Let's solve each part in two steps with explanations: --- ## (a) Gravitational Force \( F \) ### **Step 1: Write down Newton's Law of Gravitation** \[ F = \frac{G m_1 m_2}{r^2} \] Here, both satellites have mass \( M \), so \( m_1 = m_2 = M \), and they are separated by distance \( 2R \) (since each is a radius \( R \) away from the center, so from each other, it's \( 2R \)). **Explanation:** We set \( r = 2R \) because the satellites are on opposite sides of the center of mass. --- ### **Step 2: Substitute values** \[ F = \frac{G M M}{(2R)^2} = \frac{G M^2}{4R^2} \] **Explanation:** Plugging in the masses and distance gives the gravitational force between the two satellites. --- **Final Answer (a):** \[ \boxed{F = \frac{G M^2}{4R^2}} \] --- ## (b) Tangential Speed \( v \) ### **Step 1: Set gravitational force as centripetal force** For circular motion, centripetal force is provided by gravity: \[ F = \frac{M v^2}{R} \] We already found \( F = \frac{G M^2}{4R^2} \). **Explanation:** For each satellite, the gravitational force acts as the centripetal force keeping it in orbit. --- ### **Step 2: Solve for \( v \)** \[ \frac{G M^2}{4R^2} = \frac{M v^2}{R} \] \[ \frac{G M}{4R} = v^2 \] \[ v = \sqrt{\frac{G M}{4R}} \] **Explanation:** We solved for \( v \) by equating the two forces isolating \( v \). --- **Final Answer (b):** \[ \boxed{v = \sqrt{\{G M}{4R}}} \] --- Let me know if you want the remaining parts (c), (d), and (e)!

Sure! Let's continue solving the remaining parts of the question. --- ## (c) Orbital Period \( T \) ### **Step 1: Use the formula for orbital period** The orbital period for circular motion is given by: \[ T = \frac{2\pi R}{v} \] **Explanation:** This formula relates the distance traveled in one complete orbit to the tangential speed. --- ### **Step 2: Substitute \( v \)** From part (b), we found \( v = \sqrt{\frac{G M}{4R}} \). Substituting this into the orbital period formula gives: \[ T = \frac{2\pi R}{\sqrt{\frac{G M}{4R}}} \] This simplifies to: \[ T = 2\pi R \cdot \sqrt{\frac{4R}{G M}} = 2\pi \sqrt{\frac{4R^2}{G M}} = 4\pi \sqrt{\frac{R^2}{G M}} \] **Explanation:** We simplified the expression to show the relationship between the orbital period, the radius, and the mass. --- **Final Answer (c):** \[ \boxed{T = 4\pi \sqrt{\frac{R^2}{G M}}} \] --- ## (d) Total Work \( W \) ### **Step 1: Understand the work done by gravitational force** The work done by gravitational force in a closed orbit is zero because the force is always perpendicular to the displacement. **Explanation:** In circular motion, the gravitational force acts towards the center, while the displacement is tangential to the orbit. Thus, no work is done. --- ### **Step 2: Conclude the work done** \[ W = 0 \] **Explanation:** Since the gravitational force does no work during one complete orbit, the total work done is zero. --- **Final Answer (d):** \[ \boxed{W = 0} \] --- ## (e) Total Angular Momentum \( L \) ### **Step 1: Use the formula for angular momentum** The angular momentum \( L \) for a single satellite is given by: \[ L = M v R \] **Explanation:** This formula considers the mass, tangential speed, and radius of the satellite's orbit. --- ### **Step 2: Calculate total angular momentum for both satellites** Since there are two identical satellites: \[ L_{total} = 2 \cdot M v R \] Substituting \( v \) from part (b): \[ L_{total} = 2 M \left( \sqrt{\frac{G M}{4R}} \right) R = 2 M R \sqrt{\frac{G M}{4R}} = \sqrt{G M} \cdot 2 \sqrt{R^3} \] **Explanation:** We simplified the expression to find the total angular momentum considering both satellites. --- **Final Answer (e):** \[ \boxed{L_{total} = 2 \sqrt{G M} R^{3/2}} \] --- This concludes the problem with all parts solved. If you have further questions, feel free to ask!