Stand anywhere on Earth and you are standing on a carousel. The planet carries you around its axis once a day (about 1,300 km/h at the latitude of Paris) and, like any carousel, it tries to fling you outward. Gravity wins by a landslide, of course. But it doesn't win cleanly, and this essay is about the part it loses.
The outward fling, the centrifugal force, doesn't point straight up. It points away from the spin axis, the line running from pole to pole, not away from the centre of the Earth. Away from the axis and away from the centre are different directions everywhere except at the equator. That little mismatch would be enough to drag every ocean and every parcel of air down to the equator, if the Earth were actually a sphere.
01 The sphere that can't hold still
Below is a perfectly spherical, spinning Earth, seen from the side. Move your mouse (or finger) over it. At the point you touch, two forces act: gravity, pulling straight at the centre, and the centrifugal force, pushing away from the axis. Watch the centrifugal arrow as you travel: it is strongest at the equator and vanishes at the poles, because its strength depends on your distance from the axis, the dashed line the planet spins around.
Now split that centrifugal push into two parts, following the thin dashed lines. The part pointing straight off the ground just makes you slightly lighter, and the ground pushes back. But the other part lies flat along the ground, aimed at the equator. On a sphere, gravity points exactly straight down, so it has no sideways part of its own to answer with. Nothing cancels the pink arrow.
Hover or drag over the planet · then let it rain
The sideways pull vanishes at the equator (there the push points straight up, with no ground to slide along) and at the poles (no spin radius), and is strongest at 45°. In code, the split is one projection:
// forces at latitude φ on a spinning sphere of radius R
r = R * cos(φ) // your distance from the spin axis
centrifugal = ω² * r // points away from the AXIS, not the centre
// project it onto the local vertical and the local ground
straight_up = ω² * r * cos(φ) // the ground pushes back: balanced
along_ground = ω² * r * sin(φ) // toward the equator: nothing balances itPress release the rain above. Every droplet obeys that leftover force, slides downhill toward the equator and pools there. On a spherical Earth, downhill and toward the equator are the same direction everywhere. The oceans would drain from the poles and the atmosphere would sag into a belt. That is not the planet we live on, so something must be cancelling the pink arrow.
02 The planet fights back
The Earth, of course, isn't a sphere. The pull toward the equator acted on the young, molten planet too, so the planet itself flowed toward the equator, exactly like the raindrops, until it had puffed out into a bulge. The bulge changes the geometry. On a squashed planet the surface is tilted compared to the sphere: straight down to the centre and straight into the ground are no longer the same direction. Gravity now has a small sideways part of its own, and it points toward the poles.
Follow the thin dashed lines from the tips of both thick arrows down onto the ground line: gravity's sideways part (periwinkle) and the centrifugal sideways part (marigold) sit tail-to-tail, equal and opposite, at every single latitude. The bulge grew until they matched, and then the flowing stopped.
Hover or drag over the planet · rain now has nowhere to go
The teal arrow is gravity and the centrifugal force added together. That sum, called effective gravity, is the "down" a plumb line or a spirit level actually measures, and on the bulged Earth it is perpendicular to the ground everywhere. There is no downhill left to slide along, so the droplets just sit there. In code, the balance is two projections that agree:
// on the bulged Earth (ellipse, equator radius a, pole radius b)
tangent = (-a * sin(θ), b * cos(θ)) // the local ground direction
gravity_along_ground = G * (a² - b²) * sin(θ)cos(θ) // poleward
centrifugal_along_ground = -ω² * a² * sin(θ)cos(θ) // equatorward
// they cancel at EVERY latitude when ω² / G = 1 - (b/a)²
// ... which is exactly the bulge the flowing planet settled intoSea level is simply the shape on which the two pulls agree.
03 The real numbers
Everything above is exaggerated about a hundredfold so that the arrows are visible. In reality the centrifugal effect at the equator is 0.34% of gravity: you weigh about 300 g less there than at the poles (partly because the bulge itself puts you farther from the centre). The bulge amounts to 21 km of extra equatorial radius, a flattening of about 1/298. A circle squashed by a third of a percent still looks perfectly round, which is why photographs from space show a sphere.
Small, but not negligible. Because of the bulge, the summit of Ecuador's Chimborazo, not Everest, is the point on Earth farthest from the planet's centre, by more than 2 km. And "sea level" is nothing more than the surface our second animation found: the shape on which effective gravity is perpendicular to the ground everywhere, so that water has no downhill left anywhere.
A footnote for the purists: our toy model, with gravity pulling toward the centre and growing with distance, predicts a bulge about half the real size. The missing half comes from the bulge's own gravity: the extra rock around the equator pulls sideways too, tilting "down" a little further and letting the bulge grow a little fatter. Working that out properly took until Clairaut in the eighteenth century, but the balancing act is the one you just played with.
Newton reasoned his way to the bulge in the Principia without leaving his desk, by imagining two water-filled wells drilled to the centre of the Earth, one from a pole and one from the equator, and asking what it would take for them to balance. The French Academy then spent the 1730s sending expeditions to Lapland and Peru to measure the shape. Newton was right: the Earth really is slightly flung outward, and that is why the oceans stay where they are.