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
Here's the information provided: • Radius of the Earth, R = 6400 km = 6.4 × 10^6 m • Gravitational field strength at the Earth's surface, g = 9.81 N/kg • Gravitational constant, G = 6.674 × 10^-11 N m^2/kg^2 (standard value) a) Calculate the mass of the Earth Step 1: The relationship between the gravitational field strength g, the mass of the Earth M, the radius of the Earth R, and the gravitational constant G is given by the formula: g = (GM)/(R^2) Step 2: To find the mass of the Earth M, we rearrange the formula: M = (gR^2)/(G) Step 3: Substitute the numerical values into the formula: M = 9.81 N/kg × (6.4 × 10^6 m)^26.674 × 10^-11 N m^2/kg^2 M = 9.81 × (40.96 × 10^12)6.674 × 10^-11 kg M = 401.6256 × 10^126.674 × 10^-11 kg M ≈ 60.1776 × 10^23 kg M ≈ 6.02 × 10^24 kg The mass of the Earth is approximately 6.02 × 10^24 kg. b) Calculate the mean density of the Earth Step 1: The mean density is defined as the mass divided by the volume. For a sphere, the volume V is given by V = (4)/(3) R^3. = (M)/(V) = (M)/(4)3 R^3 Step 2: First, calculate the volume of the Earth using the radius R = 6.4 × 10^6 m: V = (4)/(3) (6.4 × 10^6 m)^3 V = (4)/(3) (262.144 × 10^18) m^3 V ≈ 1.098 × 10^21 m^3 Step 3: Now, substitute the mass M calculated previously and the volume V into the density formula: = 6.01776 × 10^24 kg1.098066 × 10^21 m^3 ≈ 5.480 × 10^3 kg/m^3 ≈ 5480 kg/m^3 The mean density of the Earth is approximately 5480 kg/m^3. Got more? Send 'em!