# Why is Earth not a sphere?

We've all learned at school that the Earth was a sphere. Actually, it is more nearly a slightly flattened sphere - an oblate ellipsoid of revolution, also called an oblate spheroid. This is an ellipse rotated about its shorter axis. What are the physical reasons for that phenomenon?

• I've just linked to your question in Is Earth's “pear shape” mostly J₃? – uhoh Jul 14 at 0:51
• @Uhoh: For a more detailed view on the earth's shape, don't miss out on the influence of convection, number and strength of convectional currents and layers in the mantle, etc. I recently read about considerable influence, at least over geological timescale. Don't remember where, tho, ... – a_donda Jul 14 at 11:28

Normally in the absence of rotation, the natural tenancy of gravity is to pull the Earth together in the shape of a sphere.

However the Earth in fact bulges at the equator, and the diameter across the equatorial plane is 42.72 km more than the diameter from pole to pole.

This is due to the rotation of the Earth.

As we can see in the image above, the spinning disk appears to bulge at the points on the disk furthest from the axis of rotation.

This is because in order for the particles of the disk to remain in orbit, there must be an inward force, known as the centripetal force, given by:

$$F = \frac{mv^2}{r},$$

where $F$ is the force, $m$ is the mass of rotating body, $v$ is the velocity and $r$ is the radius of particle from the axis of rotation.

If the disk is rotating at a given angular velocity, say $\omega$, then the tangential velocity $v$, is given by $v = \omega r$.

Therefore,

$$F = m\omega^2r$$

Therefore the greater the radius of the particle, the more force is required to maintain such an orbit.

Therefore particles on the Earth near the equator, which are farthest from the axis of rotation, will buldge outwards because they require a greater inward force to maintain their orbit.

Additional details for more mathematically literate now that mathjax is enabled:

The net force on an object rotating around the equator with a radius $r$ around a planet with a gravitational force of $\frac{Gm_1m_2}{r^2}$ is the centripetal force given by,

$$F_{net} = \frac{Gm_1m_2}{r^2} - N = m\omega^2r,$$ where $N$ is the normal force.

Re-arranging the above equation gives:

$$N = \frac{Gm_1m_2}{r^2} - m\omega^2r$$

The normal force here is the perceived downward force that a rotating body observers. The equation shows that the perceived downward force is lessened due to the centripetal motion. The typical example to illustrate this is there is an appearance of 0 gravity in a satellite orbiting the Earth, because in this situation the centripetal force is exactly balanced by the gravitational force. On Earth however, the centripetal force is much less than the gravitational force, so we perceive almost the whole contribution of $mg$.

Now we will examine how the perceived gravitational force differs at different angles of latitude. Let $\theta$ represent the angle of latitude. Let $F_G$ be the force of gravity.

In vector notation we will take the $j$-direction to be parallel with the axis of rotation and the $i$-direction to be perpendicular with the axis of rotation.

In the absence of the Earth's rotation,

$$F_G = N = (-\frac{Gm_1m_2}{r^2}\cos\theta)\tilde{i} + (-\frac{Gm_1m_2}{r^2}\sin\theta)\tilde{j}$$

It is easily seen that the above equation represents the perceived force of gravity in the absence of rotation. Now the centripetal force acts only in the i-direction, since it acts perpendicular to the axis of rotation.

If we let $R_{rot}$ be the radius of rotation, then the centripetal force is $m_1\omega^2R_{rot}$, which for an angle of latitude of $\theta$ corresponds to $m_1\omega^2r\cos{\theta}$

$$N = (-\frac{Gm_1m_2}{r^2} + m_1\omega^2r)\cos{\theta}\tilde{i} + (-\frac{Gm_1m_2}{r^2})\sin{\theta}\tilde{j}$$

By comparing this equation to the case shown earlier in the absence of rotation, it is apparent that as $\theta$ is increased (angle of latitude), the effect of rotation on perceived gravity becomes negligible, since the only difference lies in the $x$-component and $\cos\theta$ approaches 0 as $\theta$ approaches 90 degrees latitude. However it can also be seen that as theta approaches 0, near the equator, the $x$-component of gravity is reduced as a result of the Earth's rotation. Therefore, we can see that the magnitude of $N$ is slightly less at the equator than at the poles. The reduced apparent gravitational pull here is what gives rise to the slight bulging of the Earth at the equator, given that the Earth was not originally as rigid as it is today (see other answer).

• Assuming gravity is approximately equal over the surface of the globe, right? – naught101 Apr 16 '14 at 8:09
• @naught101 right - and gravity is equal over the surface to a sufficient approximation to approximate the shape of the planet as an oblate ellipsoid. I think that the variation beyond this would make an excellent answer in its own right :-) – Semidiurnal Simon Apr 16 '14 at 8:50
• @SimonW: The Wikipedia page of the gravity of Earth probably answers most of those outstanding issues - it seems quite comprehensive. – naught101 Apr 16 '14 at 9:00
• @naught101 also, at the poles gravity acts perpendicular to the centripetal force, as the gravitational force is directed towards the centre of gravity, while the centripetal force is directed towards the axis of rotation. – hugovdberg Apr 17 '14 at 8:57
• @hugovdberg, that's right. The greater centripetal force in the same direction as gravity along the equator causes a relative decrease in g from the perspective of a rotating observer at the equator compared to an observer at the poles. This is what gives rise the bulge. I will provide a mathematical description when mathjax is added. – Kenshin Apr 21 '14 at 7:48

Actually, the reason why the Earth is not a sphere is twofold:

1. the Earth is rotating and has been rotating for a long time
2. the Earth is not perfectly rigid, it can even be considered as a viscous fluid on long timescales

If the Earth were not rotating, it would be a sphere. If the Earth had started to rotate very recently, it wouldn't be at equilibrium, thus probably not the ellipsoid of revolution we are familiar with. Last but not least, if the Earth were perfectly rigid, it wouldn't be deformed by any process, including rotation, thus still have its initial shape.

We can consider that the Earth is a fluid in hydrostatic equilibrium (i.e. a fluid at rest) at each point, taking into account both the effect of gravity and centrifugal (pseudo) force due to rotation. Then, if we look for the shape of the Earth's surface under this condition, the solution is an ellipsoid of revolution. It is very close to the actual Earth's surface which is a good evidence that our initial assumption -- rotating fluid in hydrostatic equilibrium -- is reasonable for long timescale.

The study of this question is related to the famous Clairaut's equation from the name of the famous French scientist who published the treatise Théorie de la figure de la terre at the end of the 18th century.

NB: if we just explain the bulge at the equator referring to the effect of the centrifugal pseudo force and ignoring the hydrostatic equilibrium issue, we should conclude that the polar radius is the same with or without rotation. Though, it is smaller: about 6357 km vs 6371 km for a spherical Earth of equal volume.

• how do we know what that the polar radius will be 6371km without rotation? 6371km is the average radius of the Earth, and it is larger than the polar radius because the equatorial bulge has distorted the radius in my opinion. – Kenshin Apr 28 '14 at 23:41
• We simply know that the Earth would have the same volume (incompressibility is assumed) and would be a sphere if it were not rotating, thus a polar radius of 6371 km. 6371 km is not the mean radius of the Earth, it's as I wrote the radius of “a spherical Earth of equal volume”. – Gaialogist Apr 29 '14 at 17:07
• Better late than never: my mistake regarding the discussion about Earth radii. In very good approximation, due to the small value of Earth flattening, 6371 km is at the same time (1) the arithmetical mean radius, (2) the authalic or equal area radius and (3) the volumetric or equal volume radius. It does not change the first part of my previous comment, though: the Earth polar radius is also modified by the rotation, which is not explained in the top-voted/accepted answer. – Gaialogist Feb 6 '19 at 14:45

That the Earth is approximately an oblate spheroid is best explained by energy.

Place a marble in a bowl. No matter where you place it, it will eventually come to rest at the bottom of the bowl. This is the position that minimizes the total energy of the marble subject to the constraint of being in the bowl. Suspend a chain between two posts. When the chain comes to rest it will take on a well-known shape, that of a catenary curve. This is the shape that minimizes the energy of the chain, subject to the constraint of being suspended between the two posts.

If you place the marble away from the bottom it will roll around for a while before coming to rest. If you pull the chain away from it's catenary shape it will swing back and forth for a while before coming to rest in that stable shape. The off-center marble and out-of-plane chain have greater potential energies than they do in their stable configuration. If at all possible, nature will attempt to minimize total potential energy. It's a consequence of the second law of thermodynamics.

In the case of the Earth, that minimum energy configuration is a surface over which the sum of the gravitational and centrifugal potential energies are constant. Something that makes the Earth deviate from this equipotential surface will result in an increase in this potential energy. The Earth will eventually adjust itself back into that minimum energy configuration. This equipotential surface would be an oblate spheroid were it not for density variations such as thick and light continental crust in one place, thin and dense oceanic crust in another.

In terms of force, the quantity we call g is the gradient of the gravitational and centrifugal potential energies (specifically, $\vec g = -\nabla \Phi$). Since the Earth's surface is very close to being an equipotential surface and since that surface in turn is very close to being an oblate spheroid, gravitation at the poles is necessarily slightly more than it is at the equator.

This gravitational force will not be normal to the surface at places where the surface deviates from the equipotential surface. The tangential component of the gravitational force results in places where water flows downhill and in stresses and strains in the Earth's surface. The eventual responses to these tangential forces are erosion, floods, and sometimes even earthquakes that eventually bring the Earth back to its equilibrium shape.

Update: Why is this the right picture?

Based on comments elsewhere, a number of people don't understand why energy rather than force is the right way to look at this problem, or how the second law of thermodynamics comes into play.

There are a number of different ways to state the second law of thermodynamics. One is that a system tends to a state that maximizes its entropy. For example, put two blocks at two different temperatures in contact with one another. The cooler block will get warmer and the warmer block will get cooler until both blocks are at the same temperature, thanks to the second law of thermodynamics. That uniform temperature is the state that maximizes the entropy of this two block system.

Those two blocks only have thermal energy. What about a system with non-zero mechanical energy? Friction is almost inevitably going to sap kinetic energy from the system. That friction means the system's mechanical energy will decrease until it reaches a global minimum, if any. For a rotating, dissipative, self-gravitating body, that global minimum does exist and it is a (more or less) oblate spheroid shape.

• Have you any examples of earthquakes due to the deviation of the crust from the equipotential surface rather than tectonic stress? This example sounds weird to me... Something else: the gravitational force can be normal to the surface even when it deviates from the geoid (and not normal even when it does not deviate). – Gaialogist Apr 29 '14 at 17:00
• @Gaialogist - Regarding your second question, the geoid is the equipotential surface closest to mean sea level. Since gravitational acceleration is the gradient of the gravitational potential, the gravitational acceleration vector is necessarily normal to the geoid. It's in the math. Here's a relevant answer at math.stackexchange.com : math.stackexchange.com/questions/122222/… . – David Hammen Apr 29 '14 at 17:14
• As far as your first question is concerned, a lot of those tectonic stresses are a direct consequence of the Earth being away from hydrostatic equilibrium or an equilibrium shape. Ridge push and slab pull, for example. – David Hammen Apr 29 '14 at 17:22
• It's ok that the gravity is normal to the geoid but the surface does not have to match the geoid to have the gravity normal on it or reciprocally. Consider a surface close and parallel to the geoid but not superimposed: it may have a normal gravity; consider a surface crossing the geoid: on the crossing line the two surfaces match but the gravity is not normal to the Earth's surface. – Gaialogist Apr 29 '14 at 17:34
• For my first question, I agree with the equilibrium argument to explain any movement in (or on) the Earth. I just think it's daring to make the link between tangential components of the gravity vector and earthquakes. Maybe this point of view could even mistakenly reverse causes and consequences (the impact of tectonic structures on gravity anomalies, not the contrary)... – Gaialogist Apr 29 '14 at 17:45