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
To find the free fall acceleration on the surface of the moon, we use the formula for acceleration due to gravity: g = (GM)/(R^2) where: • G is the universal gravitational constant (6.674 × 10^-11 N m^2/kg^2) • M is the mass of the moon • R is the radius of the moon Step 1: List the given values. Mass of the moon, M = 7.40 × 10^22 kg Radius of the moon, R = 1.76 × 10^6 m Universal gravitational constant, G = 6.674 × 10^-11 N m^2/kg^2 Step 2: Substitute the values into the formula. g = (6.674 × 10^-11 N m^2/kg^2) × (7.40 × 10^22 kg)(1.76 × 10^6 m)^2 Step 3: Calculate the denominator. (1.76 × 10^6 m)^2 = (1.76)^2 × (10^6)^2 m^2 = 3.0976 × 10^12 m^2 Step 4: Perform the multiplication in the numerator. (6.674 × 10^-11) × (7.40 × 10^22) = (6.674 × 7.40) × (10^-11 × 10^22) = 49.3876 × 10^11 N m^2/kg Step 5: Divide the numerator by the denominator. g = 49.3876 × 10^11 N m^2/kg3.0976 × 10^12 m^2 g = (49.3876)/(3.0976) × 10^1110^12 m/s^2 g ≈ 15.9439 × 10^-1 m/s^2 g ≈ 1.59439 m/s^2 Step 6: Round to an appropriate number of significant figures (3 significant figures, based on the given data). g ≈ 1.59 m/s^2 The free fall acceleration on the surface of the moon is 1.59 m/s^2.