# Influences on flattening

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## Main Question or Discussion Point

I was looking at the numbers regarding the planets in our solar system, their bulge, their flattening ratio and their rotational speed. I know that rotational speed plays a role in this flattening, however what else is at play? For example, Earth's flattening ratio is nearly 1:300, whilst Mars is only about 1:135, less than half. Mars' rotation is a little slower than Earth, but not considerably, and Earth's density is less than 150% than that of Mars, so it doesn't seem like this is what's at play either, but is it? What am I missing that accounts for Earth being so much more flatter than the other planets?

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Geofleur
Gold Member
Here's an idea:

The centripetal force on a chunk of planet near the surface depends on the radius as well as the angular velocity. Consider a chunk of Earth and a chunk of Mars, each with mass $m$. The centripetal force on a chunk at radius $r$ and surface speed $v$ is $mv^2/r$. Now the surface speed and the angular velocity $\omega$ of the planet are related as $v = r\omega$. Hence, the centripetal force can also be written as $mr\omega^2$. Now take the ratio of this quantity at the surface of the Earth to that at the surface of Mars: $$\frac{m r_e \omega_e^2}{m r_m \omega_m^2}=\frac{r_e}{r_m}\left( \frac{\omega_e}{\omega_m}\right)^2.$$ Putting in the numbers for Earth and Mars, this ratio is about 2. What happens if you calculate the gravitational force minus centripetal force on each chunk? How do those forces compare? Earth's mass is about 10 times that of Mars, and I imagine that makes a difference too!

Sorry for being a simpleton, but what does $_e$ represent here? I've looked it up and I think it means eccentricity, but what kind?

Geofleur
Gold Member
e stands for Earth and m stands for Mars, so $r_e$ is the radius of the Earth and $r_m$ is that of Mars, etc.

Right okay sorry, I'll try playing it out.

with $$\frac {m r_e \omega_e^2}{m r_m \omega_m^2}=\frac{r_e}{r_m} \left( \frac {\omega_e} {\omega_m} \right ) ^2$$

Where m = 1 (one piece of Mars and one piece of Earth of equal mass)

$$\frac {(1)(6,378 km)(1,670 km/h)^2}{(1)(3,402.5 km)(868.22 km/h)^2}=\frac{6,378 km}{3,402 km} \left( \frac {1,670 km/h} {868.22 km/h} \right ) ^2 = 1.87478 \left( 1.92348 \right ) ^2$$

Geofleur said:
Putting in the numbers for Earth and Mars, this ratio is about 2. What happens if you calculate the gravitational force minus centripetal force on each chunk? How do those forces compare? Earth's mass is about 10 times that of Mars, and I imagine that makes a difference too!
$$\frac {(1)(6,378 km)(1,670 km/h)^2}{(1)(3,402.5 km)(868.22 km/h)^2}=\frac{10,651,260 N}{2,954,118.55 N}$$

$$g_e - \omega_e = 9.807 m/s^2 - 10,651,260 N$$ $$g_m - \omega_m = 3.711 m/s - 2,954,118.55 N$$

Please tell me what I've done wrong! This is the first time I've used LaTeX, let alone tried to calculate this sort of thing.

Even if I'm right, I'm not really sure what I'm meant to be taking from this as an explanation for the increase in flattening. I'm guessing that the greater number for Earth indicates the greater throw resulting in greater flattening?

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gneill
Mentor
Methinks you want to investigate the "Equilibrium of a Rotating Liquid Body". A number of great names in physics have looked at this, the figure of a rotating, self-gravitating liquid.

Hi, sorry it's been a while, I had a look at that and I feel I need a different equation. I want to know if I can measure how much a bulge will be if I know the equatorial radius, the mass of the planet and its angular velocity, but not its polar (rp) or mean (r) radius, or its density (ρ). The density depends on the planets mean radius and its smaller polar radius will affect both this and the flattening coefficient (f). Is it possible for me to figure out how much bulge is occurring and by extension how much smaller the polar radius (rp) will be by just knowing the planets mass (M), its angular velocity (ω) and equatorial radius (re)?

Bandersnatch
There's a bit to unpack here, which is probably why the thread hasn't been more active. But first things first.

The equation you'll want to be looking at is:
$$\epsilon=\frac{5}{4}\frac{\omega^2a^3}{GM}$$
$\epsilon$ is flattening expressed in decimals (so, what you get if you divide e.g. 1:300)
$\omega$ is angular velocity, $a$ is mean radius, $M$ is mass, $G$ is the gravitational constant

This equation follows from what @Geofleur was saying - there is the gravitational force from mass M of the planet compressing the entire planet into a sphere, with the strength of the compressing force falling down as a square of distance (the $a^2$ factor). While at the same time rotation induces outwards centrifugal force reducing the strength of compression perpendicular to the axis of rotation, and increasing with both angular velocity and radius (the $\omega^2a$ factor).
Details of the equation in the link above, although it's probably beyond the level of this question.

The meaning of $\epsilon$ is
$$\epsilon=\frac{a_e-a_p}{a}$$
I.e. epsilon times the mean radius gives you the difference in between equatorial and polar radii of the oblate spheroid.

If you look at eq.1 and imagine increasing or decreasing each of the variables, you should see that increasing rotation or radius increases flattening, whereas increasing mass (while holding radius constant, which translates to increasing density) causes flattening to decrease.

Furthermore, the mean radius of an oblate spheroid is given by:
$$a=\frac{2a_e+a_p}{3}$$
This is because the spheroid has three axes, and two of them are in the equatorial plane (so the equatorial radius contributes twice as much to the mean as the polar radius).

This is sufficient to replace the mean radius in the first equation with equatorial radius (since that's what you said you wanted as the input):
$$a=\frac{a_e}{1+\frac{1}{3}\epsilon}$$

Rearranging eq.1 to extract epsilon to the left side will require some tedious algebra. But we can make things simpler by cheating a bit. The cheating involves replacing the mean radius in the equation by the equatorial radius (or polar, should we want it as an input instead). This will not introduce much of an error for any regular planetary body that is not extraordinarily flattened - the few percent difference don't matter much. And anyway, this will be swamped by another error due to some underlying assumptions behind the equation (on that later).

So, the equation for flattening is eq.1 with $a_p$ instead of $a$.
Once you have $\epsilon$ calculated, you can use eqs 2 & 3 to calculate $a_e$:
$$a_e=a_p\frac{(1+\frac{1}{3}\epsilon)}{(1-\frac{2}{3}\epsilon)}$$

You should plug in the numbers and calculate epsilon for some planets with known flattening. You should find that the equation outputs flattening that is too high. E.g. approx 0.1 instead of 0.065 for Jupiter.
This is because the eq.1 assumes the planet to be composed of incompressible fluid of uniform density. Real planets have varying density, with cores denser than the outer layers, which hold the spherical shape better (this is especially pronounced in gas giants). All in all, this will always cause the output of our equation to be an overestimate.
But given the problems you've had with maths in post #6, I think you should settle for this approximation, which is not that bad anyway. You can always reduce the epsilon you get by 10-40ish % (more for larger planets), which should probably land you in a close ballpark of the actual numbers.

Now, speaking of post #6.
When you will be plugging in the numbers, you have to be more careful. For example:
with $$\frac {m r_e \omega_e^2}{m r_m \omega_m^2}=\frac{r_e}{r_m} \left( \frac {\omega_e} {\omega_m} \right ) ^2$$

Where m = 1 (one piece of Mars and one piece of Earth of equal mass)

$$\frac {(1)(6,378 km)(1,670 km/h)^2}{(1)(3,402.5 km)(868.22 km/h)^2}$$
The equation provided asked you for angular velocity, but you went and plugged in linear velocity.
Linear velocity tells you how much of a linear distance has been covered in some amount of time. It is expressed in [distance]/[time], so e.g. km/h or km/s or miles/day.
Angular velocity tells you how much of an angle has been covered in some amount of time. It is expressed in [measure of angle]/[time]. So, e.g. if you want to say that something rotates once a day, you'd say its angular velocity is a full 360 degrees ($2\pi$ in radians) per day, or however many seconds is there in a day. I.e.: $\omega=\frac{2\pi}{24*3600s}$. Since radians are treated as dimensionless, the units of angular velocity will be 1/s.

Angular velocity is very useful, and easier to use than linear velocity, since it's the same everywhere on the planet, and as long as you know the rotation period, you can calculate $\omega$ as 2 pi radians per however many seconds it takes for a full rotation.

It's extremely important to pay attention to units whenever you're doing algebraic calculations. This lets you spot any mistakes you might have made. So this:
$$\frac {(1)(6,378 km)(1,670 km/h)^2}{(1)(3,402.5 km)(868.22 km/h)^2}=\frac{10,651,260 N}{2,954,118.55 N}$$
Is really bad form. It does not follow that $km$ times $(km/h)^2$ equal Newton. And you omitted the units for mass on top of that. You can't just write what you want it to be. What you should have done here, is looked at the units on the left side of the equation, and ask yourself if $km^3/h^2$ or even $kg*km^3/h^2$ equals $N$, or anything else for that matter. If it doesn't, and you think it should, then you've made a mistake somewhere.

Furthermore, when working with numbers, all units must match. You can't have something in seconds here, and in hours there. In metres here, and kilometres there, etc. First thing you do, before plugging in any numbers, is make the units match. So, e.g. if you have or want to have Newtons anywhere, you need to have all masses in kilograms, all distances in metres, and all times in seconds - because that's how the Newton is defined (always check wikipedia if in doubt).

Make sure you've got the right values, and the right units. Then you should be fine with using the equation for flattening given above.

Lastly:
For example, Earth's flattening ratio is nearly 1:300, whilst Mars is only about 1:135, less than half. Mars' rotation is a little slower than Earth, but not considerably, and Earth's density is less than 150% than that of Mars, so it doesn't seem like this is what's at play either, but is it? What am I missing that accounts for Earth being so much more flatter than the other planets?
1:300 means that the flattening is less for Earth than the 1:135 for Mars. It means that the difference between the equatorial and polar radii is 1 part in 300 of the mean radius. Or approx 0.3% ($\epsilon=1/300=\approx0.003$). For Mars, using the ratio you provided, the flattening is over twice as much (it's less of a difference in reality, or from our equation; not sure where you got it from).
If you look at eq.1 again, you should see what causes this: Mars has roughly the same angular velocity, half the radius (which is in third power in the equation, so this counts), and 1/10th of the mass. The variables in numerator being lower cause the flattening to decrease (so, lower radius decreases flattening) while those in the denominator being lower cause the flattening to increase (so, lower mass causes higher flattening).

One can do a thing here, and express mass in eq.1 in terms of volume and density:
$$M=V\rho=\frac{4}{3}\pi a^3 \rho$$
$\rho$ being density; where we again cheat a bit by assuming the oblate spheroid's volume uses mean radius (or polar, or equatorial) only, again, with minor losses to accuracy.
This turns eq.1 into:
$$\epsilon=\frac{5}{4}\frac{\omega^2a^3}{G\frac{4}{3}\pi a^3 \rho}=\frac{15}{16}\frac{\omega^2}{G\pi \rho}$$
Which shows that, under the simplifying assumptions we were using, the degree of flattening depends only on angular velocity and density.
Mars having the same angular velocity and lower density has higher flattening.

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It's been almost a year exactly since I went away having thought about these numbers.

It turns out the project I'm working on results in a bulge of less than half a kilometre over a pole-to-pole distance of over 30,000km! Not even worth accounting for. But at least I learnt something and that made it a worthy endeavour so thank you those who replied.