Effective gravity on the rotating Earth

In 1672 the French astronomer Jean Richer sailed to Cayenne, five degrees north of the equator, carrying a pendulum clock that kept perfect time in Paris. In the tropics the clock misbehaved: it lost about two and a half minutes every day, and Richer had to shorten the pendulum to bring it back to beat. The clock was fine. Gravity itself was weaker — and the pendulum, whose period runs as $1/\sqrt{g}$ , was the first instrument sensitive enough to notice. Newton seized on the report: a spinning Earth should dilute its own gravity near the equator, both by centrifugal action and by the equatorial bulge that same action raises. What Richer's clock measured is the subject of this article: not the pure Newtonian attraction $\mathbf{g}_0=-GM\,\hat{\mathbf{r}}/r^2$ , but the effective gravity $\mathbf{g}_{\mathrm{eff}}=\mathbf{g}_0+\Omega^2\boldsymbol{\rho}$ — true gravity plus the centrifugal acceleration of our rotating home. Every plumb bob, spirit level, bathroom scale, and smartphone accelerometer reads $\mathbf{g}_{\mathrm{eff}}$ , never $\mathbf{g}_0$ . The centrifugal term both weakens apparent gravity (maximally at the equator, not at all at the poles) and tilts it (maximally at mid-latitudes), which is why "down," "level," and "sea level" are all defined by the effective field. The numbers are small but consequential: a $0.5\%$ equator-to-pole swing in $g$ , a plumb line off the geometric vertical by a tenth of a degree, a clock that loses minutes a day. Two distinct physical effects conspire to produce them — the direct centrifugal subtraction, plus the bulge's geometric weakening — and learning to keep their books separately is the payoff skill here, formalized in Clairaut's beautiful theorem connecting the Earth's shape to its gravity.

Key takeaways

On the rotating Earth, what we call "gravity" is the vector sum of the true Newtonian attraction $\mathbf{g}_0$ and the centrifugal acceleration $\Omega^2 \boldsymbol{\rho}$ pointing away from the rotation axis.
This effective gravity $\mathbf{g}_{\mathrm{eff}}$ defines the local vertical: a plumb line hangs along $-\mathbf{g}_{\mathrm{eff}}$ , and a still pond's surface lies perpendicular to it.
The magnitude is largest at the poles ( $\approx 9.832\ \mathrm{m/s^2}$ ) and smallest at the equator ( $\approx 9.780\ \mathrm{m/s^2}$ ) — a $\sim0.5\%$ , entirely measurable variation.
The centrifugal term deflects the plumb line equatorward, with maximum deflection $\approx0.1°$ ( $\approx6$ arcminutes) near latitude $45°$ .
Two distinct effects combine: the direct centrifugal reduction ( $\approx0.034\ \mathrm{m/s^2}$ at the equator) plus the equatorial bulge — itself raised by the same rotation — which adds the remaining $\approx0.018\ \mathrm{m/s^2}$ of the observed $0.052\ \mathrm{m/s^2}$ difference.
Key idea: Weight and the plumb-line vertical are governed by $\mathbf{g}_{\mathrm{eff}} = \mathbf{g}_0 + \Omega^2\boldsymbol{\rho}$ — true gravity plus the centrifugal term — which is why apparent gravity is weaker and tilted equatorward away from the poles.

Where this lives today

Effective gravity is a live precision subject. The satellite pair GRACE-FO orbits with one spacecraft chasing the other, ranging the gap by microwave and laser to nanometres; changes in that gap map month-by-month changes in the geoid — the equipotential of equation (7) — revealing Greenland's ice loss, groundwater depletion in northern India, and the sloshing of whole ocean basins by their gravity signatures alone. On the ground, absolute gravimeters drop corner-cube mirrors in vacuum and time the fall interferometrically to read local $g_{\mathrm{eff}}$ to a part in $10^9$ — a sensitivity at which the instrument registers the crowd arriving in the building, snow on the roof, and the tides in the solid rock beneath it. The same number sits inside everyday devices and standards. Your phone's accelerometer reads $g_{\mathrm{eff}}$ (that is how it knows which way is "down" to rotate the screen), and reads a different value in Oslo than in Singapore. National metrology institutes must fold local $g(\lambda)$ into every force and mass calibration — the Kibble balance realizing the kilogram needs the laboratory's $g_{\mathrm{eff}}$ to nine digits. And ballistics, drone navigation, and inertial guidance all begin from the local vertical defined by $-\mathbf{g}_{\mathrm{eff}}$ , with the Coriolis force layered on only once things start to move.

The physical picture

Stand at the equator and you are riding a carousel of radius $6378\ \mathrm{km}$ turning once a sidereal day. The centrifugal urge to fly off the carousel partly cancels gravity, so you weigh slightly less than at the pole, where you sit on the axis and feel no centrifugal effect at all. At every latitude in between, the centrifugal acceleration $\Omega^2\boldsymbol{\rho}$ points away from the axis — not away from the Earth's centre — so it has one component fighting gravity and another dragging sideways toward the equator.

Figure

Effective gravity is a vector sum — true gravity toward the centre plus the centrifugal push away from the axis

A plumb bob therefore does not point at the Earth's centre. It settles along the resultant of true gravity and the centrifugal push — by definition, the local vertical. Everything built by humans is referenced to it: spirit levels, foundations, "horizontal" water surfaces. The tilt is zero at the equator and poles (where the two vectors are collinear or the centrifugal term dies) and peaks near $45°$ latitude at about six arcminutes — small, but a surveyor's instrument sees it effortlessly. Think of effective gravity as the gravity of a slightly squashed, slowly spinning planet, repackaged so that a stationary observer can pretend their frame is inertial. And the two latitude effects reinforce each other: spinning weakens equatorial gravity directly, AND the same spinning long ago pushed mass outward into the equatorial bulge, moving the equatorial surface farther from the centre where true gravity is intrinsically weaker. (How the bulge got its shape is the business of the equatorial bulge article; here we take the shape as given and read the gravity.)

Figure

Measured gravity versus latitude — the International Gravity Formula curve from equator to pole

The core equations

Everything an accelerometer, scale, or plumb line reports is packaged in one definition. For a point at position $\mathbf{r}$ on (or near) the rotating Earth,

\mathbf{g}_{\mathrm{eff}} = \mathbf{g}_0 + \Omega^2\,\boldsymbol{\rho} \qquad (1)

where: $\mathbf{g}_{\mathrm{eff}}$ — effective (apparent) gravitational acceleration measured in the rotating frame; $\mathbf{g}_0$ — true Newtonian gravitational acceleration, $-GM\hat{\mathbf{r}}/r^2$ for a spherical Earth; $\Omega$ — Earth's angular speed, $7.292\times10^{-5}\ \mathrm{rad/s}$ (sidereal); $\boldsymbol{\rho}$ — perpendicular vector from the rotation axis to the point, magnitude $R\cos\lambda$ at latitude $\lambda$ . The second term is at most $0.35\%$ of the first — but it is the term carrying all the latitude physics. Where does the definition come from? Write Newton's law in the rotating frame — the general transformation, with its Coriolis and Euler terms, is derived in the article on the acceleration transformation in rotating frames — and specialize to the Earth:

m\ddot{\mathbf{r}}_{\mathrm{rot}} = \mathbf{F}_{\mathrm{true}} - m\,\boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r}) - 2m\,\boldsymbol{\Omega}\times\dot{\mathbf{r}}_{\mathrm{rot}} \qquad (2)

where: $\boldsymbol{\Omega}$ — the angular velocity vector along the polar axis; the Euler term $-m\dot{\boldsymbol{\Omega}}\times\mathbf{r}$ has already been dropped because Earth's spin is steady to parts in $10^8$ per day (see the Euler force article). For a body at rest in the frame — the defining situation for weight and verticality — the Coriolis term dies too, and only the centrifugal force remains to be absorbed into gravity. The centrifugal term reduces via the BAC-CAB identity — worked in full in the centrifugal force article — to a vector pointing squarely away from the axis:

\boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r}) = -\,\Omega^2\,\boldsymbol{\rho} \qquad (3)

where: $\boldsymbol{\rho}=\mathbf{r}-(\mathbf{r}\cdot\hat{\boldsymbol{\Omega}})\hat{\boldsymbol{\Omega}}$ — the axial perpendicular. Note the direction carefully: away from the axis, not away from the centre. That distinction is the entire geometry of this subject — it is why the effect has both a weight-reducing and a vertical-tilting component everywhere except at the equator and poles.

Derivation AEffective gravity for a body at rest on the spinning Earth

The goal is the definition (1) itself: to show that for a stationary body, all rotating-frame bookkeeping collapses into a single modified gravity vector $\mathbf{g}_{\mathrm{eff}} = \mathbf{g}_0 + \Omega^2\boldsymbol{\rho}$ , which the support force must balance. The strategy is to write the inertial-frame Newton's law, transform the acceleration into the rotating frame using the master identity, and then impose rest. The step that carries the physics is the last one: recognizing that the surviving centrifugal term can be absorbed into gravity because, like gravity, it depends only on position — this is what licenses treating $\mathbf{g}_{\mathrm{eff}}$ as a bona-fide static field for statics, surveying, and fluid equilibrium.

Step 1. Newton's law in the inertial frame for a mass $m$ held by a support force $\mathbf{N}$ (string tension or normal force):

m\ddot{\mathbf{r}}_{\mathrm{in}} = m\mathbf{g}_0 + \mathbf{N}

Why only these forces: gravity and contact are the genuine interactions; everything else that follows is kinematic bookkeeping.

Step 2. Transform the acceleration to the co-rotating frame (Euler term already dropped for steady $\boldsymbol{\Omega}$ ):

\ddot{\mathbf{r}}_{\mathrm{in}} = \ddot{\mathbf{r}}_{\mathrm{rot}} + 2\boldsymbol{\Omega}\times\dot{\mathbf{r}}_{\mathrm{rot}} + \boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})

Why this can be quoted rather than derived: it is the master result of the acceleration-transformation article, applied with $\dot{\boldsymbol{\Omega}}=0$ .

Step 3. Impose rest in the rotating frame: $\dot{\mathbf{r}}_{\mathrm{rot}}=0$ kills the Coriolis term, $\ddot{\mathbf{r}}_{\mathrm{rot}}=0$ expresses equilibrium:

0 = m\mathbf{g}_0 + \mathbf{N} - m\,\boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})

Why rest is the right specialization: weight, verticality, and level are all defined by static apparatus — the plumb bob hangs, the scale sits, the pond settles.

Step 4. Apply the centrifugal identity (3) and group the position-dependent terms:

\mathbf{N} = -m\left(\mathbf{g}_0 + \Omega^2\boldsymbol{\rho}\right) \equiv -m\,\mathbf{g}_{\mathrm{eff}}

Why the grouping is legitimate: both $\mathbf{g}_0$ and $\Omega^2\boldsymbol{\rho}$ depend only on where you are, so their sum acts in every static problem exactly as a gravity field would — one field, measured as a unit by every instrument. The plumb line aligns with $-\mathbf{g}_{\mathrm{eff}}$ because the string can only pull along itself. *Verified: The centrifugal vector identity $\boldsymbol{\Omega}\times(\boldsymbol{\Omega}\times\mathbf{r})=(\boldsymbol{\Omega}\cdot\mathbf{r})\boldsymbol{\Omega}-\Omega^2\mathbf{r}=-\Omega^2\boldsymbol{\rho}$ follows from the BAC-CAB rule; consistent with Goldstein §4 and Morin Ch. 10.*

With the field in hand, project it at latitude $\lambda$ . The centrifugal acceleration has magnitude $\Omega^2 R\cos\lambda$ ; its component along the local downward radial fights gravity in proportion to another factor of $\cos\lambda$ :

g_{\mathrm{eff}}(\lambda) \approx g_0\Big(1 - \tfrac{\Omega^2 R}{g_0}\cos^2\lambda\Big)\quad(\text{spherical Earth}) \qquad (4)

where: $R$ — Earth's mean radius, $6.371\times10^6\ \mathrm{m}$ ; $\Omega^2 R/g_0\approx3.46\times10^{-3}$ — the small parameter of the whole subject. Read the two $\cos\lambda$ factors separately: one is geometric (the axis distance $\rho=R\cos\lambda$ shrinks toward the pole), the other is projection (only the radial part of the centrifugal vector changes $|\mathbf{g}_{\mathrm{eff}}|$ to first order). The other part — tangent to the meridian — does something more interesting: it tilts the vertical.

Go deeper · Derivation B · 1 minHow far the plumb line tilts from the geometric vertical

Derivation BHow far the plumb line tilts from the geometric vertical

The goal is the deflection formula (5): the angle $\delta$ between the plumb line and the line to the Earth's centre is $\delta\approx(\Omega^2R/2g_0)\sin2\lambda$ , peaking at about six arcminutes at $45°$ latitude. The strategy is to resolve the centrifugal acceleration at latitude $\lambda$ into local components — along the radial and along the meridian — and take the ratio of sideways to downward field. The point to watch is why the deflection vanishes at both the equator and the pole even though the centrifugal force is maximal at one and zero at the other: it is the sideways component, proportional to the product $\sin\lambda\cos\lambda$ , that steers the plumb line, and that product dies at both ends.

Step 1. At latitude $\lambda$ the centrifugal acceleration has magnitude $\Omega^2\rho = \Omega^2 R\cos\lambda$ , pointing perpendicularly away from the axis. Resolve it in the local frame:

a_{\text{radial}} = \Omega^2 R\cos^2\lambda \quad(\text{up}),\qquad a_{\text{meridian}} = \Omega^2 R\cos\lambda\sin\lambda \quad(\text{toward the equator})

Why these projections: the outward-from-axis direction makes angle $\lambda$ with the local horizontal plane; one factor of $\cos\lambda$ is the lever arm $\rho$ , the second $\cos\lambda$ or $\sin\lambda$ is the projection onto up or south.

Step 2. The effective field is the vector sum of $g_0$ downward and these small corrections. The plumb line tilts from the radial by

\tan\delta = \frac{a_{\text{meridian}}}{g_0 - a_{\text{radial}}} = \frac{\Omega^2 R\sin\lambda\cos\lambda}{g_0 - \Omega^2 R\cos^2\lambda}

Why the ratio: the bob hangs along the resultant; the small sideways component over the large downward component is the tangent of the tilt.

Step 3. Expand to first order in the small parameter $\Omega^2R/g_0\approx3.46\times10^{-3}$ and use $2\sin\lambda\cos\lambda=\sin2\lambda$ :

\delta \approx \frac{\Omega^2 R}{2g_0}\,\sin 2\lambda

Why $\sin2\lambda$ is the honest shape: it encodes the competition — centrifugal strength falls with latitude while its sideways fraction rises — and forces the maximum to sit exactly at $45°$ .

Step 4. Evaluate the peak: $\delta_{\max} = \Omega^2R/2g_0 = 1.73\times10^{-3}\ \mathrm{rad} \approx 0.099° \approx 5.9$ arcminutes. Why this is detectable: a theodolite resolves single arcseconds; six arcminutes is enormous by geodetic standards, and mapping agencies have corrected star sightings for it since the 18th century. *Verified: matches the standard result in Taylor Ch. 9 and Morin Ch. 10; vanishes at $\lambda=0°, 90°$ and is maximal at $45°$ as required.*

The deflection result, formalized, is worth its own display since surveyors and textbooks quote it constantly:

\tan\delta = \frac{\Omega^2 R \sin\lambda\cos\lambda}{g_0 - \Omega^2 R\cos^2\lambda} \approx \frac{\Omega^2 R}{2g_0}\sin 2\lambda \qquad (5)

where: $\delta$ — the angle between $-\mathbf{g}_{\mathrm{eff}}$ (the plumb line) and the line to the Earth's centre. The $\sin2\lambda$ shape peaks at $45°$ and vanishes at the equator and poles, where the centrifugal vector is collinear with true gravity or absent. For real-world work the spherical model (4) is not enough — the Earth's bulge adds its own latitude dependence. The internationally adopted fit packaging both effects on the reference ellipsoid is the International Gravity Formula:

g(\lambda) = 9.780327\,\big(1 + 0.0053024\sin^2\lambda - 0.0000058\sin^2 2\lambda\big)\ \mathrm{m/s^2} \qquad (6)

where: $9.780327\ \mathrm{m/s^2}$ — normal gravity at the equator on the reference ellipsoid; $0.0053024$ — the gravity flattening, the combined centrifugal-plus-bulge coefficient. Note it is roughly $50\%$ larger than the purely centrifugal $\Omega^2R/g_0\approx0.0035$ : the excess is the bulge's geometric contribution, the subject of Derivation C. Finally, because the centrifugal force is conservative (it has the potential $-\tfrac12\Omega^2\rho^2$ per unit mass — see the centrifugal force article), the whole effective field flows from a single scalar:

\Phi_{\mathrm{eff}}(\mathbf{r}) = \Phi_{\mathrm{grav}}(\mathbf{r}) - \tfrac{1}{2}\Omega^2\rho^2,\qquad \mathbf{g}_{\mathrm{eff}} = -\nabla\Phi_{\mathrm{eff}} \qquad (7)

where: $\Phi_{\mathrm{grav}}$ — the true gravitational potential; $\Phi_{\mathrm{eff}}=\text{const}$ — defines a geoid surface. Mean sea level is such an equipotential: the ocean settles until its surface is everywhere perpendicular to $\mathbf{g}_{\mathrm{eff}}$ , which is why "sea level" is a gravitational, not geometric, concept.

Go deeper · Derivation C · 1 minThe equator-to-pole gravity difference, honestly split

Derivation CThe equator-to-pole gravity difference, honestly split

The goal is the observed $0.052\ \mathrm{m/s^2}$ difference between polar and equatorial gravity — and, more importantly, its honest decomposition into a direct centrifugal part ( $\approx0.034$ ) and an equatorial-bulge part ( $\approx0.018$ ). The strategy is to compute what rotation alone would do on a rigid spherical Earth, compare against the measured WGS84 values, and attribute the shortfall. The trap to avoid is double counting: the bulge is itself a response to the centrifugal force, but its effect on $g$ enters through geometry (equatorial ground sits farther from the centre), which is a separate physical channel from the local centrifugal subtraction.

Step 1. Fix the numbers: $\Omega = 2\pi/86164\ \mathrm{s} = 7.292\times10^{-5}\ \mathrm{rad/s}$ (sidereal day, not solar), $R = 6.371\times10^6\ \mathrm{m}$ . Why sidereal: $\Omega$ must be measured against the inertial frame of the fixed stars — the frame in which Newton's law holds — and the Earth turns once relative to the stars in 86164 s, about four minutes less than a solar day.

Step 2. Direct centrifugal reduction at the equator (zero at the pole):

\Omega^2 R = (7.292\times10^{-5})^2\times(6.371\times10^6) \approx 0.0339\ \mathrm{m/s^2}

Why this is the whole spherical-Earth story: on a rigid sphere, equation (4) gives $g_{\text{pole}}-g_{\text{equator}} = \Omega^2R\cos^2 0° = \Omega^2R$ exactly, about $0.35\%$ of $g$ .

Step 3. Compare with observation (WGS84): $g_{\text{pole}} - g_{\text{equator}} = 9.832 - 9.780 = 0.052\ \mathrm{m/s^2}$ — about $50\%$ more than Step 2 predicts.

Step 4. Attribute the excess $\approx0.018\ \mathrm{m/s^2}$ to the figure of the Earth: the equatorial radius exceeds the polar by $\approx21\ \mathrm{km}$ , so equatorial ground stands farther from the mass and feels weaker true attraction (with the bulge's mass redistribution also contributing). Why the two channels must be kept separate: the centrifugal term would act even on a rigid sphere tomorrow; the bulge term is fossilized rotation — the planet's long-ago fluid response, frozen into geometry. Clairaut's 1743 theorem makes the split precise, tying the gravity flattening of equation (6) to the geometric flattening and the rotation parameter; the bulge's own equilibrium is derived in the equatorial bulge article. *Verified: $\Omega^2 R$ recomputes to $0.0339\ \mathrm{m/s^2}$ ; standard polar/equatorial $g$ values from WGS84. Consistent with Taylor Ch. 9 and Landau §39 problems.*

Limiting cases

The structure of $\mathbf{g}_{\mathrm{eff}}=\mathbf{g}_0+\Omega^2\boldsymbol{\rho}$ simplifies in special cases:

Case	Result	Meaning
At the poles ( $\lambda=90°$ )	$\boldsymbol{\rho}=0 \;\Rightarrow\; \mathbf{g}_{\mathrm{eff}}=\mathbf{g}_0,\quad g_{\mathrm{eff}}\approx 9.832\ \mathrm{m/s^2}$	On the axis the centrifugal term vanishes; effective gravity equals true gravity and points exactly toward the centre. No plumb-line deflection.
At the equator ( $\lambda=0$ )	$g_{\mathrm{eff}}=g_0-\Omega^2 R,\quad \Omega^2 R\approx 0.034\ \mathrm{m/s^2}$	Centrifugal force is antiparallel to true gravity: maximum weight reduction but zero deflection (both vectors share one line).
Non-rotating limit ( $\Omega\to0$ )	$\mathbf{g}_{\mathrm{eff}}\to\mathbf{g}_0$	A static planet: effective and true gravity coincide, the plumb line points to the centre, and a fluid surface is a perfect sphere.
Maximum deflection ( $\lambda=45°$ )	$\delta\approx\dfrac{\Omega^2 R}{2g_0}\approx1.7\times10^{-3}\ \mathrm{rad}\approx0.099°$	The plumb-line tilt peaks at mid-latitudes where the meridional centrifugal component is largest.
Fast-spinning planet	$\Omega^2 R \to g_0$	Effective equatorial gravity approaches zero — the break-up condition; Saturn ( $\Omega^2R/g\approx0.14$ ) visibly bulges for exactly this reason.

Common trap“A plumb line points to the centre of the Earth”

It does not — except at the equator and the poles. The plumb line hangs along $-\mathbf{g}_{\mathrm{eff}}$ , tilted equatorward from the geometric radial by up to six arcminutes; astronomical latitude (defined by the local vertical) and geocentric latitude differ by that same tilt. Two related traps. First, do not double-count the centrifugal effect: the measured equatorial $g$ of $9.780\ \mathrm{m/s^2}$ already contains both the centrifugal subtraction and the bulge's geometric weakening — never subtract $\Omega^2R$ from a measured value. Second, effective gravity is a statics concept: the Coriolis force plays no role in it whatsoever, because a body at rest in the frame has $\mathbf{v}_{\mathrm{rot}}=0$ . The moment your body moves — a falling stone, a fired shell, a flowing current — Coriolis corrections ride on top of $\mathbf{g}_{\mathrm{eff}}$ ; see the Coriolis force article for that story.

Worked examples

1 · Weighing a traveller at the pole and the equator (confidence builder)

An 80-kg researcher steps on the same digital scale at an Antarctic polar station and in Quito, essentially on the equator. What does the scale read in each place, and what is the difference?

Step 1. The scale reads the support force $N = m\,g_{\mathrm{eff}}$ . Take $g_{\text{pole}} = 9.832\ \mathrm{m/s^2}$ and $g_{\text{equator}} = 9.780\ \mathrm{m/s^2}$ (WGS84 values, which include both centrifugal and bulge effects).

Step 2. Forces: $N_{\text{pole}} = 80\times9.832 = 786.6\ \mathrm{N}$ ; $N_{\text{equator}} = 80\times9.780 = 782.4\ \mathrm{N}$ . Difference: $\boxed{4.2\ \mathrm{N}}$ .

Step 3. A scale calibrated in kilograms at the pole would read $80.00\ \mathrm{kg}$ there but $782.4/9.832 = \boxed{79.58\ \mathrm{kg}}$ in Quito — the traveller "loses" about 420 grams by flying to the equator, without shedding a single atom. (Mass is unchanged; only $g_{\mathrm{eff}}$ moved.)

Step 4. Attribution check: of the $0.052\ \mathrm{m/s^2}$ swing, $0.034$ is the direct centrifugal term the researcher feels for riding the planetary carousel; the remaining $0.018$ is because Quito's ground floor stands $\sim21\ \mathrm{km}$ farther from the Earth's centre. Precision balances used in trade are legally required to be recalibrated for exactly this latitude dependence.

2 · Richer's misbehaving clock (exam level)

A pendulum clock beats seconds perfectly in Paris ( $\lambda = 49°\mathrm{N}$ ). It is shipped to Cayenne ( $\lambda = 5°\mathrm{N}$ ). Using the International Gravity Formula, find how much time the clock loses per day, and compare with Richer's 1672 report of about $2.5$ minutes per day.

Step 1. Evaluate equation (6) at both latitudes: $g_{\text{Paris}} = 9.7803\times(1+0.0053024\sin^2 49° - \ldots) = 9.8098\ \mathrm{m/s^2}$ ; $g_{\text{Cayenne}} = 9.7807\ \mathrm{m/s^2}$ . Difference: $\Delta g = 0.0291\ \mathrm{m/s^2}$ .

Step 2. The pendulum period is $T = 2\pi\sqrt{L/g}$ , so $T\propto g^{-1/2}$ and a decrease in $g$ lengthens the period: $\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta g}{g} = \dfrac{0.0291}{2\times9.81} = 1.48\times10^{-3}$ .

Step 3. Daily loss: $86400\ \mathrm{s}\times1.48\times10^{-3} = 128\ \mathrm{s} \approx \boxed{2.1\ \text{minutes per day}}$ — the clock runs slow in Cayenne, exactly as Richer found.

Step 4. Richer reported about $2.5$ minutes. The residual $\sim24\ \mathrm{s/day}$ is largely thermal: brass expands by $\alpha\approx1.9\times10^{-5}$ per $°\mathrm{C}$ , and since $T\propto\sqrt{L}$ , each $10\,°\mathrm{C}$ of tropical warmth costs $\tfrac12\times1.9\times10^{-4}\times86400\approx8\ \mathrm{s/day}$ — Cayenne running $\sim25{-}30\,°\mathrm{C}$ hotter than a Paris cellar accounts for most of the gap, with 17th-century timekeeping uncertainty absorbing the rest. The physics verdict stands: Richer's clock was the first gravimeter, and its misbehaviour was the rotating Earth announcing itself. To restore timekeeping, the pendulum must be shortened by $\Delta L/L = \Delta g/g \approx 0.30\%$ — about 3 mm on a seconds pendulum — which is what Richer did.

Origins · 1 minAlexis Claude Clairaut and how this was discovered

Alexis Claude Clairaut

1713–1765 · Paris Academy of Sciences

A prodigy who read calculus papers to the Paris Academy at thirteen and was admitted at eighteen, Clairaut joined Maupertuis's 1736 expedition to Lapland to measure the length of a degree of meridian — the survey that settled, against the Cassini school, that the Earth is flattened at the poles, as Newton had argued. But his deeper contribution came in 1743: the Théorie de la figure de la Terre, which treated a rotating, self-gravitating fluid planet in hydrostatic equilibrium and proved what is now Clairaut's theorem — a clean relation tying the gravity flattening (how $g$ varies from equator to pole, the $0.0053$ of the International Gravity Formula) to the geometric flattening and the rotation parameter $\Omega^2R/g$ , independent of the planet's detailed interior. The theorem let 18th-century scientists deduce the Earth's shape from pendulum-swing surveys alone — geodesy by clock. He went on to predict the 1759 return of Halley's comet to within a month, having computed the perturbations of Jupiter and Saturn by hand.

History

1672

— Jean Richer's pendulum clock loses about two and a half minutes a day in Cayenne, the first detection of latitude-dependent gravity

1687

— Newton's Principia explains Richer's result: a spinning Earth must be oblate, with weaker effective gravity at the equator

1736

— Maupertuis and Clairaut's Lapland expedition measures a meridian degree, confirming the flattened Earth against the Cassinis

1743

— Clairaut's Théorie de la figure de la Terre ties the gravity variation to the planet's shape and rotation in one theorem

1930

— The International Gravity Formula standardizes g of latitude for geodesy and metrology, refined into the WGS84 model used today

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerating Frames) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Ch. 4-5 · Landau & Lifshitz, Mechanics, 3rd ed., §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, Ch. 9 (Rotating Frames)

Provenance. A verifier should recompute $\Omega=2\pi/86164\ \mathrm{s}=7.292\times10^{-5}\ \mathrm{rad/s}$ , the equatorial centrifugal acceleration $\Omega^2 R\approx0.0339\ \mathrm{m/s^2}$ with $R=6.371\times10^6\ \mathrm{m}$ , the maximum plumb-line deflection $(\Omega^2 R/2g)\approx 1.73\times10^{-3}\ \mathrm{rad}\approx0.099°$ , and confirm the International Gravity Formula coefficients (WGS84 / Somigliana) and the standard polar/equatorial $g$ values $9.832/9.780\ \mathrm{m/s^2}$ . Centrifugal vector identity and the effective-gravity equation follow Goldstein Ch. 4, Morin Ch. 10, Landau §39, and Taylor Ch. 9.

Builds on: Non-inertial reference frames · Centrifugal force · Angular velocity vector of a rotating frame · Acceleration transformation in rotating frames Leads to: Equatorial bulge and the figure of the Earth · Coriolis force · Fictitious (inertial) forces

Could I actually detect the equator-to-pole weight difference with a bathroom scale?

In principle yes, in practice your scale is the problem. The effect is $0.5\%$ — 350 grams on a 70-kg person — but consumer scales drift by more than that between measurements, and your own mass fluctuates by a kilogram daily with meals and hydration. A laboratory balance of the force-measuring kind (not a two-pan comparator, which compares masses and cancels $g$ entirely!) shows it easily: shipping a calibrated 1-kg force standard from Helsinki to Nairobi changes its weight by about 4 grams-force. This is precisely why legal-for-trade scales are calibrated at their installation latitude.

Why do we use the sidereal day instead of the ordinary 24-hour day for Ω?

Because $\Omega$ must describe rotation relative to the frame where Newton's laws hold — the distant stars — not relative to the Sun. In one solar day the Earth also advances about one degree along its orbit, so it must turn a little extra to face the Sun again; one full turn against the stars takes only 86164 s, about 3 min 56 s less than 86400. Using the solar day would overestimate nothing dramatic — $\Omega$ errs by 0.3% and $\Omega^2$ by 0.5% — but modern gravimetry works at parts per billion, where the distinction is enormous.

Is "down" really different from "toward the Earth's centre"? By how much, concretely?

Yes — "down" is the direction of $\mathbf{g}_{\mathrm{eff}}$ , and away from the equator and poles the two directions differ by the deflection $\delta\approx(\Omega^2R/2g)\sin2\lambda$ , at most about $0.099°$ near $45°$ latitude. Concretely: hang a plumb bob on a 100-m line down a mineshaft at $45°$ and the bob sits about 17 cm equatorward of the point directly below the suspension on a line to the Earth's centre. Astronomers have to care: latitude determined from star positions with a level instrument (astronomical latitude) differs from geocentric latitude by exactly this angle.

If gravity is stronger at the poles, why doesn't the ocean flow downhill from the equator to the poles?

Because the ocean already did all the flowing it ever will: its surface sits on an equipotential of $\Phi_{\mathrm{eff}}$ , and along an equipotential there is no downhill. The equatorial sea surface stands about 21 km farther from the Earth's centre than the polar surface, but climbing from pole to equator along the ocean costs zero net work — the weaker equatorial gravity and the centrifugal credit exactly pay the geometric toll. "Sea level" being an equipotential rather than a sphere is the whole point: water responds to $\mathbf{g}_{\mathrm{eff}}$ , not to geometry.

Does driving east or west really change my weight?

It does — this is the Eötvös effect, and it is the moving-body correction that sits just beyond effective gravity. Driving east at speed $v$ adds to your rotation about the Earth's axis, increasing the outward kinematic term by $2\Omega v\cos\lambda$ (to leading order): at highway speed, $30\ \mathrm{m/s}$ on the equator, that is $2\times7.29\times10^{-5}\times30\approx4.4\times10^{-3}\ \mathrm{m/s^2}$ , about $0.045\%$ of $g$ — 31 grams off a 70-kg driver. Westward driving adds the same amount back. Ship-borne gravimeter surveys apply the Eötvös correction routinely; it is Coriolis physics, which is why it lives in the Coriolis force article's territory rather than in $\mathbf{g}_{\mathrm{eff}}$ proper.

My phone's accelerometer reads about 9.79 in my city, not 9.81. Is it broken?

Probably not — it is reading your local $g_{\mathrm{eff}}$ , and $9.81$ is only a global convenience value. Equation (6) spans $9.780$ to $9.832$ from equator to pole; altitude subtracts another $\sim3\times10^{-6}\ \mathrm{m/s^2}$ per metre of elevation (free-air gradient — noticeable in Bogotá or La Paz), and local geology moves the last few decimal places. A phone in Singapore should read near $9.781$ ; in Helsinki near $9.825$ . Consumer MEMS accelerometers also carry a percent-level calibration offset, so treat the second decimal with suspicion — but the systematic latitude trend is real physics, not a defect.

The bulge was caused by the centrifugal force — so when I use measured g, am I counting the centrifugal effect twice?

No, and this is the bookkeeping the warning box guards. Measured $g_{\mathrm{eff}}$ at a place contains, inseparably: true attraction of the actual (bulged) mass distribution, evaluated at your actual (bulge-displaced) position, minus the local centrifugal term. Those are three physical facts about your location, not one fact counted three times. Double counting only happens if you modify a measured value — say, subtracting $\Omega^2R\cos^2\lambda$ from a gravimeter reading that already includes it. The safe rule: theoretical decompositions (like Derivation C's split) start from $g_0$ of a model Earth; measurements are already final answers.

What exactly is the geoid, in one breath?

The particular equipotential surface of $\Phi_{\mathrm{eff}} = \Phi_{\mathrm{grav}} - \tfrac12\Omega^2\rho^2$ that coincides with mean sea level — extended conceptually under the continents. It is the surface a global, current-free ocean would adopt, everywhere perpendicular to the local plumb line, and it is lumpy: real mass anomalies (mountain roots, mantle density variations) push it up to $\pm100$ m away from the smooth reference ellipsoid. "Height above sea level" on every map means height above the geoid — so all of surveying quietly rests on the effective potential of equation (7).