The International Space Station orbits at a height of 400 km above Earth's surface. Estimate its orbital velocity. (R_earth = 6.37 × 10⁶ m, g = 9.8 m/s²) Step 1: Orbital radius r = R + h = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m. Step 2: Use v_o = √(GM/r). Since GM = gR², v_o = √(gR²/r) = R√(g/r). Step 3: v_o = 6.37 × 10⁶ × √(9.8 / 6.77 × 10⁶) = 6.37 × 10⁶ × √(1.447 × 10⁻⁶) = 6.37 × 10⁶ × 1.203 × 10⁻³ ≈ 7,660 m/s. Answer: ≈ 7.66 km/s. Its period is T = 2πr/v ≈ 2π × 6.77 × 10⁶ / 7660 ≈ 5550 s ≈ 93 minutes — about 16 sunrises per day for the astronauts.