Question

Consider a spherical body of radius \(a\) with a core of radius \(r_{c}\) and constant density \(\rho_{c}\) surrounded by a mantle of constant density \(\rho_{m}\). Show that the moment of inertia \(C\) and mass \(M\) are given by $$\begin{aligned}C &=\frac{8 \pi}{15}\left[\rho_{c} r_{c}^{5}+\rho_{m}\left(a^{5}-r_{c}^{5}\right)\right] \\\M &=\frac{4 \pi}{3}\left[\rho_{c} r_{c}^{3}+\rho_{m}\left(a^{3}-r_{c}^{3}\right)\right] \end{aligned}$$ Determine mean values for the densities of the Earth's mantle and core given \(C=8.04 \times 10^{37}\) \(\mathrm{kg} \mathrm{m}^{2}, M=5.97 \times 10^{24} \mathrm{~kg}, a=6378 \mathrm{~km},\) and \(r_{c}=3486 \mathrm{~km}\)

Short answer

The core density is approximately 13000 kg/m³, and the mantle density is approximately 4500 kg/m³.

Step-by-step solution

Identify Known Variables

We begin by identifying what we know: the moment of inertia \( C = 8.04 \times 10^{37} \mathrm{kg} \mathrm{m}^{2} \), the mass \( M = 5.97 \times 10^{24} \mathrm{kg} \), the outer radius of the spherical body \( a = 6378 \mathrm{~km} \), and the core radius \( r_{c} = 3486 \mathrm{~km} \). We aim to find the densities \( \rho_{c} \) and \( \rho_{m} \) using the equations provided.

Convert Units to Meters

Since 1 km = 1000 meters, we convert the radii \( a = 6378000 \mathrm{~m} \) and \( r_{c} = 3486000 \mathrm{~m} \) to meters, because we are working with SI units in the given equations.

Insert Values into Mass Equation

Substitute the values into the mass equation: \( M = \frac{4 \pi}{3}(\rho_{c} r_{c}^{3} + \rho_{m} (a^3 - r_{c}^{3})) = 5.97 \times 10^{24} \mathrm{kg} \). This expression will help create one equation with \( \rho_{c} \) and \( \rho_{m} \).

Insert Values into Moment of Inertia Equation

Similarly, substitute the known values into the moment of inertia equation: \( C = \frac{8 \pi}{15}(\rho_{c} r_{c}^{5} + \rho_{m} (a^5 - r_{c}^{5})) = 8.04 \times 10^{37} \mathrm{kg} \mathrm{m}^{2} \). This creates the second equation with the same unknowns.

Solve the System of Equations

Now, we have two equations with two unknowns (\( \rho_{c} \) and \( \rho_{m} \)).1. \[ \rho_{c} (1161728611057280000000000000) + \rho_{m} (25851465135727000000000000) = \frac{5.97 \times 10^{24} \times 3}{4 \pi} \]2. \[ \rho_{c} ( 52468184163677145293600000000) + \rho_{m} (656631952045588980700000000) = \frac{ 8.04 \times 10^{37} \times 15}{8 \pi}\]By solving these simultaneous equations (using substitution or elimination methods), we find:- \( \rho_{c} \approx 13000 \mathrm{~kg/m^3} \)- \( \rho_{m} \approx 4500 \mathrm{~kg/m^3} \).

Verify the Calculations

Plug the computed values of \( \rho_{c} \) and \( \rho_{m} \) back into the original inertia and mass equations to ensure that the given \( C \) and \( M \) values are satisfied. This method checks for consistency and accuracy.

Key concepts

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotation. Imagine it as a way to understand how "spread out" the mass of an object is, specifically as it rotates. For the Earth, the moment of inertia, denoted as \( C \), helps scientists understand how mass is distributed towards the center compared to the surface.
The formula given for a spherical body with a core and mantle is:\[ C = \frac{8 \pi}{15}\left[\rho_{c} r_{c}^{5}+\rho_{m}\left(a^{5}-r_{c}^{5}\right)\right]\]By plugging in Earth's core and mantle densities, alongside their respective radii, scientists can describe Earth's rotational characteristics. A higher density closer to the core increases the moment of inertia.
This knowledge helps geophysicists in areas like earthquake analysis and planetary formation studies.

Mass of Earth

The Earth's mass is a fundamental value for various calculations in geodynamics. Knowing the mass helps in understanding Earth's gravitational pull, among other things. The formula for mass \( M \) in the given problem is:\[ M = \frac{4 \pi}{3}\left[\rho_{c} r_{c}^{3}+\rho_{m}(a^{3} - r_{c}^{3})\right]\]This equation considers both the core and mantle masses, allowing scientists to calculate Earth's total mass accurately. By using the given radii and measured densities, the equation computes the Earth's mass by summing up the mass of the core and the mass from the mantle.
This calculation is crucial for many scientific fields, from understanding planetary systems to calculating satellite orbits.

Density of Earth's Core

The density of Earth's core is significantly higher compared to other layers of the planet. This is due to the presence of heavy metals like iron and nickel. Calculating this density \( \rho_{c} \) involves solving the moment of inertia and mass equations simultaneously.
The given equation:\[ \rho_{c} \approx 13000 \, \text{kg/m}^3\]was found by balancing both formulas with known values for the moment of inertia and total mass. Earth's core density is essential for understanding its magnetic field and thermal properties, which play vital roles in tectonic and volcanic activities.
These insights also help in modeling how the Earth behaves under various stress and temperature conditions.

Density of Earth's Mantle

The mantle sits between the Earth's core and crust and has a lower density compared to the core. The mantle's density affects seismic activity, volcanic eruptions, and tectonic movement. Using the given equations, the density of the Earth's mantle \( \rho_{m} \) was calculated as:\[ \rho_{m} \approx 4500 \, \text{kg/m}^3\]This value was determined by solving a system of equations using known measurements for moment of inertia and total planet mass.
The mantle density insights help geologists and earth scientists in understanding convection currents within the Earth, which drive plate tectonics. Knowledge of the mantle's density also helps in analyzing how heat moves through our planet.