r_{\min} = d \equiv k/E$ , the classical distance of closest approach that first sized the nucleus.

Derivation AThe deflection function from the conservation laws

We are deriving the master formula of classical scattering: the deflection angle $\theta(b,E)$ as a single radial integral, equation (4). The strategy is the standard central-force reduction — divide the angular equation of motion by the radial one to eliminate time, integrate the resulting $\mathrm{d}\varphi/\mathrm{d}r$ over the incoming leg to get the swept angle $\varphi_0$ , then convert swept angle to deflection using the mirror symmetry about the apsidal line. Two things to watch: the integrand blows up like an inverse square root at $r_{\min}$ (harmlessly — the singularity is integrable, the particle spends finite time near the apse), and the geometric bookkeeping at the end, where the "no force" baseline of $\pi$ is what turns a swept angle into a deflection.

Step 1. Eliminate time between the two conservation laws. Angular momentum gives $\dot\varphi = \ell/\mu r^2$ ; the radial equation (2) gives $\dot r$ ; their ratio is pure geometry:

\frac{\mathrm{d}\varphi}{\mathrm{d}r} = \frac{\dot\varphi}{\dot r} = \frac{\ell/r^2}{\sqrt{2\mu\big(E - U(r)\big) - \ell^2/r^2}}

Why dividing works: both rates are known functions of $r$ alone once $E$ and $\ell$ are fixed, so the orbit's shape decouples from its schedule — we never need to know when the particle is where, only where it goes.

Step 2. Integrate over the incoming leg. Let $\varphi_0$ be the polar angle swept as the particle comes from $r = \infty$ down to $r_{\min}$ :

\varphi_0 = \int_{r_{\min}}^{\infty} \frac{(\ell/r^2)\,\mathrm{d}r}{\sqrt{2\mu\big(E - U(r)\big) - \ell^2/r^2}}

Why the endpoint is safe: near $r_{\min}$ the radicand vanishes linearly, so the integrand diverges as $(r - r_{\min})^{-1/2}$ — an integrable singularity. Physically the apse hogs time, not angle.

Step 3. Use the mirror symmetry. The orbit is symmetric about the apsidal line, so the outgoing leg sweeps another $\varphi_0$ : the angle between the two asymptotes, measured around the force center, is $2\varphi_0$ .

Step 4. Convert to deflection. With no force the velocity direction of a passing particle rotates by exactly $\pi$ relative to the center (in on one side, out the other, same line). Each swept angle $\varphi_0$ replaces part of that straight-through geometry, so the net turning of the velocity is

\theta(b,E) = \pi - 2\int_{r_{\min}}^{\infty} \frac{(\ell/r^2)\,\mathrm{d}r}{\sqrt{2\mu\big(E - U(r)\big) - \ell^2/r^2}} \qquad (4)

where: $\ell = b\sqrt{2\mu E}$ carries the aim, and $r_{\min}$ is the outermost root of (3). Why the sanity checks work: switch the force off ( $U = 0$ ) and the integral evaluates to $\arcsin(b/r)\big|_{r_{\min}=b}^{\infty} = \pi/2$ , giving $\theta = \pi - \pi = 0$ — no force, no deflection. Head-on repulsion ( $b \to 0$ , so $\ell \to 0$ ) kills the integrand and gives $\theta = \pi$ — straight back. *The two limits bracket the formula correctly: $U=0 \Rightarrow \theta=0$ (no force, no deflection) and $b=0 \Rightarrow \theta=\pi$ (head-on, straight back).*

Go deeper · Derivation B · 2 minThe Coulomb deflection function from the hyperbola

Derivation BThe Coulomb deflection function from the hyperbola

We are deriving the single most consequential deflection function in physics: $\tan(\theta/2) = |k|/(2Eb)$ for the potential $U = k/r$ — the formula that measured the nucleus and steers spacecraft. Rather than grinding through the integral (4), the strategy is to exploit what we already own from the Kepler problem: an $E > 0$ orbit in a $1/r$ potential is a hyperbola with known eccentricity, and a hyperbola's asymptote directions are elementary geometry. We find the half-angle $\varphi_0$ between the apsidal line and an asymptote, then apply $\theta = \pi - 2\varphi_0$ . Watch how the sign of $k$ drops out at the end — repulsive and attractive centers bend trajectories to opposite sides, but produce the same magnitude of deflection at the same aim.

Step 1. Recall the orbit's eccentricity. For $U = k/r$ with energy $E$ and angular momentum $\ell$ , the conic has

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

and $E > 0$ forces $e > 1$ : a hyperbola. Why the sign of $k$ is already invisible: only $k^2$ enters — the orbit's shape parameter cannot tell attraction from repulsion.

Step 2. Find where the asymptotes point. The conic sends $r \to \infty$ where its denominator vanishes; for either sign of $k$ this happens at polar angles (measured from the apsidal line) satisfying

\cos\varphi_0 = \frac{1}{e}

Why: for the repulsive branch $1/r \propto (e\cos\varphi - 1)$ , which has its zero at $\cos\varphi = 1/e$ ; the attractive branch $1/r \propto (1 + e\cos\varphi)$ reaches zero at $\cos\varphi = -1/e$ , an angle $\pi - \varphi_0$ — the mirror-image geometry, same opening between the asymptotes.

Step 3. Convert to the scattering angle with $\theta = \pi - 2\varphi_0$ :

\sin\frac{\theta}{2} = \sin\Big(\frac{\pi}{2} - \varphi_0\Big) = \cos\varphi_0 = \frac{1}{e} \;\;\Rightarrow\;\; \cot\frac{\theta}{2} = \sqrt{e^2 - 1}

Step 4. Insert the eccentricity and the aim. With $e^2 - 1 = 2E\ell^2/\mu k^2$ and $\ell^2 = 2\mu Eb^2$ from (1):

\cot^2\frac{\theta}{2} = \frac{2E\,(2\mu Eb^2)}{\mu k^2} = \left(\frac{2Eb}{k}\right)^2

Taking the positive root,

\tan\frac{\theta}{2} = \frac{|k|}{2Eb} \qquad (5)

where: $|k| = Z_1Z_2e^2/4\pi\varepsilon_0$ for two charges, or $GMm$ for gravity. Equivalently $\tan(\theta/2) = d/2b$ with $d = |k|/E$ the head-on closest-approach distance — the deflection is order one exactly when the aim is comparable to $d$ . Why this is the door to everything downstream: inverting to $b(\theta) = (|k|/2E)\cot(\theta/2)$ and feeding it to the cross-section Jacobian yields the Rutherford formula — carried out in the companion article on the differential cross-section. *The small-angle limit $\theta \approx |k|/Eb$ agrees with the impulse approximation (7) below, and the head-on limit $b\to0$ gives $\theta\to\pi$ as required.*

Go deeper · Derivation C · 1 minThe hard sphere by pure reflection geometry

Derivation CThe hard sphere by pure reflection geometry

We are deriving the deflection function of an impenetrable sphere of radius $R$ — the simplest scatterer there is, and the sanity anchor for the whole subject. No integral is needed: the "orbit" is two straight lines joined by one specular bounce, so the strategy is elementary geometry at the point of contact — relate the impact parameter to the angle of incidence, then apply the reflection law. The point to watch is the range of validity: the geometry only exists for $b \le R$ ; beyond that the particle misses entirely and $\theta = 0$ identically, the signature of a strictly finite-range interaction.

Step 1. Locate the contact. A particle aimed with impact parameter $b \le R$ strikes the sphere where its straight path meets the surface; if $\alpha$ is the angle between the incoming ray and the local inward normal (which points at the center), the right triangle formed by the center, the contact point, and the aim line gives

\sin\alpha = \frac{b}{R}

Step 2. Reflect. Specular bouncing preserves the angle to the normal, so the velocity turns by the exterior angle:

\theta = \pi - 2\alpha

Why this is the same $\pi - 2\varphi_0$ structure: the bounce is the degenerate limit of an apsidal turn — all the swept angle happens at one point, and $\alpha$ plays the role of $\varphi_0$ .

Step 3. Eliminate $\alpha$ to get the deflection function and its inverse:

\theta(b) = 2\arccos\frac{b}{R}, \qquad b(\theta) = R\cos\frac{\theta}{2} \quad (b \le R) \qquad (6)

where: head-on ( $b = 0$ ) gives $\theta = \pi$ , a grazing hit ( $b = R$ ) gives $\theta = 0$ , and for $b > R$ there is no deflection at all. Why physicists keep this example loaded: it is the cleanest possible contrast with Coulomb — a deflection function with a hard endpoint versus one with an infinite tail — and its cross-section works out to the geometric shadow $\pi R^2$ , the benchmark every intuition about "effective target area" is calibrated against (see the differential cross-section article). *The deflection falls monotonically from $\pi$ to $0$ as $b$ runs from $0$ to $R$ — the green curve in the second figure above.*

When the deflection is small — large $b$ , or high energy — the master integral simplifies into a form you can use on any potential without solving for the orbit at all. Treat the trajectory as the undeflected straight line $z \in (-\infty, \infty)$ at transverse distance $b$ , integrate the transverse force along it, and divide the accumulated transverse momentum by the forward momentum:

\theta(b) \approx \frac{|\Delta p_\perp|}{p} = -\frac{1}{2E}\int_{-\infty}^{\infty} \frac{\partial U\big(\sqrt{b^2 + z^2}\big)}{\partial b}\,\mathrm{d}z \qquad (7)

where: the derivative acts on the impact parameter inside $r = \sqrt{b^2+z^2}$ . This is the impulse (eikonal) approximation, valid when $\theta \ll 1$ , i.e. when $U(b) \ll E$ . For Coulomb it gives $\theta \approx |k|/Eb$ — exactly the small-angle limit of the exact result (5), a reassuring closure of the loop. Its real value is generality: for a screened potential $U = (k/r)e^{-r/a}$ , or any messy numerical $U(r)$ , equation (7) delivers the far-field deflection with one integral.

Which formula answers which question

you want	use	notes
aim → angular momentum	$\ell = b\sqrt{2\mu E}$ , eq. (1)	the only place $b$ enters the dynamics
distance of closest approach	eq. (3), outermost root	equals $U^{-1}(E)$ only when $b = 0$
exact deflection, any $U(r)$	master integral, eq. (4)	one quadrature; integrable endpoint singularity
Coulomb / gravity deflection	$\tan(\theta/2) = \vert k\vert/2Eb$ , eq. (5)	same $\vert\theta\vert$ for attraction and repulsion
hard sphere	$b = R\cos(\theta/2)$ , eq. (6)	zero deflection beyond $b = R$
weak deflection, any potential	impulse approximation, eq. (7)	valid when $U(b) \ll E$

Limiting cases

Case	Result	Meaning
no force, $U = 0$	$\theta = 0$	integral (4) evaluates to exactly $\pi/2$ per leg — the baseline
head-on repulsion, $b = 0$	$\theta = \pi$ , $r_{\min} = U^{-1}(E)$	straight back; Coulomb: $r_{\min} = k/E$ , Rutherford's ruler for the nucleus
wide aim, $b \to \infty$	Coulomb $\theta \approx \vert k\vert/Eb \to 0$ but never $0$	infinite range: every aim scatters a little — the forward divergence of the cross-section
high energy, fixed $b$	$\theta \to 0$ as $1/E$	fast particles punch through — why higher beam energies probe smaller structures
hard sphere, $b > R$	$\theta = 0$ exactly	finite range: sharp geometric shadow
attractive $U = -\alpha/r^2$ , $\ell^2 < 2\mu\alpha$	capture: spiral into the center	the centrifugal barrier loses to the potential — no asymptote, $\theta$ undefined

Common trapThe scattering angle is the polar angle the particle sweeps around the center

It is not, and conflating the two is the most common structural error in this subject. The swept polar angle from entry to exit is $2\varphi_0$ — always between $0$ and... well, for Coulomb it approaches $\pi$ for a grazing pass. The scattering angle is the turn of the velocity vector, $\theta = \pi - 2\varphi_0$ : a barely-deflected particle sweeps nearly $\pi$ of polar angle while its velocity turns hardly at all. Related traps: (1) $r_{\min}$ is set by the effective potential — solving $U(r_{\min}) = E$ is correct only for $b = 0$ ; at finite aim the centrifugal barrier turns the particle sooner. (2) All formulas here live in the center-of-mass frame with the reduced mass $\mu$ ; comparing with a lab experiment on a stationary target requires the kinematic transformation $\tan\theta_{\mathrm{lab}} = \sin\theta/(\cos\theta + m_1/m_2)$ — only for a heavy target do the two angles coincide. (3) For attractive potentials, watch for capture: if the trajectory has no turning point, the deflection function simply does not exist at that $(b, E)$ .

Worked examples

1 · An alpha particle grazes a gold nucleus (confidence builder)

A 5.0 MeV alpha particle ( $Z_1 = 2$ ) scatters off a gold nucleus ( $Z_2 = 79$ ). Using $e^2/4\pi\varepsilon_0 = 1.44\ \mathrm{MeV\,fm}$ , find the head-on distance of closest approach, then the deflection for an aim of $b = 45.5$ fm, and the aim required for a $90°$ deflection.

Step 1. Coulomb strength: $k = Z_1Z_2\,e^2/4\pi\varepsilon_0 = 2 \times 79 \times 1.44 = 227.5\ \mathrm{MeV\,fm}$ .

Step 2. Head-on closest approach: $d = k/E = 227.5/5.0 = \boxed{45.5\ \mathrm{fm}}$ — the alpha stops 45.5 fm from the center, far outside the $\sim7$ fm gold nucleus, which is why the pure Coulomb formula works at this energy. (Raise $E$ until $d$ reaches the nuclear surface and the scattering starts deviating from Coulomb — that onset measured the nuclear radius.)

Step 3. Deflection at $b = 45.5$ fm from (5): $\tan(\theta/2) = d/2b = 45.5/91.0 = 0.50$ , so $\theta/2 = 26.6°$ and $\boxed{\theta = 53.1°}$ .

Step 4. Aim for $90°$ : $b = (d/2)\cot(45°) = d/2 = \boxed{22.8\ \mathrm{fm}}$ . Read the scale of the result: to be deflected through a right angle, the alpha must be aimed within about 23 fm of a nucleus — a target roughly $10^{-9}$ the area of the atom. Large-angle scattering is rare precisely in proportion to how small the nucleus is; the rate of rare events is the measurement.

2 · Designing a Jupiter gravity assist (exam level)

A probe approaches Jupiter ( $GM_J = 1.267\times10^{17}\ \mathrm{m^3/s^2}$ , $R_J = 7.15\times10^7$ m) with $v_\infty = 10.0$ km/s and impact parameter $b = 3.0\times10^9$ m. Find the deflection angle, the perijove (closest approach), and the speed change the maneuver produces in the planet's frame.

Step 1. For gravity, $|k| = GM_Jm$ and $E = \tfrac12 mv_\infty^2$ , so the mass cancels in (5): $\tan(\theta/2) = GM_J/(v_\infty^2\,b) = 1.267\times10^{17}/(10^8 \times 3.0\times10^9) = 0.422$ .

Step 2. Deflection: $\theta/2 = \arctan(0.422) = 22.9°$ , so $\boxed{\theta = 45.8°}$ — the probe's direction swings by almost half a right angle without burning a gram of propellant.

Step 3. Perijove. Eccentricity from Step 3 of the Coulomb derivation: $e = 1/\sin(\theta/2) = 1/0.389 = 2.57$ . The hyperbola's scale is $a_h = GM_J/v_\infty^2 = 1.267\times10^9$ m, and closest approach for an attractive hyperbola is $r_{\min} = a_h(e - 1) = 1.267\times10^9 \times 1.57 = \boxed{2.0\times10^9\ \mathrm{m} \approx 28\,R_J}$ — comfortably above the radiation belts; the design is flyable.

Step 4. Speed change in Jupiter's frame. The speed at infinity is unchanged ( $|\mathbf{v}_{\mathrm{out}}| = |\mathbf{v}_{\mathrm{in}}| = v_\infty$ — gravity does zero net work over the symmetric encounter), but the direction turns by $\theta$ , so the vector change is $|\Delta\mathbf{v}| = 2v_\infty\sin(\theta/2) = 2 \times 10.0 \times 0.389 = \boxed{7.8\ \mathrm{km/s}}$ .

Step 5. Where the free lunch comes from. In the Sun's frame, Jupiter's own orbital velocity ( $\approx13$ km/s) adds vectorially to both legs; leaving on the right side of the planet, the probe can pocket a heliocentric speed gain of order $|\Delta\mathbf{v}|$ , paid for by an immeasurably small slowing of Jupiter. The deflection function is the steering law of the entire outer Solar System program.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · Manchester, then the Cavendish

Rutherford was the supreme experimentalist of the atomic age — a New Zealand farm boy who won the 1908 Nobel Prize in Chemistry for unraveling radioactivity, then did his greatest work afterwards. At Manchester in 1909 he set Geiger and the undergraduate Marsden to count alpha particles bounced off gold foil; the one-in-eight-thousand that came backwards stunned him — "as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you." His 1911 response was this article's mathematics: treating the atom's positive charge as a point and solving the hyperbolic orbit, he derived the deflection law $\tan(\theta/2) = k/2Eb$ and from it an angular distribution that Geiger and Marsden then confirmed point by point — the nucleus, discovered by quadrature. He went on to achieve the first artificial nuclear transmutation (1919) and to direct the Cavendish Laboratory through the golden years that split the atom and found the neutron.

History

1687

— Newton's Principia solves motion under inverse-square forces for all conic sections; the unbound hyperbolic branch waits two centuries for a job.

1909

— Geiger and Marsden observe alpha particles backscattered from gold foil, at roughly 1 in 8000 — impossible for a diffuse "plum pudding" of charge.

1911

— Rutherford publishes the point-charge scattering analysis: the hyperbolic-orbit deflection law and the angular law it implies, matching the data — the nuclear atom.

1961

— Michael Minovitch computes that hyperbolic planetary flybys can add heliocentric energy for free, founding gravity-assist mission design.

1974

— Mariner 10 flies the first gravity assist, using Venus to reach Mercury; Voyager's Grand Tour follows the same mathematics across the outer planets.

Going further

Texts. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §3.10–3.11 (scattering in central fields) · Landau & Lifshitz, Mechanics, 3rd ed., §18–19 · Taylor, Classical Mechanics, ch. 14 · Morin, Introduction to Classical Mechanics, ch. 7 (central forces). Key results collected here for reference: Rutherford $\mathrm{d}\sigma/\mathrm{d}\Omega=(k/4E)^2\csc^4(\theta/2)$ ; hard-sphere total cross-section $\sigma=\pi R^2$ with isotropic distribution giving $\mathrm{d}\sigma/\mathrm{d}\Omega = R^2/4$ ; eccentricity $e=\sqrt{1+2E\ell^2/\mu k^2}$ ; angular momentum $\ell=b\sqrt{2\mu E}$ . Worked-example numbers on this page ( $d = 45.5$ fm and $\theta = 53.1°$ for the alpha particle; $\theta = 45.8°$ , $r_{\min} \approx 28R_J$ , $|\Delta v| = 7.8$ km/s for the Jupiter flyby) follow directly from the boxed formulas.

Builds on: The Effective Potential · Angular Momentum Conservation in Central Fields · Reduction to the Radial Equation · Classification of Orbits by Energy · The Two-Body Problem and Reduced Mass. Leads to: The Differential Cross-Section · The Kepler Problem · Conic-Section Orbits · The Orbit Equation and Binet's Equation.

Why is the no-force baseline π and not 0 in the formula θ = π − 2φ0?

Because $\varphi_0$ measures the swept polar angle around the center, not the turning of the velocity. Watch a force-free particle pass a reference point: it comes in from infinity on one side and leaves to infinity on the other along the same straight line, and its position vector, seen from that point, rotates through exactly $\pi$ ( $\varphi_0 = \pi/2$ per leg — the integral in (4) gives precisely this with $U = 0$ ). So a swept angle of $2\varphi_0 = \pi$ means zero deflection, and every bit by which the potential drags $\varphi_0$ below $\pi/2$ (repulsion, which hurries the turn) shows up as positive deflection $\theta = \pi - 2\varphi_0$ . A strongly attractive potential can do the opposite — hold the particle and swing it through more than $\pi$ of polar angle, making the formal $\theta$ negative (the particle is deflected to the other side); the observable angle is then $|\theta|$ folded into $[0, \pi]$ .

Do attractive and repulsive Coulomb centers really deflect by the same angle at the same aim?

The same magnitude, yes — and it surprises everyone once. The eccentricity $e = \sqrt{1 + 2E\ell^2/\mu k^2}$ contains only $k^2$ , and the asymptote geometry then gives the same opening angle either way; the trajectories are quite different (the repulsive orbit's apse is on the near side, the attractive one bends around the far side, turning the opposite way), but $|\theta|(b)$ is identical. So a purely classical scattering experiment measuring only deflection magnitudes cannot tell the sign of the interaction — alpha particles on nuclei and electrons on nuclei would trace the same $|\theta|(b)$ law. Where the difference does show: the attractive case reaches much smaller $r_{\min}$ at the same aim (it is pulled in, not pushed out), so it probes deeper into the target and fails the pure-Coulomb assumption sooner; and quantum interference (Mott scattering) can distinguish the signs where classical mechanics cannot.

Is r min just the radius where U(r) = E?

Only for a perfectly head-on shot. At any finite impact parameter the particle must also carry angular momentum $\ell = b\sqrt{2\mu E}$ , and the turning point is where the effective potential — $U(r)$ plus the centrifugal term $\ell^2/2\mu r^2$ — climbs to $E$ , equation (3). The centrifugal barrier always turns the particle earlier (larger $r_{\min}$ ) than the bare potential would. The error is largest exactly where it matters: for grazing encounters, $r_{\min} \approx b$ , set almost entirely by the centrifugal term with the potential a bystander. Quick Coulomb check: at $b = d/2$ (the $90°$ aim), solving (3) gives $r_{\min} = (d/2)(1 + \sqrt2) \approx 1.21\,d$ — noticeably farther out than the head-on $d$ , even though this is already a hard collision.

The integrand of the deflection integral blows up at r min — how is the integral finite?

The divergence is an inverse square root, and $\int \mathrm{d}x/\sqrt{x}$ converges at zero. Near the turning point the radicand $2\mu(E - U) - \ell^2/r^2$ vanishes linearly in $r - r_{\min}$ (its derivative there is the effective radial force, generically nonzero), so the integrand grows like $(r - r_{\min})^{-1/2}$ — integrable. Physically: $\dot r \to 0$ at the apse, so the particle spends a long time there, but $\dot\varphi = \ell/\mu r^2$ stays finite, so it does not accumulate unbounded angle. Numerically the singularity still deserves respect: substitute $r = r_{\min}\,(1 + u^2)$ or integrate in the variable $\sqrt{r - r_{\min}}$ to tame it before throwing quadrature at equation (4). The exception proving the rule: if the effective-force derivative also vanishes at the turning point (orbiting), the divergence sharpens to logarithmic and $\theta$ genuinely diverges — the particle spirals many times around.

Where exactly does the energy for a gravity assist come from, if the planet-frame speed is unchanged?

From the planet's orbital motion — legally, via a frame change. In Jupiter's rest frame the encounter is perfectly elastic: same $v_\infty$ in and out, only the direction turns by $\theta$ . But the Sun's frame sees $\mathbf{v}_{\mathrm{helio}} = \mathbf{v}_{\mathrm{planet}} + \mathbf{v}_{\mathrm{probe/planet}}$ , and rotating the second vector changes the magnitude of the sum: pass behind the planet and you leave with the planet's velocity added more constructively — a net heliocentric speed gain up to $2v_\infty\sin(\theta/2)$ . Momentum and energy balance exactly: the probe's gain is the planet's loss, but a tonne of spacecraft slowing $10^{27}$ kg of Jupiter produces a velocity change of order $10^{-24}$ m/s — unmeasurable forever. Voyager 2's grand tour extracted tens of km/s this way; the deflection function (5) with attractive $k$ is the entire steering law.

When can I trust the straight-line impulse approximation, quantitatively?

The honest criterion is $U(b)/E \ll 1$ — the potential energy at the distance of nearest aimed approach must be small against the kinetic energy — which guarantees both that the deflection is small and that the true path hugs the straight line used in the integral. For Coulomb this reads $\theta \approx d/b \ll 1$ : trustworthy for aims many times the head-on distance $d$ . Two refinements worth knowing. First, the errors are second order: the approximation reproduces the exact Coulomb small-angle law $\theta = k/Eb$ with corrections of relative order $(d/b)^2$ , so it is better than it has any right to be. Second, it fails structurally, not just numerically, near features of the deflection function — rainbow extrema, orbiting — where the true path departs qualitatively from a straight line. It is a far-field tool: superb in the tail, blind near the drama.

How do I compare these center-of-mass formulas with a lab experiment where the target starts at rest?

Two conversions, both mechanical. First, kinematics: the CM-frame angle $\theta$ maps to the lab angle via $\tan\theta_{\mathrm{lab}} = \sin\theta/(\cos\theta + m_1/m_2)$ , where $m_1$ is the projectile and $m_2$ the target. For $m_1 \ll m_2$ (alphas on gold: ratio $\approx 1/49$ ) the correction is a percent-level detail and $\theta_{\mathrm{lab}} \approx \theta$ — one reason Rutherford's analysis could ignore it. For comparable masses it is dramatic: equal masses give $\theta_{\mathrm{lab}} = \theta/2$ with nothing ever scattered backward in the lab. Second, dynamics: all formulas here use the reduced mass $\mu = m_1m_2/(m_1+m_2)$ and the relative energy $E = \tfrac12\mu v_\infty^2$ — for a light target, $\mu$ differs substantially from $m_1$ and forgetting the swap shifts every energy in the problem. The angular distribution also picks up a Jacobian under the angle map, handled in the cross-section article.