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:
$$\begin{equation}
\epsilon=\frac{5}{4}\frac{\omega^2a^3}{GM}
\end{equation}$$
(more about it
here)
##\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
$$\begin{equation}
\epsilon=\frac{a_e-a_p}{a}
\end{equation}$$
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:
$$
\begin{equation}
a=\frac{2a_e+a_p}{3}
\end{equation}$$
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:
JGHunter said:
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 let's you spot any mistakes you might have made. So this:
JGHunter said:
$$ \frac {(1)(6,378 km)(1,670 km/h)^2}{(1)(3,402.5 km)(868.22 km/h)^2}=\frac{10,651,260 N}{2,954,118.55 N} $$
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:
JGHunter said:
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:
$$
\begin{equation}
\epsilon=\frac{5}{4}\frac{\omega^2a^3}{G\frac{4}{3}\pi a^3 \rho}=\frac{15}{16}\frac{\omega^2}{G\pi \rho}
\end{equation}$$
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.