# Why does an irregular moon point its longest axis towards its parent planet?

I'm reading the book Ice Worlds of the Solar System. Page 62, it states that an irregular moon should point its longest axis towards its planet. Why is this the more stable configuration?

• If its axis points towards the planet, then after revolving 180 degrees on its orbit, won't it have the opposite rotational angular momentum? – Keith McClary Nov 11 '20 at 6:08

Rotation of a rigid body around the major axis (the principal axis with the highest moment of inertia, which will tend to be close to the longest axis of the satellite) minimises the kinetic energy for a given angular momentum. As a quick way to visualise this, write the equation for rotational kinetic energy $$T$$ in terms of angular momentum $$L$$ and moment of inertia $$I$$:

$$T = \frac{L^2}{2I}$$

At constant $$L$$, $$T \propto I^{-1}$$, therefore picking the axis with the highest moment of inertia will give the lowest kinetic energy.

As a satellite is not perfectly rigid nor perfectly elastic, deformations will tend to dissipate energy but angular momentum is conserved, thus leading it to evolve into this state.

Some lecture notes on the relevant dynamics can be found here, with the relevant result (the "major axis theorem") given in §4.11.

Any freely rotating body that is not perfectly rigid, will lose kinetic energy and whist its angular momentum remains constant in space, it moves with respect to the body until the body is rotating about its major axis.

This explains the rotation, likewise orienting the long axis towards the planet minimises the gravitational potential energy. As an example to demonstrate this, consider a rod of length $$2r$$ of linear density $$\rho$$ with its centre located at a distance $$R$$ from a mass $$M$$, with the rod at angle $$\theta$$ to the radius vector. Neglecting variations in the gravitational field perpendicular to the radius vector, the potential energy is

\begin{aligned}V &= \int_{-r}^{r} \frac{-GM\rho}{R+x \cos \theta} dx \\ &= -\frac{GM\rho}{\cos \theta} \log \left( \frac{R+r \cos \theta}{R-r \cos \theta} \right) \\ &= -2GM\rho\left( \frac{r}{R} + \frac{1}{3} \left( \frac{r}{R} \right)^3 \cos^2 \theta + \ldots \right) \end{aligned}

where the last line uses the MacLaurin series for $$\log (1+x)$$.

All the terms in the sequence in brackets are positive and proportional to increasing even powers of $$\cos \theta$$, so the potential energy decreases as $$\cos \theta$$ increases. The minimum potential energy is therefore when $$\cos \theta = 1$$, i.e. when the rod is aligned with the radius vector.

The state of rotating around the major axis with the major axis aligned with the radius vector (long axis pointing towards the planet) is therefore the minimum energy state.

• Does this yet explain why the moon's major axis points towards the planet about which it orbits? At the moment it explains why the moon's rotational axis will over time tend to coincide with its major axis, but there's more energy to be dissipated by one or both of those axes eventually pointing inward towards the central body. – uhoh Nov 10 '20 at 23:57