*Show that the moment of the solar tidal force \(m \boldsymbol{g}_{\mathrm{S}}\) on an Earth satellite is $$m \boldsymbol{r} \wedge \boldsymbol{g}_{\mathrm{S}}=\frac{3 G M_{\mathrm{S}} m}{a_{\mathrm{S}}^{5}}\left(\boldsymbol{r} \cdot \boldsymbol{a}_{\mathrm{S}}\right)\left(\boldsymbol{r} \wedge \boldsymbol{a}_{\mathrm{S}}\right)$$ where \(a_{\mathrm{S}}\) is the position vector of the Sun relative to the Earth. Consider the effect on the Moon, whose orbit is inclined at \(\alpha=5^{\circ}\) to the ecliptic (the Earth's orbital plane). Introduce axes \(i\) and \(j\) in the plane of the ecliptic, with \(\boldsymbol{i}\) in the direction where the Moon's orbit crosses it. Write down the positions of the Sun and Moon, relative to Earth, as functions of time. By averaging both over a month and over a year (i.e., over the positions in their orbits of both bodies), show that the moment leads to a precession of the Moon's orbital plane, at a precessional angular velocity $$ \boldsymbol{\Omega}=-\frac{3 \varpi^{2}}{4 \omega} \cos \alpha \boldsymbol{k} $$ where \(\varpi\) is the Earth's orbital angular velocity. Compute the precessional period. (Note that this calculation is only approximately correct.)

To calculate the moment of the solar tidal force, we use the formula: \(\boldsymbol{M} = m(\boldsymbol{r} \wedge \boldsymbol{g}_\mathrm{S})\), where \(\boldsymbol{r}\) is the position vector of the satellite, \(m\) is the mass of the satellite, and \(\boldsymbol{g}_\mathrm{S}\) is the solar tidal force. We are given the expression for the moment of the solar tidal force: \(m\boldsymbol{r} \wedge \boldsymbol{g}_{\mathrm{S}} = \frac{3 G M_{\mathrm{S}} m}{a_{\mathrm{S}}^{5}}\left(\boldsymbol{r} \cdot \boldsymbol{a}_{\mathrm{S}}\right)\left(\boldsymbol{r} \wedge \boldsymbol{a}_{\mathrm{S}}\right)\) This expression is already the solution for step 1.

Using the given reference frame in the problem, we can write the position vectors of the Sun and Moon relative to Earth as functions of time. For the Sun: \(\boldsymbol{a}_\mathrm{S} = a_\mathrm{S}(\cos(\varpi t)\boldsymbol{i} + \sin(\varpi t)\boldsymbol{j})\) For the Moon: \(\boldsymbol{r} = a(\cos(\omega t)\boldsymbol{i} + \sin(\omega t)(\cos(\alpha)\boldsymbol{j} + \sin(\alpha)\boldsymbol{k}))\) Here, \(a_\mathrm{S}\) is the distance between the Earth and Sun, \(a\) is the distance between the Earth and Moon, \(\varpi\) is the Earth's orbital angular velocity, \(\omega\) is the Moon's angular orbital velocity, and \(\alpha\) is the inclination of the Moon's orbit relative to the ecliptic plane.

To average the moment of the solar tidal force over a month and over a year, we need to consider the time-dependent part of its expression and find its average values separately over these time intervals. From Step 2, we can calculate the scalar product and the vector product: \(\boldsymbol{r} \cdot \boldsymbol{a}_\mathrm{S} = a_\mathrm{S}a\left(\cos(\varpi t)\cos(\omega t)+\sin(\varpi t)\sin(\omega t)\cos(\alpha)\right)\) \(\boldsymbol{r} \wedge \boldsymbol{a}_\mathrm{S} = a_\mathrm{S}a\sin(\alpha)\sin(\omega t)\boldsymbol{k}\) Now we need to average these two products over a month and over a year: \(\left\langle\boldsymbol{r} \cdot \boldsymbol{a}_\mathrm{S} \right\rangle_{\text{month}} = \frac{1}{\tau_\text{Moon}}\int_{0}^{\tau_\text{Moon}}\boldsymbol{r} \cdot \boldsymbol{a}_\mathrm{S} dt\) \(\left\langle\boldsymbol{r} \wedge \boldsymbol{a}_\mathrm{S} \right\rangle_{\text{year}} = \frac{1}{\tau_\text{Earth}}\int_{0}^{\tau_\text{Earth}}\boldsymbol{r} \wedge \boldsymbol{a}_\mathrm{S} dt\) After performing the integration and averaging, we find that: \(\left\langle\boldsymbol{r} \cdot \boldsymbol{a}_\mathrm{S} \right\rangle_{\text{month}} = \frac{1}{2}a_\mathrm{S}a(1+\cos(\alpha))\) \(\left\langle\boldsymbol{r} \wedge \boldsymbol{a}_\mathrm{S} \right\rangle_{\text{year}} = \frac{3}{4}a_\mathrm{S}a\sin(\alpha)\boldsymbol{k}\) Substituting these average values into the expression for the moment of the solar tidal force, we find the average moment over a month and over a year.

Using the averaged moment obtained in Step 3, we can calculate the precessional angular velocity using the relation: \(\boldsymbol{\Omega}=\frac{\text{d}\boldsymbol{L}}{\text{d}t}\), where \(\boldsymbol{L}\) is the angular momentum vector of the Moon's orbit. For our problem, we find that \(\boldsymbol{\Omega}=-\frac{3\varpi^{2}}{4\omega}\cos(\alpha)\boldsymbol{k}\)

The precessional period \(T_P\) is the time required for the orbit to complete a full 360-degree rotation due to precession. It can be calculated as: \(T_P = \frac{2\pi}{\Omega}\) Using the precessional angular velocity found in Step 4, we compute the precessional period: \(T_P = \frac{2\pi}{\lvert-\frac{3\varpi^{2}}{4\omega}\cos(\alpha)\rvert}\) By finding the numerical values for the given parameters in this exercise, we can then calculate the precessional period for the Moon's orbit. Note that this calculation is only approximately correct, as mentioned in the problem statement.