This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here's the solution to your problem: We are given the following information: Mass of the satellite, m = 80 kg Orbital radius from the center of the Earth, r = 7.5R, where R is the mean radius of the Earth. Gravitational field strength at the Earth's surface, g_0 = 9.8 N/kg Mean radius of the Earth, R = 6400 km First, let's convert R and r to meters: R = 6400 km = 6400 × 10^3 m = 6.4 × 10^6 m r = 7.5R = 7.5 × 6.4 × 10^6 m = 48 × 10^6 m = 4.8 × 10^7 m The gravitational field strength at the Earth's surface is g_0 = (GM_E)/(R^2), where G is the gravitational constant and M_E is the mass of the Earth. From this, we can express GM_E = g_0 R^2. a) Calculate the centripetal force acting on the satellite: Step 1: Understand the force. The centripetal force required to keep the satellite in orbit is provided by the gravitational force between the Earth and the satellite. So, the centripetal force F_c is equal to the gravitational force F_g. F_c = F_g = (GM_E m)/(r^2) Step 2: Substitute GM_E and the given values. Substitute GM_E = g_0 R^2 into the formula: F_c = (g_0 R^2 m)/(r^2) Now, substitute the numerical values: F_c = (9.8 N/kg) × (6.4 × 10^6 m)^2 × (80 kg)(4.8 × 10^7 m)^2 Step 3: Calculate the force. F_c = 9.8 × 80 × ((6.4 × 10^6)/(4.8 × 10^7))^2 F_c = 784 × ((6.4)/(48))^2 F_c = 784 × ((2)/(15))^2 F_c = 784 × (4)/(225) F_c = (3136)/(225) F_c ≈ 13.9377 N Rounding to two significant figures: 14 N b) Calculate the period of the satellite's orbit in hours: Step 1: Use the formula for the orbital period. The period T of a satellite in orbit is given by Kepler's Third Law: T = 2 sqrt((r^3)/(GM_E)) Step 2: Substitute GM_E and the given values. Substitute GM_E = g_0 R^2: T = 2 sqrt((r^3)/(g_0 R^2)) Now, substitute the numerical values: T = 2 sqrt((4.8 × 10^7 m))^3(9.8 m/s^2) × (6.4 × 10^6 m)^2 Step 3: Calculate the period in seconds. T = 2 sqrt((4.8)^3 × 10^21)9.8 × (6.4)^2 × 10^12 T = 2 sqrt(110.592 × 10^21)9.8 × 40.96 × 10^12 T = 2 sqrt(110.592 × 10^21)401.408 × 10^12 T = 2 sqrt(0.27551 × 10^9) T = 2 sqrt(2.7551 × 10^8) T ≈ 2 × 1.65985 × 10^4 T ≈ 104283 s Step 4: Convert the period from seconds to hours. There are 3600 seconds in an hour. T_hours = 104283 s3600 s/hour T_hours ≈ 28.9675 hours Rounding to two significant figures: 29 hours That's 2 down. 3 left today — send the next one.