Dzhanibekov effect on the Earth

Question

Take an asymmetrical object and spin it in space and eventually it will flip on its own. A perfect sphere doesn’t have to worry about this, however the Earth is NOT a perfect sphere. My question is, is the Earth asymmetrical enough that somewhere down the line there is a possibility of the Earth flipping through the Dzhanibekov effect. Or alternatively could the gravity of the moon substitute for enough asymmetry to cause it happen instead. Don’t just say it won’t happen, show me the math or an example showing the Earth is spherical enough that it is in no way a concern.

Thank you kindly.

Edit: I wish to be shown how you get your answer. In effect, I am asking a mathematical question and would like to see the problem worked out. The same way if I asked the answer to another mathematical equation, I shouldn’t have to just accept 42 as the answer if I’m specifically asking to be shown how you got there. Step by step please. Please reopen my question.

Answer 1

Lets look at the equations

Euler Equations of Motion

The Euler equations governing the rotational motion of a rigid body in the body frame are:

\begin{align} I_1 \dot{\omega}_1 - (I_2 - I_3) \omega_2 \omega_3 &= 0, \\ I_2 \dot{\omega}_2 - (I_3 - I_1) \omega_3 \omega_1 &= 0, \\ I_3 \dot{\omega}_3 - (I_1 - I_2) \omega_1 \omega_2 &= 0. \end{align}

where $~( I_1, I_2, I_3 )~$ are the principal moments of inertia, and $~( \omega_1, \omega_2, \omega_3 )~$ are the angular velocity components in the body frame.

Linearization for Small Perturbations

Assuming that $~\omega_1 ~, \omega_2 ~$ are small perturbations around a dominant rotation about the third principal axis, with $$I_3 \dot{\omega}_3 =0\quad\Rightarrow\quad \omega_3 = \Omega, \quad \text{(constant)} $$

and linearize the first two equations:

\begin{align} I_1 \dot{\omega}_1 &= (I_2 - I_3) \Omega \omega_2, \\ I_2 \dot{\omega}_2 &= (I_3 - I_1) \Omega \omega_1. \end{align}

Matrix Form and Eigenvalues

Writing in matrix form:

$$ \begin{bmatrix} I_1 & 0 \\ 0 & I_2 \end{bmatrix} \begin{bmatrix} \dot{\omega}_1 \\ \dot{\omega}_2 \end{bmatrix} = \begin{bmatrix} 0 & (I_2 - I_3) \Omega \\ (I_3 - I_1) \Omega & 0 \end{bmatrix} \begin{bmatrix} \omega_1 \\ \omega_2 \end{bmatrix}. $$

Dividing by the moments of inertia:

$$ \begin{bmatrix} \dot{\omega}_1 \\ \dot{\omega}_2 \end{bmatrix} = \begin{bmatrix} 0 & \frac{I_2 - I_3}{I_1} \Omega \\ \frac{I_3 - I_1}{I_2} \Omega & 0 \end{bmatrix} \begin{bmatrix} \omega_1 \\ \omega_2 \end{bmatrix}. $$

To find the characteristic equation, we solve:

$$ \det \begin{bmatrix} -\lambda & \frac{I_2 - I_3}{I_1} \Omega \\ \frac{I_3 - I_1}{I_2} \Omega & -\lambda \end{bmatrix} = 0. $$ Expanding the determinant:

$$ \lambda^2 - \left( \frac{(I_2 - I_3)(I_3 - I_1)}{I_1 I_2} \right) \Omega^2 = 0. $$

Solving for $~ \lambda ~$:

$$ \lambda = \pm \Omega \sqrt{\frac{(I_2 - I_3)(I_3 - I_1)}{I_1 I_2}}. $$

Stability Analysis

• If $~(I_2 - I_3)(I_3 - I_1) > 0 ~$, then $~ \lambda ~$ is real, leading to exponential growth or decay (unstable rotation).
• If $~ (I_2 - I_3)(I_3 - I_1) < 0 ~$, then $~ \lambda ~$ is purely imaginary, leading to oscillations (stable rotation).

This explains why rotation about the largest or smallest principal inertia axis is stable, while rotation about the intermediate axis is unstable (the Dzhanibekov ( Tennis Racket ) Theorem).

for the earth

The Earth is an oblate spheroid, meaning $~I_3 > I_1 \approx I_2 $ leading to oscillations (stable rotation).