Final answer:
The radius of the Earth if compressed to the average nuclear density would be approximately 200 meters. This is calculated by using the Earth's mass and applying the volume formula of a sphere in conjunction with the nuclear density.
Step-by-step explanation:
The question asks what the radius of the Earth would be if its matter were compressed to the density of an atomic nucleus. The average nuclear density is given as 1.8 × 10⁴⁴ grams per cubic centimeter, which is equivalent to 1.8 × 10ⁱ⁷ kilograms per cubic meter. To find the radius of the Earth at this density, we must use the formula for the volume of a sphere (V = (4/3)πR³) and the fact that mass (m) equals density (ρ) times volume (V), thus m = ρV. The Earth's mass does not change, only its density does. With the given average density of the Earth (5.5 g/cm³), we can calculate its mass and then solve for the new radius using the compressed nuclear density.
Let's perform these calculations step by step:
- First, convert Earth's average density from g/cm³ to kg/m³:
5.5 g/cm³ = 5500 kg/m³ - Then, we calculate the Earth's mass using the current average radius (R) and density (ρ):
Mass of Earth (m) = ρ × Volume
m = 5500 kg/m³ × (4/3)π(6400000 m)³ - Using this mass, now we solve for the new radius (R') with the nuclear density (ρ'):
m = ρ' × Volume with nuclear density
m = 1.8 × 10ⁱ⁷ kg/m³ × (4/3)πR'³
R' = √[³√(m / (1.8 × 10ⁱ⁷ kg/m³) / ((4/3)π))]
After solving the above equations with appropriate values, we find that if the Earth were compressed to average nuclear density, its radius would be approximately 200 meters.