# Question about Earth's equatorial bulge-derivation

1. Sep 30, 2011

### glw1722

How does one go about calculating the expected equatorial bulge of the earth? When I do the calculation I do not get the expected increase in diameter. I get a difference in diameters (D| equator-D|poles) 21.9 km when it should be 42.7km. My calculation equates the reduced acceleration at the equator (9.85-0.0339 m/s^2) to a gravitational acceleration at an increased radius (r|pole + h, where h is the increased radius) and I derive h. If I had gotten an h larger than expected I would have said the difference is due to the earth's stiffness, but I don't even get that. I have what I think is a possible answer but before I go there I thought I'd see if anybody really knows other than my guesses.

2. Sep 30, 2011

### xts

I can't understand why you tried to equate those. If at all, you could try to equate potentials, but it also would be wrong, as the reference frame is not inertial.

The simplest way I know to calculate the bulge is:
Calculate forces acting on a droplet of water, somewhere at lattitude φ. The centrifugal force is $m\omega^2r$, its horizontal component is $m\omega^2r\sin\varphi$. This force must be compensated by some slope: $$m\omega^2r\sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}$$ As the first level approximation you may substitute mean values for $r$ and $g$ and integrate this to obtain $$r(\varphi)\approx r(\pi/2)+\frac{\omega^2\bar{r}^2}{\bar{g}}\cos\varphi \qquad \qquad r(0) - r(\pi/2) \approx\frac{\omega^2\bar{r}^2}{\bar{g}} \approx 21.9\,{\rm km}$$ - that is a result for radius, so the difference in diameter is twice bigger.

Last edited: Sep 30, 2011
3. Oct 2, 2011

### RedX

Shouldn't your r in the LHS of:

$$m\omega^2r\sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}$$

be r*cos(phi), or distance from the rotation axis? If you do that, then you get the original poster's value of half the bulge.

4. Oct 2, 2011

### xts

Oh, yes, you are perfectly right: $\cos \varphi$, but it doesn't change the final value.
No...
$$\int_0^{\pi/2}\cos\varphi\,d\varphi = \int_0^{\pi/2}\sin\varphi\,d\varphi=1$$
But, of course, you are right, I gave wrong formula for the shape of the bulge, should be then
$$r(\varphi) \approx r(0) - \frac{\omega^2\bar r ^2}{\bar g} \sin\varphi$$

5. Oct 2, 2011

### RedX

I thought the formula should be:

$$m\omega^2(r \cos \varphi) \sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}$$

but maybe I'm wrong.

You need the centripetal force along the tangential direction, so $m\omega^2r \cos \varphi$ gets multiplied by $\sin\varphi$.

When you integrate:

$$m\omega^2(r \cos \varphi) \sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}$$

you get half the value for the difference in radius.

6. Oct 2, 2011

### xts

Yeah... (I am :grumpy: in a shame)... You are right... I must calculate it again... Tomorrow...

remember implication law: $T\implies T$, but also $F\implies T$, and in this case it seems that I gave wrong justification for right result... [ blushing more than emoticons available here may express ]

Last edited: Oct 2, 2011
7. Oct 2, 2011

### RedX

If you can get the right answer, let us know! I've been trying for hours and I've given up. I teach a lab course and I'm trying to give my students some information on centripetal acceleration, but I can't figure out how to calculate the bulge - like the original poster I get half the value.

8. Oct 3, 2011

### xts

Our calculations would be correct if the whole Earth mass was condensed in its centre. But as the mass is distributed over the whole depth, then the bulge causes that gravity vector does not point towards Earth centre. This is a zero order correction to our calculation - can't be neglected. Clairaut presented formulae for gravity of ellipsoid, as you solve them the final flattening may vary from $\frac{1}{2}\omega^2 r^2/g$ for mass concentrated in the centre to over three times more for uniform density.

Unfortunately I can't find any on-line English textbook explaining those calculations by Clairaut and Laplace...

9. Oct 3, 2011

### RedX

Thanks. I appreciate it. I wish I could just slip in a factor of 2 on

$$m\omega^2(r \cos \varphi) \sin\varphi = -mg \frac{1}{2}\frac{1}{r}\frac{dr}{d\varphi}$$

and say that the 2 comes from considering mass spread over the entire ellipsoid. That would produce the right answer. But unfortunately, as you say, that factor could be as low as 1/6 for uniform density (producing a bulge of 3 times more than when the mass is concentrated at the center). I guess the earth is somewhere in between mass all at the center and mass uniformly distributed, so let's just say it's exactly in between those two cases, so that it's exactly between: $\frac{1}{2}\omega^2 r^2/g$ and $\frac{3}{2}\omega^2 r^2/g$, which would be...$\omega^2 r^2/g$

10. Oct 3, 2011

### RedX

Actually, according to Wikipedia:

http://en.wikipedia.org/wiki/Clairaut's_theorem

the gravity is modified by:

$$g[1+(\frac{5m}{2}-f)\sin^2 \varphi]$$

where m is the ratio of the centrifugal force to gravity at the equator (which should be really small), and f is proportional to the difference in semi-axis of the ellipsoid (which should also be really small). So it seems that this new value of gravity is not different by zeroth order, and you should really be able to just use "g". Maybe I'm reading the article wrong.

11. Oct 3, 2011

### xts

It is not exactly 2 for Earth, but close to it. Anyway - the exact value for uniform density is not exactly 3 - it is a bit more, as a sum of some series. Yes - we have more dense core, surrounded by ligther mantle - it leads to something close to 2.

It is very far from my major point of interest (actually - it was you, who made me to search for some theory...) so I am not pretending to be an authority. I found lots of theory in geophysics, making our calculations in reverse: we may measure actual flattening, and Earth momentum of inertia, then utilize those to find the density distribution inside Earth.

12. Oct 3, 2011

### xts

Maybe I was wrong calling it zero-order, maybe I should say first-order...
What I meant was that flattening goes to the equation in the same order as $m$. So in order to calculate flattening, you may use iteration:
1. calculate flattening as we did previously;
2. calculate again, using modified gravity (esp. its direction);
3. and again, and again...
On each iteration the correction stays on the same order of magnitude - it is convergent, but not very fast, and the final result may be several times bigger than first step.

13. Oct 3, 2011

### RedX

Thanks for your help. I appreciate it. This is not my area either, but I teach lab courses to pay for my tuition, and I wanted to give my students examples of some good applications of centripetal acceleration and I thought what could be a more grand example of centripetal acceleration than the spinning earth itself! So in a way you helped my 48 students, and not just me!

14. Oct 3, 2011

### xts

I am not quite sure if that argumentation about iterative approach was true ;) Better don't use it with students...

Laplace's one was integration from centre of Earth: you start from very small, spherical Earth, then you build next infinitesimal layer of some density, using gravity from previous one to find its shape, and so on.
On every stage the correction due to flatteness of inner layer has the same order of magnitude, as $m$ - so the final result is much different from Newton's (and ours) approach with central mass distribution.

15. Oct 8, 2011

### glw1722

I've been following and it is interesting how you both have developed various solutions. Let me throw this out to get your thoughts, could it be possible that when the earth first formed and was hot and very malleable and was spinning faster and this gave rise to the bulge which became frozen in? Perhaps it has settled some since then so the rotation one would calculate might appear somewhat reduced from the actual birth rotation. I get an ~17 hour rotation.

16. Oct 8, 2011

### RedX

I'm not that familiar with stellar formation, so you might be right. According to this website:

http://www.josleys.com/show_gallery.php?galid=313

the earth is malleable because of its liquid core and tectonic plates on the surface (quite interestingly, according to that website, if the earth spins fast enough, it'll look like a pear...who would have thought?).

The way I explained it to my students is that I showed the derivation discussed here, assuming all the mass of the planet is at its center, to get 1/2 w^2 r^2/g. Then I made the comment that only spherical shapes allow you to assume that all of the mass is at the center, and if you have an ellipsoid gravity points a little more towards the semi-major axis (you can convince yourself of this by just drawing an ellipse and cutting it in half through the origin to the point you want to calculate the field). Because gravity is pointing more towards the equator in an ellipse than a circle, the bulge on a spinning ellipsoid has to be a little greater. So you get a higher value: 3/2 w^2 r^2/g.

Then I argued that most of the mass of the earth however is at its center anyways, so you take the average: w^2r^2/g. Ironically, if you just guessed the result from dimensional analysis, you'd get that result.