Answer · confidence: partial

Does Earth's spin move ground faster at the equator than in the middle?

Research question: How much does the ground under a launch site move because of Earth's spin, and how does that change between a city in the middle of a continent and a site on the equator?

The facts do not give the ground's speed or total displacement at either site, so they cannot state how far a launch site moves from Earth's spin. The records show that the rotation's effect depends on latitude: a Foucault pendulum's plane completes a full turn in 24 hours only at the poles ² and takes 48 hours at 30° latitude ⁵. The Coriolis parameter also depends on latitude through the sine of latitude ³, and on the equator the Coriolis effect on north-south motion is zero ⁷. Gravity is affected too, since the spin's centrifugal correction is zero at the poles and about -0.3% at the equator ⁴, and surface gravity is about 0.5% higher at the poles than at the equator ⁶. A mid-continent city falls between these extremes, as the standard gravity value corresponds to a sea-level site at 45° latitude ¹⁰.

Checked 2026-10-18. Only documented and supported facts are used.

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Sources

[1] supported

Earth completes one rotation per sidereal day, and because of this the Coriolis effect is imperceptible for everyday motions and matters only over large distances and long times or where high precision is needed.

"The Earth completes one rotation for each sidereal day , so for motions of everyday objects the Coriolis force is imperceptible; its effects become noticeable only for motions occurring over large distances and long periods of time, such as large-scale movement of air in the atmosphere or water in the ocean, or where high precision is important, such as artillery or missile trajectories."

Coriolis force

[2] supported

A Foucault pendulum's plane completes a full turn in 24 hours only at the poles.

"Mais le plan n'effectue un tour complet en 24 heures qu'aux pôles."

Pendule de Foucault

[3] supported

The Coriolis parameter f, which sets how strongly Earth's rotation deflects moving bodies horizontally, depends on latitude through f = -2 sin(latitude) × Ω_T.

"ϕ = latitude f = − 2 sin ⁡ ( ϕ ) Ω T"

Force de Coriolis

[4] supported

The Earth's rotation adds an outward centrifugal correction to gravity, and this correction is zero at the poles and reaches about -0.3% at the equator.

"La correction, nulle aux pôles, atteint -0,3 % sur l'équateur"

Pesanteur

[5] supported

At a latitude of 30°, a Foucault pendulum's plane would take 48 hours to complete one full turn.

"Le sinus de 30° valant 1/2, un pendule de Foucault implanté à une latitude de 30° effectuerait un tour complet en 48 heures ."

Pendule de Foucault

[6] supported

Surface gravity is not uniform: it is higher at the poles than at the equator, by about 0.5%.

"elle est notamment plus élevée aux pôles qu'à l' équateur (d'environ 0,5 %)"

Pesanteur

[7] supported

On Earth's equator, a body traveling east is deflected upward, while the Coriolis effect on a body moving horizontally north or south is zero at the equator.

"On Earth's equator, a body traveling east will deviate upwards."

Coriolis force

[8] supported

Because the Coriolis effect is computed from Earth's rotation, the sidereal day used for the calculation lasts 23 hours, 56 minutes and 4.09 seconds.

"il faut utiliser la durée du jour sidéral T s i d e r a l qui correspond à 23 h 56 min 4,09 s ."

Force de Coriolis

[9] supported

The rotation correction is an outward acceleration directed perpendicular to the axis of the poles, added to the gravitational acceleration.

"La rotation de la Terre sur elle-même entraîne une correction consistant à ajouter à l'accélération de la gravité une accélération d’entraînement axifuge, dirigée perpendiculairement à l'axe des pôles"

Pesanteur

[10] supported

The standard gravity value of about 9.81 m/s² corresponds to sea level at a latitude of 45°, which is roughly the situation of a mid-latitude site rather than the equator.

"Cette valeur correspond à la pesanteur sur un ellipsoïde idéal approchant le niveau de la mer et à 45° de latitude ."

Pesanteur

[11] supported

During the First World War, shells fired by the Paris guns at a range of 110 km were deflected by nearly 1,600 metres by the Coriolis force.

"les obus tirés par les Pariser Kanonen , appelés Grosse Bertha par les Français, qui pilonnaient Paris à 110 kilomètres de distance, étaient déviés de près de 1 600 mètres par la force de Coriolis"

Force de Coriolis