# Equatorial bulge

An equatorial bulge is a difference between the equatorial and polar diameters of a planet, due to the centrifugal force of its rotation. A rotating body tends to form an oblate spheroid rather than a sphere. The Earth has an equatorial bulge of 42.77 km (26.58 mi): that is, its diameter measured across the equatorial plane (12,756.27 km (7,926.38 mi)) is 42.77 km more than that measured between the poles (12,713.56 km (7,899.84 mi)). An observer standing at sea level on either pole, therefore, is 21.36 km closer to Earth's centrepoint than if standing at sea level on the equator. The value of Earth's radius may be approximated by the average of these radii.

An often-cited result of Earth's equatorial bulge is that the highest point on Earth, measured from the center outwards, is the peak of Mount Chimborazo in Ecuador, rather than Mount Everest. But since the ocean, like the Earth and the atmosphere, bulges, Chimborazo is not as high above sea level as Everest is.

Basing this on centripetal force, the relationship $F_c = Mv^2 / R$ applies. Viewing the globe as a series of rotating discs, the mass M and radius R at the poles get very small at the same time and thus produce a smaller force for the same velocity. Moving towards the equator, while R gets much bigger, M increases quicker than R thus producing a greater force at the equator. This may be because Earth’s core is included in the cross sectional disc at the equator; the density of the Earth's core is significantly higher than that of the Earth's outer layers, so it contributes more to the mass of the disc. There is a bulge in the water envelope of the oceans surrounding earth. So the fact that water is a fluid and the Earth experiences its greatest centrifugal force at the equator and the fact that the greatest bulge of that water envelope occurs at the equator demonstrates that centrifugal force of earth’s rotation is helping to produce that bulge independent of tides. Sea level at the equator is 21.36 km higher than sea level at the poles in terms of their distances from the center of the planet.

## The equilibrium as a balance of energies

File:Equatorial bulge model.png
Fixed to the vertical rod is a spring metal band. When stationary the spring metal band is circular in shape. The top of the metal band can slide along the vertical rod. When spun, the spring-metal band bulges at its equator and flattens at its poles in analogy with the Earth.

Gravity tends to contract a celestial body into a sphere, the shape for which all the mass is as close to the center of gravity as possible. Rotation causes a distortion from this spherical shape; a common measure of the distortion is the flattening (sometimes called ellipticity or oblateness), which can depend on a variety of factors including the size, angular velocity, density, and elasticity.

To get a feel for the type of equilibrium that is involved, imagine someone seated in a spinning swivel chair, with weights in their hands. If the person in the chair pulls the weights towards them, they are doing work and their rotational kinetic energy increases. The increase of rotation rate is so strong that at the faster rotation rate the required centripetal force is larger than with the starting rotation rate.

Something analogous to this occurs in planet formation. Matter first coalesces into a slowly rotating disk-shaped distribution, and collisions and friction convert kinetic energy to heat, which allows the disk to self-gravitate into a very oblate spheroid.

As long as the proto-planet is still too oblate to be in equilibrium, the release of gravitational potential energy on contraction keeps driving the increase in rotational kinetic energy. As the contraction proceeds the rotation rate keeps going up, hence the required force for further contraction keeps going up. There is a point where the increase of rotational kinetic energy on further contraction would be larger than the release of gravitational potential energy. The contraction process can only proceed up to that point, so it halts there.

As long as there is no equilibrium there can be violent convection, and as long as there is violent convection friction can convert kinetic energy to heat, draining rotational kinetic energy from the system. When the equilibrium state has been reached then large scale conversion of kinetic energy to heat ceases. In that sense the equilibrium state is the lowest state of energy that can be reached.

The Earth's rotation rate is still slowing down, though gradually, by about two thousandths of a second per rotation every 100 years.[1] Estimates of how fast the Earth was rotating in the past vary, because it is not known exactly how the moon was formed. Estimates of the Earth's rotation 500 million years ago are around 20 modern hours per "day".

The Earth's rate of rotation is slowing down mainly because of tidal interactions with the Moon and the Sun. Since the solid parts of the Earth are ductile, the Earth's equatorial bulge has been decreasing in step with the decrease in the rate of rotation.

## Differences in gravitational acceleration

File:Forces oblate spheroid2.gif
The forces at play in the case of a planet with an equatorial bulge due to rotation.
Red arrow: gravity
Green arrow, the normal force
Blue arrow: the resultant force

The resultant force provides required centripetal force. Without this centripetal force frictionless objects would slide towards the equator.

In calculations, when a coordinate system is used that is co-rotating with the Earth, the vector of the fictitious centrifugal force points outward, and is just as large as the vector representing the centripetal force.

Because of a planet's rotation around its own axis, the gravitational acceleration is less at the equator than at the poles. In the 17th century, following the invention of the pendulum clock, French scientists found that clocks sent to French Guiana, on the northern coast of South America, ran slower than their exact counterparts in Paris. Measurements of the acceleration due to gravity at the equator must also take into account the planet's rotation. Any object that is stationary with respect to the surface of the Earth is actually following a circular trajectory, circumnavigating the Earth's axis. Pulling an object into such a circular trajectory requires a force. The acceleration that is required to circumnavigate the Earth's axis along the equator at one revolution per sidereal day is 0.0339 m/s². Providing this acceleration decreases the effective gravitational acceleration. At the equator, the effective gravitational acceleration is 9.7805 m/s2. This means that the true gravitational acceleration at the equator must be 9.8144 m/s2 (9.7805 + 0.0339 = 9.8144).

At the poles, the gravitational acceleration is 9.8322 m/s2. The difference of 0.0178 m/s2 between the gravitational acceleration at the poles and the true gravitational acceleration at the equator is because objects located on the equator are about 21 kilometers further away from the center of mass of the Earth than at the poles, which corresponds to a smaller gravitational acceleration.

In summary, there are two contributions to the fact that the effective gravitational acceleration is less strong at the equator than at the poles. About 70 percent of the difference is contributed by the fact that objects circumnavigate the Earth's axis, and about 30 percent is due to the non-spherical shape of the Earth.

The diagram illustrates that on all latitudes the effective gravitational acceleration is decreased by the requirement of providing a centripetal force; the decreasing effect is strongest on the equator.

## Satellite orbits

The fact that the Earth's gravitational field slightly deviates from being spherically symmetrical also affects the orbits of satellites. The principal effect is to cause nodal precession, so that the plane of the orbit does not remain fixed in inertial space. Smaller effects include deviation of orbits away from pure ellipses.[citation needed] This is especially important in the case of the trajectories of GPS satellites.

## Other celestial bodies

Generally any celestial body that is rotating (and that is sufficiently massive to draw itself into spherical or near spherical shape) will have an equatorial bulge matching its rotation rate. Saturn is the planet with the largest equatorial bulge in the Solar System (11808 km, 7337 miles). However, Haumea is the dwarf planet with the largest equatorial bulge, indeed greater than that of Saturn. This gives it the shape of a tri-axial ellipsoid.

The following is a table of the equatorial bulge of some major celestial bodies of the Solar System:

Body Equatorial diameter Polar diameter Equatorial bulge Flattening ratio
Earth 12,756.27 km 12,713.56 km 42.77 km 1:298.2575
Mars 6,805 km 6,754.8 km 50.2 km 1:135.56
Ceres 975 km 909 km 66 km 1:14.77
Jupiter 143,884 km 133,709 km 10,175 km 1:14.14
Saturn 120,536 km 108,728 km 11,808 km 1:10.21
Uranus 51,118 km 49,946 km 1,172 km 1:43.62
Neptune 49,528 km 48,682 km 846 km 1:58.54

## Mathematical expression

The flattening coefficient $f$ for the equilibrium configuration of a self-gravitating spheroid, composed of uniform density incompressible fluid, rotating steadily about some fixed axis, for a small amount of flattening, is approximated by:[2]

$f = \frac{a_e-a_p}{a} = {5 \over 4} {\omega^2 a^3 \over G M} = {15 \pi \over 4} {1 \over G T^2 \rho}$

where $a_e = a(1+{ f \over 3} )$ and $a_p = a(1-{ 2f\over 3})$ are respectively the equatorial and polar radius, $a$ is the mean radius, $\omega = {2 \pi \over T}$ is the angular velocity, $T$ is the rotation period, $G$ is the universal gravitational constant, $M \simeq {4 \over 3} \pi \rho a^3$ is the total body mass, and $\rho$ is the body density.

## References

1. Hadhazy, Adam. "Fact or Fiction: The Days (and Nights) Are Getting Longer". Scientific American. Retrieved 5 December 2011.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>
2. "Rotational Flattening". utexas.edu.<templatestyles src="Module:Citation/CS1/styles.css"></templatestyles>