Simplest expression for gravitational flattening of Earth-like planet, from zero rotation rate to Earth rate, to within 10 percent error

Question

Richard Fitzpatrick discusses the rotational flattening of a celestial body. A model that assumes homogenous density arrives at a difference between the Earth's equatorial and polar radii of 27.5 kilometer, significantly larger than the actual difference of about 21.4 kilometer.

My goal is to create a plot that gives flattening as a function of angular velocity of an Earth-size, Earth-mass celestial body. For what I will be using it for an error under 10% is sufficient.

For the rotation rate: the range of interest is from zero rotation rate to at most double the rotation rate of the actual Earth. (Most texts that discuss rotation of celestial bodies focus on extremely high rotation rates. It is surprisingly hard to find discussion of Earth-like rotation rate.)

A crude approach would be to assume that at at any rotation rate (close to actual Earth rotation rate) the error of the homogenous-density-model will be in the same ratio. So then, to get to within an error of 10%, it would be enough to always apply an adjustment factor of 21.4/27.5

Can anyone confirm or disconfirm that?

(Most material I found aims for far higher accuracy, and the formula's are complicated. For my particular purpose that level of accuracy is overkill.)

Answer 1

After equation 6.56 Fitzpatrick concludes that equation 6.43 is accurate and any inaccuracy between the Earth's equatorial and polar radii is due to the earths quadrupole moment $J_2$.

So it's recommended that you try to find the relation of $J_2$ to $\Omega$ for the earth.

As it's so similar in size to $\zeta$ , equation 6.37, $J_2$ might also be proportional to $\Omega^2$.

If so you should be ok using your adjustment factor.

Answer 2

Let's see how this works out for Mars, ok? We have for the red planet:

$M_M=6.417\times10^{23}$kg$= 0.107 M_\oplus$

$R_M=3389.5$km$=0.532R_\oplus$

$T_M=24.6229$hr=$1.029 T_\oplus$

With these we can start computing. The critical radius is $R_{c,M}=215$km, which is why Mars is spherical. The dimensionless parameters are $\zeta_M = 0.001523=1.32\zeta_\oplus$ and $\tilde{\mu}_M=60=0.69\tilde{\mu}_\oplus$, where I've assumed that the shear modulus of the rock on Mars is identical to that on Earth. I think its safe to assume the yield stress for the Mars rock will also be identical to that of Earth, so $\zeta_{c,M}=0.013$. As on Earth, $\zeta_M>\zeta_{c,M}$, so the Mars rock will flow like a liquid until the surface stress vanishes, and $\epsilon_M=\frac{15}{4}\zeta_M=0.00571$. The observed value for Mars is $\epsilon_M^{obs}=0.00589$, which is much closer to the theoretical value. Now sure, Mars doesn't have a molten core like the Earth, so treating it like an elastic body is a lot more accurate.

For exoplanets similar to Earth, lets fall back on $\epsilon=\frac{3}{2}(J_2+\zeta)$, which fits the Earth data exquisitely. The ratio of observed to theoretical values is

$\frac{\epsilon^{obs}}{\epsilon}=\frac{2}{5}\left(1+\frac{J_2}{\zeta}\right)=\frac{2}{5}\left(1+0.94\alpha \right)$ with $\alpha=\left(\frac{\Omega_\oplus}{\Omega}\right)^2\frac{\rho}{\rho_\oplus}\frac{J_2}{J_{2,\oplus}}$.

Note that your use of $\frac{21.4}{27.5}\approx 0.778$, and the calculated value, $\frac{2\times1.94}{5}\approx 0.776$, for Earth( $\alpha=1$) are almost equal. So as long as the rotational period, average density, and quadrupole (which depends on the precise radial density distribution) don't differ much from Earth's, multiplying by your ratio will give a very good approximation for the observed value.

Hope that helps!