# Step 1: Calculate the Orbital Speed The orbital speed \( v \) for a circular orbit is given by: \[ v = \sqrt{\frac{GM}{r}} \] where: - \( G = 6.674 \times 10^{-11} \) N·m²/kg² (gravitational constant) - \( M = 6.42 \times 10^{23} \) kg (mass of Mars) - \( r = \) orbital radius from Mars' center **Calculate \( r \):** - Mars radius = 3,390 km - Orbit altitude = 4,000 km - \( r = 3,390 + 4,000 = 7,390 \) km = \( 7,390,000 \) m **Plug in the values:** \[ v = \sqrt{\frac{6.674 \times 10^{-11} \times 6.42 \times 10^{23}}{7,390,000}} \] \[ v = \sqrt{\frac{4.283 \times 10^{13}}{7,390,000}} \] \[ v = \sqrt{5.796 \times 10^6} \] \[ v \approx 2,407 \text{ m/s} \] **Explanation:** We used Newton's law of gravitation to find the force that provides the centripetal acceleration for circular motion, leading to the formula for orbital speed. --- # Step 2: Calculate the Orbital Period The period \( T \) is the time for one full orbit: \[ T = \frac{2\pi r}{v} \] **Plug in the values:** \[ T = \frac{2\pi \times 7,390,000}{2,407} \] \[ T = \frac{46,409,000}{2,407} \] \[ T \approx 19,282 \text{ seconds} \] Convert seconds to hours: \[ \text{Hours} = \frac{19,282}{3,600} \approx 5.36 \text{ hours} \] **Explanation:** We divided the orbital circumference by the calculated speed to get the time for one full revolution and then converted seconds to hours. --- # Final Answers - **Orbital speed:** \( \boxed{2,407 \text{ m/s}} \) - **Orbital period:** \( \boxed{5.36 \text{ hours}} \)