Question

In the motion called precession of the equinoxes, the Earth's axis of rotation precesses about the perpendicular to its orbital plane with a period of 2.58 104 yr. How would you calculate the torque on the Earth that is causing this precession? ?

The angular momentum vector of a precessing gyroscope sweeps out a cone as shown in the figure below. The angular speed of the tip of the angular momentum vector, called its precessional frequency, is given by `ωp = τ/L`, where `τ` is the magnitude of the torque on the gyroscope and `L` is the magnitude of its angular momentum.

Answer 1

Torque :
`\tau = (dL)/(dt) = 2/5 MR^2\sin\delta(2\pi)/T_s.(2\pi)/T_p=(8\pi^2)/5\frac{MR^2\sin\delta}{T_sT_p}`
`=2.166\times10^{22}\quad N.m`

Explanation:

Torque is defined as the rate at which the angular momentum vector changes.
`\vec{\tau_{}} \equiv \frac{d\vec{L_{}}}{dt}`

In the case of Earth's gyration, the angular momentum's magnitude remains the same and only its direction changes due to the torque.

The angular momentum vector `\vec{L_{}}` can be resolved into a longitudinal component (parallel to precession axis) and a transverse component (perpendicular to the precession axis). Of these two components the longitudinal component is a constant while the transverse component is changing with time.
`\vec{L}(t) = \vec{L_l} + \vec{L_{t}}(t); \qquad \frac{d\vec{L_{}}}{dt} = \frac{d\vec{L_t}}{dt}`

If `\delta` is Earth's axial tilt, that would be the apex angle of the precession cone. Magnitude of the longitudinal and transverse components of the angular momentum vector can then be written as :
`L_l = L\cos\delta; \qquad L_t = L\sin\delta;`

If `\vec{L_t}` sweep an infinitesimal angle `d\phi` in an infinitesimal time `dt` on the precession plane (base of precession cone), then
`dL_t = L_td\phi; \qquad \frac{dL_t}{dt}= L_t(d\phi)/(dt)= L_t\omega_p`,
where `\omega_p` is the angular frequency of precession.

So the magnitude of the torque is :
`\tau = (dL)/(dt) = L_t\omega_p = L\sin\delta\omega_p`;
One can write the spin angular momentum in terms of the moment-of-inertia (`I`) and the angular frequency of spin (`\omega_s`) as
`L=I\omega_s`. If we assume the Earth to be a homogeneous sphere of mass `M` and radius `R`, then `I = 2/5 MR^2`

`\tau = (2/5MR^2\omega_s)\sin\delta\omega_p = 2/5MR^2(\sin\delta)\omega_p\omega_s`

The angular frequencies of spin and precession can be written in terms of the the spin time period and precession time period as :
`\omega_s = (2\pi)/T_s; \qquad \omega_p=(2\pi)/T_p;`

`\tau = 2/5 MR^2\sin\delta(2\pi)/T_s.(2\pi)/T_p=(8\pi^2)/5\frac{MR^2\sin\delta}{T_sT_p}`

Numerical Evaluation:
`\delta = 23.44^o; \qquad T_s=1 day=8.64\times10^4\quad sec;`
`T_p = 2.58\times10^4 \quad Yrs = 8.136\times10^{11}\quad s`
`M=5.972\times10^{24}\quad kg; \qquad R=6.371\times10^6\quad km`
`\tau = 2.166\times10^{22}\quad N.m`