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Here's the solution to your question: First, let's list the given values: Mass of satellite, m = 60 kg Gravitational field strength at Earth's surface, g = 9.8 N/kg Mean radius of the Earth, R = 6400 km = 6.4 × 10^6 m Distance of satellite from Earth's center, r_s = 5.7R a) The centripetal force acting on the satellite Step 1: Determine the formula for the gravitational force. The gravitational force provides the centripetal force for the satellite. The gravitational field strength g' at a distance r_s from the center of the Earth is related to the surface gravitational field strength g by the formula: g' = g ((R)/(r_s))^2 The centripetal force F_c acting on the satellite is then F_c = m g'. Substituting r_s = 5.7R: F_c = m · g ((R)/(5.7R))^2 = m · g ((1)/(5.7))^2 = (m g)/((5.7)^2) Step 2: Substitute the given values and calculate the centripetal force. F_c = 60 kg × 9.8 N/kg(5.7)^2 F_c = (588)/(32.49) F_c ≈ 18.09786 N Rounding to two significant figures: F_c ≈ 18 N b) The period of the satellite in orbit Step 1: Calculate the orbital radius of the satellite. r_s = 5.7R = 5.7 × (6.4 × 10^6 m) r_s = 3.648 × 10^7 m Step 2: Calculate the orbital velocity of the satellite. The centripetal force is also given by F_c = (m v^2)/(r_s). Equating this to the gravitational force m g': (m v^2)/(r_s) = m g' v^2 = g' r_s Substitute g' = g ((R)/(r_s))^2: v^2 = g ((R)/(r_s))^2 r_s = (g R^2)/(r_s) v = sqrt((g R^2)/(r_s)) Substitute the values: v = sqrt(9.8 m/s)^2 × (6.4 × 10^6 m)^23.648 × 10^7 m v = sqrt(9.8 × 40.96 × 10^12)3.648 × 10^7 v = sqrt(1.09999 × 10^7 m)^2/s^2 v ≈ 3316.6 m/s Step 3: Calculate the period of the satellite. The period T is the time taken for one complete orbit, given by T = (2 r_s)/(v). T = 2 × (3.648 × 10^7 m)3316.6 m/s T ≈ (229219440)/(3316.6) T ≈ 69110.8 s Rounding to two significant figures: T ≈ 6.9 × 10^4 s That's 2 down. 3 left today — send the next one.