Deflection of falling bodies

In 1679 Newton wrote to Robert Hooke with a proposal for making the Earth's rotation visible: drop a stone from a high place. Common sense said a spinning Earth should leave the stone behind, landing it to the west; Newton saw it the other way — the top of a tower rides a bigger circle than its base, so the stone starts with extra eastward speed and should land slightly east of the plumb line. He even sketched the spiral it would trace. The effect is maddeningly small — millimetres for any drop a person can arrange — and it took over a century before Guglielmini dropped balls inside a Bologna tower, Benzenberg down a Hamburg church spire, and finally Ferdinand Reich, in 1831, dropped pellets 158.5 metres down a Freiberg mineshaft and measured a mean eastward deflection of 28.3 mm against a predicted 27.6 mm — the first laboratory-grade proof that the ground beneath us turns. The falling stone is the cleanest solvable problem in rotating-frame mechanics. One writes Newton's law with the Coriolis force, expands in the small parameter $\omega t$ , and integrates: out falls the eastward drift $\tfrac13\omega g\cos\lambda\,t^3$ , a second-order equatorward whisper, and a set of sign lessons ( $\cos\lambda$ here, $\sin\lambda$ for the Foucault pendulum) that carry over to every projectile, missile, and atmospheric flow. Better still, the same answer can be re-derived in the inertial frame from nothing but angular-momentum conservation — and comparing the two routes settles exactly why the plausible-looking shortcuts get the answer wrong by factors of $\tfrac12$ and $\tfrac32$ .

Key takeaways

A body dropped from rest above a rotating Earth does not fall along the local plumb line: the Coriolis force $-2m\,\boldsymbol{\omega}\times\mathbf{v}$ deflects it, chiefly toward the east.
To leading order the eastward deflection from height $h$ is $x_E=\tfrac{1}{3}\,\omega g\,\cos\lambda\,t^3=\tfrac{1}{3}\,\omega\cos\lambda\sqrt{\tfrac{8h^3}{g}}$ , growing as $h^{3/2}$ .
The effect is tiny but measured: $2.2$ cm for a 100 m drop at the equator; Reich's 1831 Freiberg mineshaft experiment found $28.3$ mm where theory predicts $27.6$ mm.
The inertial-frame picture: the release point rides a larger circle than the ground, and conservation of angular momentum about Earth's axis makes the falling body outrun the ground beneath it — hence eastward, in both hemispheres.
At second order there is a far smaller equatorward deflection $y_S=-\tfrac16\omega^2 g\sin\lambda\cos\lambda\,t^4$ , plus corrections from the plumb-line tilt of effective gravity.
Key idea: Free fall + the Coriolis force $-2m\,\boldsymbol{\omega}\times\mathbf{v}$ , solved by iteration in $\omega$ : the leading effect is an eastward deflection $x_E=\tfrac{1}{3}\omega g\cos\lambda\,t^3$ , maximal at the equator and zero at the poles.

Where this lives today

The falling-body deflection is the "hello world" of inertial navigation. A submarine or aircraft inertial navigation system is, mathematically, an object integrating $\ddot{\mathbf r}=\mathbf g_{\text{eff}}-2\boldsymbol\omega\times\dot{\mathbf r}+\dots$ from accelerometer and gyro data over hours — precisely equation (1) — and if the software mishandled the Coriolis term, position errors would accumulate at exactly the $\tfrac13\omega g t^3$ rate computed here: centimetres in seconds, kilometres in hours. Long-range ballistics inherits the same physics with the vertical velocity changing sign in flight, and artillery fire-control tables have carried Coriolis columns since the First World War. The modern precision frontier is quantum. Atom interferometers measure $g$ by dropping clouds of ultracold atoms a few tens of centimetres and reading the phase of matter waves — and even over such tiny falls the Coriolis deflection shifts the interference phase enough to matter at the $10^{-9}\,g$ level. State-of-the-art gravimeters cancel it with tip-tilt mirrors that counter-rotate the reference laser against the Earth's rotation during each drop. Facilities like the Bremen drop tower, running catapulted capsules through 9 seconds of free fall in an evacuated 120 m tube, operate in exactly the regime of Worked Example 2 — where the eastward drift is a design parameter, not a curiosity. Newton's 1679 thought experiment is now a calibration line item in quantum gravimetry.

The physical picture

Imagine releasing a ball from the top of a tall tower. The ball starts with the eastward speed of the tower top, which is farther from Earth's axis than the base and therefore moving faster eastward. As it falls it keeps — indeed, slightly grows — that tangential motion while the ground beneath moves more slowly, so it lands a little to the east. Seen from the rotating Earth, the same story is told by a sideways force: the Coriolis force acting on the downward velocity points east. Two frames, one landing spot.

Figure

One drop, two stories — an eastward force in the rotating frame, an angular-momentum race in the inertial frame

The $\cos\lambda$ factor is geometric: only the component of $\boldsymbol\omega$ perpendicular to the (vertical) velocity produces a horizontal Coriolis force. At the equator $\boldsymbol\omega$ points north — fully horizontal, at right angles to a vertical fall — giving the maximal eastward push. At a pole $\boldsymbol\omega$ is vertical, parallel to the fall, and $\boldsymbol\omega\times\mathbf v=0$ : no deflection at all. This is exactly complementary to the Foucault pendulum, whose horizontal swing velocities couple to the vertical component $\omega\sin\lambda$ — the falling stone and the pendulum together measure both components of the Earth's angular velocity vector without ever looking at the sky.

Figure

Only the horizontal part of the Earth's rotation deflects a vertical fall — and it pushes east

Setting up the problem

The starting point is Newton's second law in the Earth-fixed rotating frame, with the full roster of fictitious forces. Its derivation and the sorting of the terms belong to the acceleration-transformation and Coriolis force articles; here we take the result and put it to work:

m\ddot{\mathbf r}=\mathbf F+m\mathbf g_{\text{eff}}-2m\,\boldsymbol\omega\times\dot{\mathbf r}-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)-m\,\dot{\boldsymbol\omega}\times\mathbf r \qquad (1)

where: $\mathbf r,\dot{\mathbf r}$ — position and velocity measured in the rotating (Earth-fixed) frame; $\boldsymbol\omega$ — Earth's angular velocity, $|\boldsymbol\omega|\approx7.292\times10^{-5}\ \mathrm{s^{-1}}$ ; $-2m\,\boldsymbol\omega\times\dot{\mathbf r}$ — the Coriolis force, our deflecting agent; $-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)$ — the centrifugal force, which we absorb into $\mathbf g_{\text{eff}}$ once and for all (that bookkeeping, and the plumb-line tilt it produces, is the effective-gravity article's business); $-m\,\dot{\boldsymbol\omega}\times\mathbf r$ — the Euler force, negligible since $\dot{\boldsymbol\omega}\approx0$ for the Earth. For a body in free fall the applied force is $\mathbf F=0$ , leaving the clean working equation $\ddot{\mathbf r}=\mathbf g-2\,\boldsymbol\omega\times\dot{\mathbf r}$ , with $g\approx9.81\ \mathrm{m/s^2}$ understood to point along the local plumb line. Next, geography. Set up local axes at the drop site: $\hat{\mathbf x}$ east, $\hat{\mathbf y}$ north, $\hat{\mathbf z}$ up. At geographic latitude $\lambda$ the planet's angular velocity vector, which points along the polar axis, has no east component and splits between north and up:

\boldsymbol\omega=\omega\,(0,\ \cos\lambda,\ \sin\lambda) \qquad (2)

where: $\omega\cos\lambda$ — the horizontal (northward) component; $\omega\sin\lambda$ — the vertical component. For a falling body the velocity is vertical, so only the horizontal component $\omega\cos\lambda$ can generate a horizontal force — the origin of every $\cos\lambda$ in this article, and the reason the effect is strongest at the equator. Writing out the cross product with $\dot{\mathbf r}=(\dot x,\dot y,\dot z)$ turns the vector equation into three coupled scalar ones:

\ddot x=2\omega(\dot y\sin\lambda-\dot z\cos\lambda),\qquad \ddot y=-2\omega\,\dot x\sin\lambda,\qquad \ddot z=-g+2\omega\,\dot x\cos\lambda \qquad (3)

where: the east acceleration is driven by the fall speed $\dot z$ (this is the main event); the north equation is fed only by the eastward velocity that the first equation creates (so it is second order in $\omega$ ); the vertical equation carries a Coriolis correction to $g$ that is likewise second order in its effect on position. The natural expansion parameter is $\omega t$ — for a hundred-metre drop, $\omega t\sim3\times10^{-4}$ , which is why iteration converges brutally fast.

Derivation AThe eastward deflection by iteration in the rotation rate

We are deriving the leading deflection of a body dropped from rest at height $h$ : the eastward drift of equations (4) and (5). The strategy is perturbative iteration in the small parameter $\omega t$ : solve the equations of motion (3) with $\omega=0$ (ordinary free fall), feed that zeroth-order solution into the Coriolis terms to get the first-order correction, and stop — the neglected terms are smaller by another factor of $\omega t\sim10^{-4}$ . Watch two things: only the vertical velocity is nonzero at zeroth order, so of the three Coriolis couplings only one fires; and the initial conditions ( $x=\dot x=0$ ) kill both constants of integration, which is what produces the clean $t^3$ .

Step 1. Zeroth order: switch the rotation off. With $\ddot{\mathbf r}^{(0)}=(0,0,-g)$ and release from rest at height $h$ :

\dot{\mathbf r}^{(0)}=(0,\,0,\,-gt),\qquad \mathbf r^{(0)}=(0,\,0,\,h-\tfrac12 g t^2)

Why start here: the Coriolis force needs a velocity to act on; free fall supplies the dominant velocity, and every Coriolis effect is then a small correction proportional to $\omega$ .

Step 2. First order: insert the zeroth-order velocity into the east equation of (3). With $\dot y^{(0)}=0$ and $\dot z^{(0)}=-gt$ :

\ddot x^{(1)}=2\omega(\dot y^{(0)}\sin\lambda-\dot z^{(0)}\cos\lambda)=2\omega g t\cos\lambda

Why only this coupling survives: the body falls straight down at zeroth order, so the $\dot y\sin\lambda$ term has nothing to bite on; the entire first-order force is the horizontal component of $\boldsymbol\omega$ crossed with the fall velocity — pointing east, with strength growing linearly in time as the fall speeds up.

Step 3. Integrate twice with $x(0)=\dot x(0)=0$ :

\dot x^{(1)}=\omega g t^2\cos\lambda,\qquad x^{(1)}=\tfrac{1}{3}\,\omega g t^3\cos\lambda

x_E(t)=\tfrac{1}{3}\,\omega g\cos\lambda\,t^3 \qquad (4)

where: $x_E$ — eastward displacement from the plumb line after fall time $t$ . The $t^3$ is the signature of a force that itself grows linearly in time, integrated twice.

Step 4. Convert to drop height using the zeroth-order fall time $t=\sqrt{2h/g}$ , so $t^3=(2h/g)^{3/2}$ :

x_E=\tfrac{1}{3}\,\omega\cos\lambda\,\sqrt{\frac{8h^3}{g}} \qquad (5)

where: $h$ — release height. Note the $h^{3/2}$ scaling: doubling the drop nearly triples the deflection, which is why the measurable experiments all used mineshafts and church spires rather than lecture halls. For $h=100$ m at the equator, $t=4.52$ s and $x_E=2.2$ cm — the standard benchmark value. The sign is eastward (positive $\hat{\mathbf x}$ ) at every latitude since $\cos\lambda\ge0$ in both hemispheres.

The same answer without fictitious forces

A result this clean deserves a second, independent derivation — and the inertial-frame route is where the physical understanding actually lives. It also arbitrates between two plausible shortcuts that give wrong answers, a rite of passage for every student of this problem.

Go deeper · Derivation B · 2 minThe inertial-frame route: angular momentum arbitrates the naive estimates

Derivation BThe inertial-frame route: angular momentum arbitrates the naive estimates

We are re-deriving the eastward deflection (4) in the inertial frame, where there is no Coriolis force — only gravity, pointing at Earth's center. The conserved quantity that replaces the force bookkeeping is the angular momentum about the polar axis: central gravity exerts no torque about that axis, so $L_z$ is constant during the fall. The strategy is to track the angular race between the falling body and the ground beneath it, because — and this is the subtle point — the race starts even: at release the body co-rotates with the tower, so its angular velocity equals the ground's, and the eastward lead builds up only as the body descends to a smaller distance from the axis while conserving $L_z$ . Linear-velocity intuitions fail precisely because they miss this angular start.

Step 1. Set up in the equatorial projection. Let $\rho$ be the distance from the rotation axis; the body is released co-rotating at $\rho_0=(R+h)\cos\lambda$ with tangential speed $v_\phi(0)=\omega\rho_0$ , so its conserved angular momentum per unit mass is $L_z=\omega\rho_0^2$ . Why $L_z$ is conserved: gravity points at the center of the Earth, so its moment arm about the polar axis is zero; during a seconds-long fall nothing else touches the body.

Step 2. As the body falls, its distance from the axis shrinks: $\rho(t)=\rho_0-\delta(t)$ with $\delta(t)=\tfrac12 g t^2\cos\lambda$ (the vertical drop, projected onto the direction away from the axis). Conservation gives its angular velocity an edge over the ground's:

\dot\phi_{\text{body}}=\frac{L_z}{\rho^2}=\frac{\omega\rho_0^2}{(\rho_0-\delta)^2}\approx\omega\Big(1+\frac{2\delta}{\rho_0}\Big)

Why the factor 2 appears: $v_\phi=L_z/\rho$ grows as $\rho$ shrinks (the skater pulling in her arms), AND the same tangential speed corresponds to a larger angular rate at smaller radius. Each effect contributes $\omega\delta/\rho_0$ ; the Coriolis factor of 2 in disguise.

Step 3. Integrate the angular lead over the fall and convert to arc length at radius $\rho_0$ :

x_E=\rho_0\int_0^t\big(\dot\phi_{\text{body}}-\omega\big)\,\mathrm dt'=2\omega\int_0^t\delta(t')\,\mathrm dt'=2\omega\cdot\frac{g\cos\lambda}{2}\cdot\frac{t^3}{3}=\tfrac13\,\omega g\cos\lambda\,t^3

Exactly equation (4) — no Coriolis force anywhere in sight.

Step 4. Now grade the shortcuts. Shortcut one says: "the top has excess speed $\omega h\cos\lambda$ over the ground; multiply by the fall time" — giving $\omega h t\cos\lambda=\tfrac12\omega g t^3\cos\lambda$ , too big by $\tfrac32$ . It fails because at release the body has zero angular lead (same $\dot\phi$ as the tower); the linear speed surplus only converts into ground-relative gain as the body comes down to smaller radius. Shortcut two says: "the body keeps its initial tangential velocity $\omega\rho_0$ " — that gives a lead rate of only $\omega\delta/\rho_0$ and hence $\tfrac16\omega g t^3\cos\lambda$ , too small by 2. It fails because it freezes $v_\phi$ and forgets the angular-momentum speed-up of Step 2. The perturbative and angular-momentum routes agree with each other and with experiment; the shortcuts bracket them from either side. The agreement between the rotating-frame and inertial-frame derivations is itself a powerful consistency check; the factor-of-2 anatomy in Step 2 — half from the growth of $v_\phi$ and half from the smaller radius — maps directly onto the two-contribution structure of the Coriolis term in the acceleration transformation.

The second-order whisper, and the numbers

Iterating once more exposes the next effect in line. The first-order eastward velocity $\dot x^{(1)}=\omega g t^2\cos\lambda$ now feeds the north equation of (3), producing a deflection that is second order in $\omega$ — directed toward the equator:

y_S(t)=-\tfrac{1}{6}\,\omega^2 g\sin\lambda\cos\lambda\,t^4 \qquad (6)

where: $y_S$ — north–south displacement; the minus sign means southward in the Northern Hemisphere (equatorward in both, since $\sin\lambda$ flips sign across the equator). Relative to the eastward deflection it is smaller by a factor of order $\omega t\sim10^{-4}$ : micrometres where the eastward effect is centimetres, hopelessly below the noise of any drop experiment.

Go deeper · Derivation C · 1 minThe second-order equatorward deflection

Derivation CThe second-order equatorward deflection

We are deriving equation (6), the north–south companion to the eastward drift. The strategy is one more turn of the iteration crank: take the first-order eastward velocity from Derivation A, insert it into the north equation $\ddot y=-2\omega\dot x\sin\lambda$ , and integrate twice. Since $\dot x^{(1)}$ is already proportional to $\omega$ , everything here is $\mathcal O(\omega^2)$ — which both explains why the effect is minuscule and warns us to be consistent: at this order one should in principle also track the variation of the centrifugal correction with height, but that shifts the vertical motion, not the leading north–south term computed here.

Step 1. Take the first-order eastward velocity from Derivation A:

\dot x^{(1)}=\omega g t^2\cos\lambda

Step 2. Insert it into the north equation of (3):

\ddot y^{(2)}=-2\omega\,\dot x^{(1)}\sin\lambda=-2\,\omega^2 g\,t^2\sin\lambda\cos\lambda

Why the sign: an eastward velocity crossed with the vertical component of $\boldsymbol\omega$ pushes toward the equator — the same right-of-velocity rule that turns Northern-Hemisphere winds, applied to the Coriolis-generated velocity itself.

Step 3. Integrate twice with $y(0)=\dot y(0)=0$ : the first integration gives $\dot y^{(2)}=-\tfrac23\omega^2 g t^3\sin\lambda\cos\lambda$ , the second gives

y^{(2)}=-\tfrac{1}{6}\,\omega^2 g\,t^4\sin\lambda\cos\lambda

Why $t^4$ : a source growing as $t^2$ , integrated twice, always lands on $t^4/12$ ; the 2 in front reduces it to $t^4/6$ .

Step 4. Size it. The ratio to the eastward deflection is $|y_S/x_E|=\tfrac12\omega t\sin\lambda$ — about $1.6\times10^{-4}$ for Reich's 158.5 m fall at $\lambda=51°$ , giving $y_S\approx4.4\ \mathrm{\mu m}$ versus 27.6 mm eastward. No dropped-ball experiment has credibly measured it. The $\sin\lambda\cos\lambda$ structure and $\mathcal O(\omega^2)$ scaling are a direct consequence of the iteration; direct double integration of $-2\omega^2 g t^2$ gives $-t^4/6$ , confirming the coefficient unambiguously — note that some sources quote a different prefactor, but the integration is straightforward.

Limiting cases

Behaviour of the eastward deflection $x_E=\tfrac13\omega g\cos\lambda\,t^3$ in the natural limits:

Case	Result	Meaning
Equator	$\lambda=0$ : $x_E=\tfrac{1}{3}\,\omega g\,t^3$	Maximum deflection: $\boldsymbol\omega$ is horizontal and perpendicular to the vertical fall.
Pole	$\lambda=\pm 90°$ : $x_E=0$	$\boldsymbol\omega\parallel\mathbf v$ so the Coriolis force vanishes; no horizontal deflection.
Small height	$x_E=\tfrac{1}{3}\omega\cos\lambda\sqrt{8h^3/g}\ \propto\ h^{3/2}$	Deflection grows faster than linearly in drop height — deep shafts win.
Slow rotation ( $\omega\to 0$ )	$x_E=\mathcal O(\omega)$ , $y_S=\mathcal O(\omega^2)\to 0$	Inertial-frame limit: straight-down free fall recovered.
Thrown upward instead	net deflection $-\tfrac43\,\omega\cos\lambda\,v^3/g^2$ (west)	For a full up-and-down flight the sign reverses — see the FAQ.

Common trapThe Earth does not simply rotate out from under the stone

The most natural guess — "the Earth turns east beneath the falling stone, so it lands west" — has the wrong sign. The stone is not left behind, because at release it shares the tower's eastward motion; there is no "wind of rotation" blowing past it. What decides the outcome is the difference in motion between release height and ground: the top of the tower rides a larger circle, so the stone arrives with an eastward surplus and lands east. Two bookkeeping traps compound the confusion: measure the deflection from the local plumb line (which already includes the centrifugal tilt of effective gravity), not from the true radial direction; and keep perturbation orders separate — eastward is $\mathcal O(\omega)$ , equatorward is $\mathcal O(\omega^2)$ , and mixing them manufactures spurious terms.

Worked examples

1 · Reich's mineshaft, 1831 (confidence builder)

Ferdinand Reich dropped pellets $h=158.5\ \mathrm{m}$ down a mineshaft at Freiberg, Saxony ( $\lambda=51°\mathrm{N}$ ). Predict the eastward deflection and compare with his measured mean of $28.3\ \mathrm{mm}$ .

Step 1. Fall time: $t=\sqrt{2h/g}=\sqrt{2\times158.5/9.81}=5.68\ \mathrm{s}$ , so $t^3=183.7\ \mathrm{s^3}$ .

Step 2. Insert into equation (4) with $\cos 51°=0.629$ : $x_E=\tfrac13\times7.292\times10^{-5}\times9.81\times183.7\times0.629=\boxed{27.6\ \mathrm{mm\ east}}$ .

Step 3. Compare: Reich's 106 drops averaged $\boxed{28.3\ \mathrm{mm}}$ — agreement at the 3% level, spectacular for 1831 and the first precision confirmation of the rotating-frame prediction. (The scatter of individual drops was several millimetres; only the deep, draft-free shaft and the large number of repetitions made the mean trustworthy.)

Step 4. Sanity-check the neglected terms: $\omega t=7.292\times10^{-5}\times5.68=4.1\times10^{-4}$ , so second-order effects are parts in ten thousand of 27.6 mm — micrometres, invisible. The perturbative truncation is airtight.

2 · A 400-metre super-tower (exam level)

A ball is dropped (in vacuum, ideally a drop tube) from $h=400\ \mathrm{m}$ . Find the eastward deflection at the equator and at $\lambda=30°\mathrm{N}$ , and the equatorward deflection at $30°\mathrm{N}$ . Could the east deflection survive in real air?

Step 1. Fall time: $t=\sqrt{2\times400/9.81}=9.03\ \mathrm{s}$ ; $t^3=736.4\ \mathrm{s^3}$ .

Step 2. Equator ( $\cos\lambda=1$ ): $x_E=\tfrac13\times7.292\times10^{-5}\times9.81\times736.4=\boxed{17.6\ \mathrm{cm\ east}}$ — the $h^{3/2}$ law at work: four times Reich's height, six times his deflection.

Step 3. At $\lambda=30°$ : multiply by $\cos30°=0.866$ : $x_E=\boxed{15.2\ \mathrm{cm}}$ .

Step 4. Equatorward term at $30°$ from equation (6), with $t^4=6650\ \mathrm{s^4}$ and $\sin30°\cos30°=0.433$ : $|y_S|=\tfrac16\times(7.292\times10^{-5})^2\times9.81\times0.433\times6650=\boxed{0.025\ \mathrm{mm}}$ — four orders of magnitude below the eastward signal, exactly the $\tfrac12\omega t\sin\lambda$ ratio predicted.

Step 5. Air check: a falling sphere reaches its terminal-velocity regime within a couple hundred metres, capping $\dot z$ and breaking the $t^3$ growth; crosswinds of even $0.1\ \mathrm{m/s}$ displace the ball by metres over 9 s. In open air the centimetre-scale Coriolis signal is buried; you need an evacuated drop tube or a sheltered shaft — which is precisely why the historical experiments lived in mineshafts and why modern drop towers pump their tubes down to millipascals.

Origins · 1 minCarl Friedrich Gauss and how this was discovered

Carl Friedrich Gauss

1777–1855 · Göttingen observatory

Gauss enters this story as the theorist on call for an experiment. When Johann Benzenberg prepared his 1802 falling-ball measurements in the tower of St. Michael's church in Hamburg, he asked the 25-year-old Gauss to work out what deflection a rotating Earth actually predicts. Gauss's calculation — equivalent to the iteration performed above — gave 8.7 mm eastward for the 76 m drop, and Benzenberg's scattered but persistent results averaged close to 9 mm; Olbers, refereeing the exchange, declared the Earth's rotation demonstrated by experiment. It was a characteristic Gauss cameo: the supreme calculator quietly supplying the decisive number, decades before Coriolis systematized the forces or Foucault made rotation a public spectacle. Gauss never left the subject of the turning, curving Earth — he directed the Göttingen observatory for half his life, ran the geodetic survey of Hanover (inventing the heliotrope to do it), and built the theory of curved surfaces that the survey demanded, the same differential geometry that would one day carry general relativity.

History

1679

— Newton, in correspondence with Hooke, proposes the eastward deflection of a dropped body as a direct test of the Earth's rotation.

1791

— Guglielmini drops balls inside Bologna's Asinelli tower; his eastward result (roughly double the ~11 mm prediction) is suggestive but crude.

1802

— Benzenberg's drops in St. Michael's tower, Hamburg, average about 9 mm east against Gauss's predicted 8.7 mm.

1831

— Reich's 106 drops down a 158.5 m Freiberg mineshaft give 28.3 mm east versus 27.6 mm predicted — the definitive measurement.

1851

— Foucault's pendulum makes the Earth's rotation visible to a lay audience, retiring the falling-ball experiment as proof and leaving it as pedagogy.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerating frames) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §4.9–4.10 · Landau & Lifshitz, Mechanics, 3rd ed., §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, Ch. 9 Key results to keep in mind: the eastward formula $x_E=\tfrac13\omega g\cos\lambda\,t^3$ and its equivalent $h^{3/2}$ form; the equatorial benchmark $x_E\approx 2.2\,\mathrm{cm}$ for $h=100\,\mathrm{m}$ (using $\omega=7.292\times10^{-5}\,\mathrm{s^{-1}}$ , $g=9.81\,\mathrm{m/s^2}$ ); the second-order southward term, obtained by double integration of $\ddot y=-2\omega\dot x^{(1)}\sin\lambda$ , giving $y_S=-\tfrac16\omega^2 g\sin\lambda\cos\lambda\,t^4$ (some sources quote a different coefficient, but the direct integration is unambiguous); and the agreement between the inertial angular-momentum derivation and the rotating-frame result.

Builds on: Coriolis force · Time derivatives in rotating frames · Effective gravity on the rotating Earth · Angular velocity vector of a rotating frame Leads to: Fictitious (inertial) forces · Centrifugal force · Euler force

Intuitively the Earth rotates eastward under the falling stone — should it not land to the west?

This is the single most common wrong turn, and it fails because the stone is not released from rest in space — it is released from rest on the tower, already moving east at the tower top's full speed of hundreds of metres per second. Nothing "slides out from under it." The real effect is a small difference: the release point, being farther from Earth's axis, moves slightly faster eastward than the landing point, and angular-momentum conservation even amplifies that surplus as the stone descends. The stone therefore leads the ground, landing east. The westward intuition would be right only for an object that did not share the Earth's rotation at release — say, a body dropped from a hovering point fixed in inertial space, which is not how towers work.

Why does the falling stone care about cos λ when the Foucault pendulum cares about sin λ?

Because they probe perpendicular components of the same vector. A falling stone's velocity is vertical, and a vertical velocity is only turned by the horizontal part of $\boldsymbol\omega$ , which has magnitude $\omega\cos\lambda$ — maximal at the equator, zero at the poles. A pendulum bob's velocity is horizontal, and horizontal velocities are steered by the vertical component $\omega\sin\lambda$ — zero at the equator, maximal at the poles. The two experiments are perfectly complementary: run both at one location and you have measured both components of the Earth's angular velocity vector, hence its magnitude and direction, without a single astronomical observation. The Coriolis force article develops the $\sin\lambda$ side of the story.

Has the tiny southward deflection ever actually been measured?

Not convincingly. At second order in $\omega$ it is around $10^{-4}$ of the eastward deflection — micrometres for any practical drop — while release jitter, residual air currents, and electrostatic effects perturb real drops at the level of millimetres. Several nineteenth-century experimenters (including Guglielmini, whose southward claim helped motivate later work) reported southward offsets far larger than theory allows, which is now understood as systematic error; the anomalously large claims are a classic cautionary tale about finding what you expect. The eastward effect, by contrast, has been solid since Benzenberg and Reich: first order in $\omega$ , centimetre-scale in deep shafts, and reproducible.

Does the deflection depend on the mass or material of the falling body?

Not at all, and the reason is worth internalizing: the Coriolis term, like gravity, produces the same acceleration for every body — the $m$ in $-2m\,\boldsymbol\omega\times\mathbf v$ cancels the $m$ in $m\ddot{\mathbf r}$ . The deflection is pure kinematics of the rotating frame, which is exactly what "fictitious force" means. What does depend on the body is air drag: a light or fluffy object reaches terminal velocity early, caps its $\dot z$ , and breaks the $t^3$ accumulation, shrinking and blurring the effect. That is why Reich used dense pellets in a still mineshaft and why a feather is hopeless even though its ideal deflection would be identical.

What happens if I throw the ball straight up instead of dropping it?

The net deflection reverses: the ball lands to the west of the launch point. On the way up the velocity points up, and $-2\boldsymbol\omega\times\mathbf v$ points west; on the way down it points east. The legs do not cancel, because the westward push acts early and its accumulated westward velocity then operates over the entire remaining flight. Carrying out the same iteration with $\dot z=v-gt$ gives a net displacement $-\tfrac43\,\omega\cos\lambda\,v^3/g^2$ — westward, and four times larger in magnitude than the eastward deflection of a ball merely dropped from the trajectory's peak height. Artillery officers know this asymmetry well: up-and-over trajectories need different Coriolis corrections from plunging ones.

Why were the historical experiments in Italy and Germany rather than near the equator where the effect is biggest?

Because $\cos\lambda$ is cheap and quiet air is expensive. Moving from $51°$ N to the equator only buys a factor $1/\cos51°\approx1.6$ in signal, while a deep, draft-free mineshaft or an enclosed tower buys orders of magnitude in noise reduction — wind, temperature currents, and release disturbances swamp millimetre signals in the open. The $h^{3/2}$ scaling also favours whatever deep vertical spaces you have: Reich's 158.5 m shaft at $51°$ N out-signals a 50 m equatorial tower several times over. Precision experiments are won by controlling systematics, not by maximizing the ideal signal — a lesson as true for atom interferometers today as for Reich's pellets.

Is the deflection formula exact, or an approximation — and when would it fail?

Equation (4) is the leading term of an expansion in $\omega t$ , and for terrestrial drops it is superbly accurate: with $\omega t\lesssim10^{-3}$ , the neglected terms are parts in a thousand of an already-small deflection. It would fail when $\omega t$ ceases to be small — falls lasting a substantial fraction of the rotation period. Drop something down an evacuated shaft through the whole Earth, or track a satellite's descent over hours, and you must solve the full coupled equations (3) (or better, work in the inertial frame from the start); the trajectory then bends into the loops and rosettes Newton sketched for Hooke. The formula also presumes constant $g$ over the fall; for $h\ll R$ that correction enters at order $h/R$ , still negligible against drag in any real experiment.