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Here's the information provided: • Mass of the satellite, m = 80 kg • Orbital radius, r = 5R, where R is the mean radius of the Earth. • Gravitational field strength at the Earth's surface, g = 9.81 N/kg • Radius of the Earth, R = 6400 km = 6.4 × 10^6 m a) Calculate the centripetal force Step 1: The centripetal force required to keep the satellite in orbit is provided by the gravitational force. The gravitational force is given by Newton's Law of Universal Gravitation: F_g = (GMm)/(r^2) where G is the gravitational constant and M is the mass of the Earth. Step 2: We know that the gravitational field strength at the Earth's surface is g = (GM)/(R^2). From this, we can express GM as GM = gR^2. Step 3: Substitute this expression for GM into the gravitational force formula. Since the centripetal force F_c is equal to the gravitational force F_g: F_c = ((gR^2)m)/(r^2) Given that the orbital radius is r = 5R, substitute this value: F_c = (gR^2m)/((5R)^2) F_c = (gR^2m)/(25R^2) The R^2 terms cancel out: F_c = (gm)/(25) Step 4: Now, substitute the numerical values: F_c = 9.81 N/kg × 80 kg25 F_c = (784.8)/(25) F_c = 31.392 N The centripetal force is 31.392 N. b) Calculate the period of the satellite in orbit Step 1: For a satellite in a circular orbit, the centripetal force is equal to the gravitational force. The centripetal force can also be expressed as F_c = (mv^2)/(r), where v is the orbital speed of the satellite. So, we have: (mv^2)/(r) = (GMm)/(r^2) Simplifying m and r: v^2 = (GM)/(r) Step 2: The orbital speed v is also related to the period T by the relation v = (2 r)/(T). Squaring this gives: v^2 = (4^2 r^2)/(T^2) Step 3: Equate the two expressions for v^2: (4^2 r^2)/(T^2) = (GM)/(r) Rearrange to solve for T^2: T^2 = (4^2 r^3)/(GM) Step 4: Substitute GM = gR^2 and r = 5R into the equation for T^2: T^2 = (4^2 (5R)^3)/(gR^2) T^2 = (4^2 (125R^3))/(gR^2) T^2 = (500^2 R)/(g) Step 5: Take the square root to find T: T = sqrt((500^2 R)/(g)) T = sqrt((500R)/(g)) Step 6: Now, substitute the numerical values: T = sqrt(500 × 6.4 × 10^6 m)9.81 N/kg T = sqrt((3200 × 10^6)/(9.81)) T = sqrt(326197757.39) T = × 18060.9456 T ≈ 56797.8 s The period of the satellite is approximately 56797.8 s. 3 done, 2 left today. You're making progress.