Let's work through the four parts of the problem based on the schematic of the planet-moon system orbiting the star.

Problem Interpretation:

• Objects:
  • $m_p$: Mass of the planet.
  • $m_m$: Mass of the moon.
  • $M_s$: Mass of the star.
• Position Vectors:
  • $\vec{r}_p$: Position vector from the star to the planet.
  • $\vec{r}_m$: Position vector from the star to the moon.
  • $\vec{r}_{cm}$: Position vector from the star to the center of mass of the planet-moon system.

(i) Force on the moon $m_m$ due to gravity of planet $m_p$:

The gravitational force exerted by the planet $m_p$ on the moon $m_m$ is given by Newton's law of gravitation:

$\vec{F}_{m_m} = - \frac{G m_p m_m}{R^2} \hat{R}$ Where:

• $G$ is the gravitational constant,
• $R$ is the distance between the planet and the moon,
• $\hat{R}$ is the unit vector pointing from the moon to the planet.

(ii) Position vector of the center of mass $\vec{r}_{cm}$:

The center of mass of the planet-moon system is found using the following formula:

$\vec{r}_{cm} = \frac{m_p \vec{r}_p + m_m \vec{r}_m}{m_p + m_m}$ Here, $\vec{r}_p$ and $\vec{r}_m$ are the position vectors of the planet and moon from the star, respectively. This vector gives the location of the center of mass of the system with respect to the star.

(iii) Trajectory of the center of mass and linear velocity:

Since the center of mass of the system moves in a circular orbit around the star with constant angular velocity $\Omega$, the motion can be described as:

1. Trajectory: $\vec{r}_{cm}(t) = r_{cm} \left( \cos(\Omega t) \hat{i} + \sin(\Omega t) \hat{j} \right)$ Where $r_{cm}$ is the distance of the center of mass from the star, and $\Omega$ is the angular velocity.

2. Linear velocity: The linear velocity of the center of mass is given by: $v_{cm} = \Omega r_{cm}$ The velocity vector can be written as: $\vec{v}_{cm}(t) = -r_{cm} \Omega \sin(\Omega t) \hat{i} + r_{cm} \Omega \cos(\Omega t) \hat{j}$ This describes the velocity of the center of mass at any given time.

(iv) Angular momentum of the center of mass:

The angular momentum $\vec{L}_{cm}$ of the center of mass relative to the star can be expressed as:

$\vec{L}_{cm} = M_{tot} \vec{r}_{cm} \times \vec{v}_{cm}$ Where $M_{tot} = m_p + m_m$ is the total mass of the planet-moon system. Since the motion is circular and planar, the direction of the angular momentum will be perpendicular to the plane of motion (in the $\hat{k}$-direction) and its magnitude will be:

$L_{cm} = M_{tot} r_{cm}^2 \Omega$ The direction of the angular momentum is along the $z$-axis (out of the plane of motion), so:

$\vec{L}_{cm} = M_{tot} r_{cm}^2 \Omega \hat{k}$

Would you like more details on any of these steps? Here are five questions to explore further:

1. How does the gravitational force between two bodies vary with distance?
2. How can you interpret the trajectory equation of a body moving in a circular orbit?
3. What is the significance of the center of mass in a multi-body system?
4. How does angular momentum relate to the rotational motion of a system?
5. What is the physical meaning of the cross-product when computing angular momentum?

Tip: In orbital mechanics, the angular momentum of a body or a system remains conserved if no external torque is acting on it.