# 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?

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Possible duplicates: physics.stackexchange.com/q/8074/2451 and physics.stackexchange.com/q/10670/2451 – Qmechanic♦ Nov 23 '12 at 16:43

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The answer is Yes. – Serg Nov 23 '12 at 17:34

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@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 '12 at 17:40

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@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 '12 at 18:11

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@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. – Keith Thompson Nov 23 '12 at 19:27
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## 2 Answers

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 from 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 here.

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Is this outwards force due to rotation perhaps the centrifugal force? – Mack Nov 23 '12 at 18:24

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Yes, it is the centrifugal force. – Emilio Pisanty Nov 23 '12 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. – Eduardo Guerras Valera Nov 24 '12 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. – John Rennie Nov 24 '12 at 14:24

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.

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