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.
ScanSolve AI Answer
Step 1: Equate gravitational force to centripetal force for circular orbit. Gravitational force: F_g = (G M m)/(R^2) Centripetal force: F_c = (m v^2)/(R) Set F_g = F_c (both in N or kg · m/s^2): (G M m)/(R^2) = (m v^2)/(R) Cancel m: (G M)/(R^2) = (v^2)/(R) Multiply both sides by R: (G M)/(R) = v^2 (m^2/s^2) Step 2: Express orbital speed in terms of period T. v = (2 R)/(T) (m/s) Substitute into v^2: (G M)/(R) = ( (2 R)/(T) )^2 = (4 ^2 R^2)/(T^2) Step 3: Solve for R. Multiply both sides by R T^2: G M T^2 = 4 ^2 R^3 R^3 = (G M T^2)/(4 ^2) (m^3) R = ( (G M T^2)/(4 ^2) )^1/3 Step 4: Substitute numerical values (include units). Period: T = 1 day = 24 × 60 × 60 = 86400 s Earth's gravitational parameter: G M = 3.986 × 10^14 m^3 s^-2 Compute T^2: T^2 = (86400)^2 = 7.46496 × 10^9 \ s^2 Compute G M T^2: G M T^2 = 3.986 × 10^14 × 7.46496 × 10^9 = 2.9755 × 10^24 \ m^3 Compute 4 ^2: ^2 ≈ 9.8696, 4 ^2 ≈ 39.4784 Compute fraction: (G M T^2)/(4 ^2) = 2.9755 × 10^2439.4784 = 7.538 × 10^22 \ m^3 Compute cube root: R = [3]7.538 × 10^22 = 4.22 × 10^7 \ m Verification: (4.22 × 10^7)^3 = 4.22^3 × 10^21 ≈ 75.1 × 10^21 = 7.51 × 10^22 \ m^3 (matches closely). 4.22 × 10^7 \ m