Use information on the endpapers of this book to calculate the average density of the Earth. Where does the value fit among those listed in Table 14.1 in Chapter 14? Look up the density of a typical surface rock like granite in another source and compare it with the density of the Earth
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Reference
The reference for this problem is "Serway & Jewett Physics For Scientists And Engineers With Modern Physics 9ed". This comprehensive textbook covers a wide range of physics topics, including mechanics, thermodynamics, electromagnetism, and modern physics. It is designed for advanced undergraduate science and engineering students, providing in-depth explanations and problem-solving techniques for various physical phenomena.
Question
1. (a) Use information on the endpapers of this book to calculate the average density of the Earth. (b) Where does the value fit among those listed in Table 14.1 in Chapter 14? Look up the density of a typical surface rock like granite in another source and compare it with the density of the Earth
Explain Question
This question is divided into two parts: a) Calculate the average density of the Earth using information provided in the book's endpapers. This requires finding the Earth's mass and volume, then using the density formula. b) Compare the calculated density to values in Table 14.1 of Chapter 14. Additionally, research the density of granite (a typical surface rock) from an external source and compare it to Earth's average density. To solve this, we need: 1. Earth's mass and radius (from the book's endpapers) 2. Formula for the volume of a sphere 3. Density formula 4. Table 14.1 from Chapter 14 5. Density of granite from an external source
Explain Solve
Step 1: Calculate Earth's volume - Use the formula for the volume of a sphere: V = (4/3)πr³ - Earth's radius (r) = 6.37 × 10⁶ m - V = (4/3)π(6.37 × 10⁶ m)³ = 1.08 × 10²¹ m³ Step 2: Calculate Earth's density - Use the density formula: ρ = m / V - Earth's mass (m) = 5.98 × 10²⁴ kg - ρ = (5.98 × 10²⁴ kg) / (1.08 × 10²¹ m³) = 5.52 × 10³ kg/m³ Step 3: Compare with Table 14.1 - The calculated density (5.52 × 10³ kg/m³) falls between the densities of aluminum and iron. Step 4: Compare with granite - Typical granite density: 2600-2700 kg/m³ - Earth's average density (5.52 × 10³ kg/m³) is significantly higher than granite, indicating denser materials in Earth's interior.
Summary
To solve this problem, we calculated Earth's volume using its radius, then used its mass to determine the average density. We found that Earth's density is 5.52 × 10³ kg/m³, which is between aluminum and iron densities. This value is much higher than typical surface rocks like granite, suggesting that Earth's interior contains denser materials.
Solve
P1.1 (a) Modeling the Earth as a sphere, we find its volume as 4/3 πr³ = 4/3 π(6.37 × 10⁶ m)³ = 1.08 × 10²¹ m³. Its density is then ρ = m/V = (5.98 × 10²⁴ kg) / (1.08 × 10²¹ m³) = 5.52 × 10³ kg/m³. (b) This value is intermediate between the tabulated densities of aluminum and iron. Typical rocks have densities around 2000 to 3000 kg/m³. The average density of the Earth is significantly higher, so higher-density material must be down below the surface.
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