# rotational oblateness

I am trying to compute the amount of oblateness that is caused by planetary rotation. I picture the force of gravity added to the centrifugal force caused by the rotation of the planet as follows:

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That is, at the point in question, at latitude $\phi$, the distance from the axis of rotation is $r\cos(\phi)$. Thus, the centrifugal force would be $\omega^2r\cos(\phi)$ in a direction perpendicular to the axis of rotation. The radial and tangential components would be $\omega^2r\cos^2(\phi)$ and $\omega^2r\cos(\phi)\sin(\phi)$, respectively.

My assumption is that the surface of the planet would adjust so that it would be perpendicular to the effective $g$; that is, the sum of the gravitational and centrifugal forces. This would lead to the equation $$\frac{\mathrm{d}r}{r\,\mathrm{d}\phi}=-\frac{\omega^2r\cos(\phi)\sin(\phi)}{g-\omega^2r\cos^2(\phi)}$$ We can make several assumptions here, and I will assume that $\omega^2r$ is small compared to $g$. Thus, we get $$\int_{\text{eq}}^{\text{np}}\frac{\mathrm{d}r}{r^2} =-\frac{\omega^2}{g}\int_0^{\pi/2}\cos(\phi)\sin(\phi)\,\mathrm{d}\phi$$ which leads to $$\frac1{r_{\text{np}}}-\frac1{r_{\text{eq}}} =\frac{\omega^2}{2g}$$ and $$1-\frac{r_{\text{np}}}{r_{\text{eq}}} =\frac{\omega^2r_{\text{np}}}{2g}$$ However, numerical evaluation and Wikipedia seem to indicate that this should be twice what I am getting. That is, $$1-\frac{r_{\text{np}}}{r_{\text{eq}}} =\frac{\omega^2r^3}{Gm} =\frac{\omega^2r}{g}$$ What am I doing wrong?

• I see that this is related to Why is the Earth so fat?. One possibility for the error in my computation might be the direction of gravity generated by an oblate spheroid. However, this would seem to be dependent on the mass distribution inside the spheroid. The factor of $2$ between my estimate and Wikipedia's seems to discount that the mass distribution would be the cause. – robjohn Jun 29 '13 at 22:26
• Well, aren't you forgetting about the southern half of the earth in your integral? Or are these calculations always done for a hemishere? – Wojciech Morawiec Jun 29 '13 at 23:52
• @WojciechMorawiec: By symmetry, integrating through the southern hemisphere should bring $r_{\text{eq}}$ back to $r_{\text{np}}$. – robjohn Jun 29 '13 at 23:56
• @robjohn The factor of two seems to be explained in the answer to the question you referenced - due to the bad approximation of initially treating the Earth as a sphere in your calculation. – Will Jun 30 '13 at 2:57
• @Will: if that were the case, then I would expect the error to be dependent on the density map of the oblate spheroid. The formula in Wikipedia is not specifically for the Earth, yet it is twice what I got, seemingly independent of the density. – robjohn Jun 30 '13 at 4:02

As is well-known from Newton's shell theorem, the gravitational field $g(r)=\frac{GM}{r^2}$ outside a spherically symmetric mass-distribution is the same as if the total mass $M$ sat in the center.

It seems that OP wants to calculate the oblateness of Earth under the simplifying assumption that the backreaction (which the re-distributed mass has on Earth's gravitational field) can be ignored. In other words, we assume that the gravitational field is given by just the monopole contribution $g(r)=\frac{GM}{r^2}$, and we neglect higher multipole moments in a multipole expansion.

I) This is what Mark Eichenlaub did in this Phys.SE post. To compare let us replace the latitude $\phi$ with the polar angle $\theta=\frac{\pi}{2}-\phi$. The total potential energy is a sum of the gravitational monopole potential energy and the centrifugal potential energy

$$\tag{1} U~=~- \frac{GM}{r}-\frac{(\omega r \sin \theta)^2}{2}.$$

The point is now that (in this idealized model) the surface of Earth is an equipotential surface. ("Else the water in the oceans would rush to re-distribute itself.") Comparing Equator and the North pole leads to

$$\tag{2} g(a)h~\approx~\frac{GM}{b}-\frac{GM}{a}~\stackrel{(1)}{=}~\frac{(\omega a)^2}{2}~>~0,$$

where $a$ and $b$ are the equatorial and polar radius of the Earth, respectively; and $h:=a-b\ll a$ is the sought-for height difference. Equation (2) leads precisely to Mark Eichenlaub's monopole result for $h$, which is $\frac{2}{5}$ smaller than the quadrupole result.

II) On the other hand, if we differentiate eq. (1), we get precisely OP's force equilibrium formula

$$\tag{3} 0~=~\mathrm{d}U~=~\left(g(r)-(\omega\sin \theta)^2 r\right)\mathrm{d}r -(\omega r )^2\sin \theta \cos \theta \mathrm{d}\theta.$$

At this point OP ignores the radial dependence of $g(r)$, and treats it as a constant $g$. This model corresponds to a total potential energy

$$\tag{4} V~=~ gr-\frac{(\omega r \sin \theta)^2}{2}.$$

Comparing Equator and the North pole leads to

$$\tag{5} gh~=~g b-ga~\stackrel{(4)}{=}~-\frac{(\omega a)^2}{2}~<~0,$$

which predicts an prolate Earth rather than an oblate Earth.

III) Next OP assumes that one of the centrifugal terms $(\omega\sin \theta)^2 r \ll g$ in eq. (3) is small and should be ignored. This mean that eq. (3) is no longer a perfect differential. However an integrating factor is $\frac{1}{r^2}$, so this leads to a first integral

$$\tag{6} W~=~ -\frac{g}{r}-\frac{(\omega \sin \theta)^2}{2}.$$

Comparing Equator and the North pole leads to

$$\tag{7} \frac{gh}{ab}~=~\frac{g}{b}-\frac{g}{a}~\stackrel{(6)}{=}~\frac{\omega^2}{2}~>~0,$$

which remarkably reproduces Mark Eichenlaub's monopole result (2). In other words, two not-so-small-approximations by OP have cancelled out.

• Thanks for your answer. I will have to take some time to digest it. I am also glad that you provided a link to Mark Eichenlaub's question. I will read the answers there as well. – robjohn Nov 19 '13 at 23:22

Problem is with the assumption:

... I will assume that $\omega^{2}r$ is small compared to $g$

This means: centrifugal force at the equator is negligible.
And this cannot be true, because the planet would be perfect sphere.

(Note: word "small" was used as "negligible" when constructing the equation)

• I am including the contribution from centrifugal force; that's the $\omega^2r\cos(\phi)$. It is the contribution from the redistribution of mass that I was assuming was negligible. Evidently, it is not. – robjohn Oct 7 '13 at 17:50
• I meant centrifugal force at the equator (added to the answer). Centrifugal force at the equator is proportional to $\omega^{2}r$ and gravitational is proportional to $g$. – user1086737 Oct 7 '13 at 20:14
• I am assuming that $\omega^2r$ is small compared to $g$, not that it is insignificant. If it were insignificant, there would be no oblateness at all. This means that second order terms in $\omega^2r$ will be even smaller compared to $g^2$. – robjohn Oct 7 '13 at 21:35
• If $\omega^2r$ is small but not insignificant - can $\omega^2r\cos^2(\phi)$ be omitted from the integral? – user1086737 Oct 7 '13 at 21:40
• I am looking at the first order contributions. Since $\omega^2r/g\doteq0.003433$, approximating the denominator with $g$ rather than $g-\omega^2r\cos^2(\phi)$ will alter the result by at most $0.34\%$. That is much smaller than the difference seen. – robjohn Oct 7 '13 at 21:51