# Does Earth's Rotation Affect Its Shape?

The question I am working on is, "Consider the following.

(a) Find the angular speed of Earth's rotation about its axis. rad/s

(b) How does this rotation affect the shape of Earth?"

I am fully capable of solving part (a); however, I am not sure how to describe the effect earth's rotation on its shape. I tried to search my textbook for the answer, but could not find anything. Is there an actual effect on the shape?

• Possible duplicates: physics.stackexchange.com/q/8074/2451 and physics.stackexchange.com/q/10670/2451 Nov 23, 2012 at 16:43
– Serg
Nov 23, 2012 at 17:34
• @EMACK, definitely the rotation gives the Earth the shape of a disk, rather than a sphere. The distance to the Earth center is less on the poles than it's on the equator. It also affects the gravitation g that is measured at sea level on poles or equator.
– Serg
Nov 23, 2012 at 17:40
• @EMACK, The disk shape is a trade-off between gravitational force to the Earth's center, and centrifugal force due to rotation. The centrifugal force is zero on poles and the largest on the equator. But you can find much more detailed and numeric answer on the link posted by Qmechanic.
– Serg
Nov 23, 2012 at 18:11
• @EMACK: I suggest updating your title to reflect exactly what you're asking. Your current title is ambiguous; you could be asking whether the Earth's rotation affects its shape, or how it does so, or why it does so. Nov 23, 2012 at 19:27

The Earth is mostly fluid. This may seem a strange claim but the rock in the mantle behaves like an extremely viscous fluid, which is why continental drift can happen.

Anyhow, if you imagine a stationary drop of liquid it will form a sphere. This is a bit of a cheat because small drops form spheres due to surface tension not gravity, but the end results are similar. If you start the drop rotating the water at the "equator" is going to feel an outwards force due to the rotation, so the drop will change shape and get bigger around the equator while the poles flatten. This shape is known as an oblate spheroid, and indeed it's the shape of the Earth because the Earth behaves like a rotating fluid drop.

To try and calculate the change of shape is a little messy, but luckily someone has done all the hard work for you and you can find the results: Thayer Watkins: The shape of a rotating fluid mass.

• Is this outwards force due to rotation perhaps the centrifugal force?
– Mack
Nov 23, 2012 at 18:24
• Yes, it is the centrifugal force. Nov 23, 2012 at 19:14
• It is a very simple problem, it is not hard at all! Just add the gravitational and centrifugal potential and equal it to a constant. Nov 24, 2012 at 7:33
• Whoever downvoted, please would you explain why you thought the answer wasn't helpful. I'm not complaining about the downvote, but I can't improve my answers if you won't tell me what's wrong with them. Nov 24, 2012 at 14:24
• "[...] the rock in the mantle behaves like an extremely viscous fluid, which is why continental drift can happen" I would rather say that continental drift happens because the "rock" floats on "lava". However, I'm just guessing.
– dani
May 31, 2015 at 14:37

I don't think it is difficult to derive analytically the shape of the Earth. Simply look for the shape of the surfaces of equal potential.

The geometrical symmetry reduces the calculation to a 2-dimensional problem. Assume the rotation axis is vertical. The potential is the sum of the gravitational plus centrifugal:

$\Phi=\Phi_{g}+\Phi_{c}=-\frac{GM_{(x,y)}}{\sqrt{x^2+y^2}}+\frac{\omega^{2}}{2}x^{2}=-\frac{GM_{r}}{r}+\frac{\omega^{2}}{2}r^{2} \cos^{2} l$

The angle $l$ is the same as the latitude, and $M_{r}$ is the mass enclosed by a spherical surface (but please see footnote) at the point, i.e $M_{r}=\frac{4}{3}\pi \rho r^{3}$ by assuming a constant density model. Therefore,

$\Phi= -\frac{4}{3} G \pi \rho r^{2} +\frac{\omega^{2}}{2}r^{2} \cos^{2} l = r^{2}(\frac{\omega^{2}}{2}\cos^{2} l -\frac{4}{3} G \pi \rho)$

Thus, the family of curves of constant (negative) potential $\Phi=-C^{2}$ is:

$-C^{2} = r^{2}(\frac{\omega^{2}}{2}\cos^{2} l -\frac{4}{3} G \pi \rho) = r^{2}(A^{2} \cos^{2} l -B^{2})$

Let's go back to rectangular coordinates, to see that this is indeed an ellipse:

$C^{2} = r^{2}(B^{2} - A^{2} \cos^{2} l) = (x^{2}+y^{2})(B^{2} - A^{2} \frac{x^{2}}{x^{2}+y^{2}}) = (x^{2}+y^{2})B^{2} - A^{2} x^{2}$

$C^{2} = (B^{2} - A^{2}) x^{2} + B^{2} y^{2}$

For that equation to be an ellipse, $B^{2} - A^{2}$ must be positive. This is natural, otherwise (see how we defined $A$ and $B$) the angular speed $\omega$ would make the centrifugal force stronger than the gravitational force. The semiaxis are then $1/B$ for the vertical direction, and $1/\sqrt{B^{2} - A^{2}}$, i.e. bigger, in the horizontal direction. Note too, that $A=0$ for $\omega = 0$, that is, you recover the spherical shape if there is no rotation.

Thus, an Earth with constant density that rotates as a rigid solid can be approximated by an ellipsoid shape, whose dimension along the rotation axis is smaller.

Additionally, we probably don't need the interior of the Earth to be molten, for the hydrostatic equilibrium assumption to be valid. It could be completely cold and solid and the model still would hold, because at that size scales, relative small deviations of matter distribution from the constant potential surfaces give rise to enormous shear stress that rocks, no matter how hard and solid, cannot resist. That is why the liquid model is a valid approximation (but I have not done any numbers on this).

NOTE: We have assumed that any point belongs to a spherical surface that is completely full of matter, therefore the potential gravitational energy is the same as if all matter inside that sphere were located at the Earth centre. If the Earth were much more flattened, this approximation would not be valid.

The explanation of ovality of earth surface is quite illustrative but needs further detailing to predict the geometry of the real surface. Because of the wide variations in the density of the earth from point to point the proportion of gravitational attraction and the centrifugal force over different elements of mass $$dm=\rho\,dv$$ varied from point to point. If this fact is taken in to considerations, the surface of earth shall not be a mathematical surface but the portion of earth having higher density will pop up to form mountains and the portion with lower density will depress down to accommodate the condensed water of globe. We named them sea and ocean.

This also implies that the area beneath the Ocean will have lighter materials like coal, petroleum etc. and those beneath the mountains will have heavy elements like gold, platinum etc. This and the alubium deposits (at some places over flakes of shale) clearly implies that the whole water of the present oceans had condensed and flown through valleys within few days to form rivers and seas, when the Earth Environmental Temperatures had fallen down to condensation temp.

Let us generate data of density variation and carry out the calculations to predict the real geometry of earth surface or vice versa.

• Thank you for the answer and welcome to physics SE! The question explicitely asks about the influence of rotation on the shape of the earth. Please elaborate on this topic. Otherwise this may fail to answer the question. May 31, 2015 at 15:01