It has been known since ancient times that the Earth's axis of rotation precesses. That is, the axis itself rotates around the line perpendicular to the ecliptic plane, i.e., the plane containing the Earth's orbit around the Sun. The ancient Greek astronomer Hipparchus concluded that the annual angular displacement of the axis was approximately 45'' (arcseconds), which would imply that the period of axial precession is around 29000 years. Modern measurements indicate that the period is approximately 25800 years. In this problem, you are asked to investigate this phenomenon using Newtonian mechanics.
You may need the following constants:
The Sun and the Moon exert non-zero torques on the Earth because of the Earth's non-spherical shape, giving rise to its axial precession. The main reason behind the Earth's non-spherical shape is the centrifugal force caused by the Earth's rotation about its axis. The tectonic plates located on the Earth's surface have deformed over millions of years to minimise the stresses within them. Therefore, as an approximation, let us model the Earth as a large liquid droplet of uniform density whose shape is determined by centrifugal and gravitational forces. In this model, the Earth's surface is an oblate spheroid (ellipsoid of revolution) characterised by the polar radius $R_p$ and the equatorial radius $R_e$ (see Figure A.1).
The difference between the equatorial and polar radii of the Earth, $h_\textrm{max}=R_e-R_p $ is much smaller than the average radius $R=(R_e+R_p)/2$. Up to a dimensionless constant, the value of $h_\textrm{max}$ can be expressed in terms of the angular speed of the Earth's rotation $\omega$, its mass $M_E$ and average radius $R$ as
$$h_\textrm{max}\propto G^{-1} \omega^\beta M_E^\gamma R^\delta,$$
where $G$ is the gravitational constant, and $\beta$, $\gamma$ and $\delta$ are constant exponents.
Regardless of whether you were able to find $h_{\textrm{max}}$ in part A.2., use the empirical value $h_{\textrm{max}}=21~\textrm{km}$ in the following questions.
To see why the Sun exerts a non-zero torque (with respect to the centre of the Earth) on our planet, consider Figure B.1 below. The difference in distance from the Sun causes the gravitational force $F_1$ to be greater than its counterpart $F_2$.
The magnitude of this torque acting on the Earth varies continuously during the year. In the position shown in Figure B.1, the torque is maximal. A quarter of a year later, due to symmetry, the torque becomes zero. After half a year, it reaches the maximum again. Then after three-quarters of the year it is zero once again, and so on. Since the period of axial precession is much larger than one year, this time-dependent torque can be approximated by its one-year average.
To calculate the average torque exerted by the Sun on the Earth, let us first determine the time average of the gravitational field generated by the Sun in the vicinity of the Earth. This average can be calculated as the field of a uniformly dense ring, a Sun ring, whose mass equals the mass of the Sun $M_S$ and whose radius equals the average distance between the Sun and the Earth $d_{SE}$ (see Figure B.2).
Let our cylindrical coordinate system have the origin at the centre of the Earth, and let the $z$ axis be perpendicular to the ecliptic plane (i.e. the plane of the ring). The axis of rotation of the Earth makes an angle of $\alpha=23.5^\circ$ with the $z$ axis.
In this section, you are asked to determine the torque exerted on the Earth due to the gravitational field. For simplicity, consider the Earth as a rigid body with a homogeneous mass distribution. Let us model the rotational ellipsoid as a sphere of radius $R_e$ (the equatorial radius of Earth), with the excess regions at the poles removed (see Figure C.1).
It can be shown that the torque acting on the excess regions is equivalent to the torque acting on two point masses, each with a mass equal to $2m/5$, positioned at the endpoints $A$ and $B$ of the polar diameter (see Figure C.1).
The Earth's axis of rotation moves very slowly around the $z$ axis in a conical motion. That is, it precesses.
The value obtained in Part D is much larger than the observed value. The reason for this is that so far we have only considered the torque exerted by the Sun, and neglected the effect of the Moon. In the following calculations, assume that the Moon's orbit is in the ecliptic plane, and that the orbit of the Moon around the Earth is a circle of radius $d_{ME}$. Let us denote the mass of the Moon by $M_M$ and the period of precession in this modified model by $T_2$.