Problem 103 - Thomas Calculus

Question

Density of center of a planet \(A\) planet is in the shape of a sphere of radius \(R\) and total mass \(M\) with spherically symmetric density distribution that increases linearly as one approaches its center. What is the density at the center of this planet if the density at its edge (surface) is taken to be zero?

Step-by-Step Solution

Answer

The density at the center is \( \frac{3M}{\pi R^3} \).

Step 1: Understanding the Problem

We need to find the density at the center of a spherical planet given a specific density distribution. The density is zero at the surface (radius = R) and increases linearly towards the center.

Step 2: Mathematical Representation of Density

Since the density increases linearly from the surface to the center, we'll express the density function as \( \rho(r) = \alpha (R - r) \), where \( \rho(r) \) is the density at a radius \( r \), and \( \alpha \) is a constant we'll determine.

Step 3: Find Total Mass of the Sphere

The total mass \( M \) of the sphere is given by integrating the density over the volume of the sphere: \[ M = \int_0^R \rho(r) \cdot 4\pi r^2 \, dr = 4\pi \alpha \int_0^R (R - r) r^2 \, dr. \]

Step 4: Solving the Integral

Calculate the integral from the previous step:\[ \int_0^R (R - r) r^2 \, dr = \int_0^R (Rr^2 - r^3) \, dr. \]This gives:\[ = \left[ \frac{Rr^3}{3} - \frac{r^4}{4} \right]_0^R = \frac{R^4}{3} - \frac{R^4}{4} = \frac{R^4}{12}. \]

Step 5: Setting the Expression for Total Mass

Substitute the integral into the expression for mass:\[ M = 4\pi \alpha \cdot \frac{R^4}{12} = \frac{\pi \alpha R^4}{3} \].Now, solve for \( \alpha \):\[ \alpha = \frac{3M}{\pi R^4}. \]

Step 6: Find Density at the Center

Substitute \( r = 0 \) in the density function, as the center is at \( r = 0 \):\[ \rho(0) = \alpha R = \frac{3M}{\pi R^4} \cdot R = \frac{3M}{\pi R^3}. \]

Step 7: Conclusion

Hence, the density at the center of the planet can be expressed as \( \frac{3M}{\pi R^3} \).

Key Concepts

Spherical Symmetry

Spherical symmetry is a concept where a distribution looks the same when viewed from any direction around a central point. In our context, it represents how the density of a planet changes with distance from its center. This planet, being a sphere, has a density that varies only with radius and not with angles. When a body has spherical symmetry, it simplifies calculations related to mass and other properties. Only one variable, the radius, needs to be considered.
This symmetry allows us to use radial functions to describe properties like mass and density. The simplicity provided by spherical symmetry is especially useful in problems that involve integrating over the volume of a sphere. Essentially, it's much easier to calculate when you're only dealing with changes that depend on how far away something is from the center, rather than also needing to consider different directions.

Linear Density Function

A linear density function means that the density changes at a constant rate as you move from the edge of the sphere to its center. In this problem, the density function is expressed as \( \rho(r) = \alpha (R - r) \). Here, \( R \) is the radius of the sphere, \( r \) is the distance from the center, and \( \alpha \) is a constant that scales the density. This form of the function indicates that density decreases linearly as you move outward from the center to the edge.
The function starts with a maximum density at the center (when \( r = 0 \)) and diminishes until it reaches zero at the surface (when \( r = R \)). The linear property of this function simplifies calculations, allowing us to use straightforward algebra to determine the distribution throughout the planet.

Mass Integration

Mass integration involves calculating the total mass of an object by integrating its density over its volume. To find the total mass \( M \) of the sphere, we use the density function and integrate this over the volume of the sphere. The volume integration for a spherical body involves the expression \( V = \int_0^R \rho(r) \, 4\pi r^2 \ dr \), where \( 4\pi r^2 \) represents the surface area of a spherical shell.
By integrating the density function over the volume, we effectively sum up tiny mass elements spread throughout the volume of the sphere. This integral provides us with the total mass, which is a fundamental property needed to determine the density constant \( \alpha \) and further calculate other quantities of interest, such as density at specific points.

Integral Calculus

Integral calculus is a branch of mathematics used to find quantities like mass, area, and volume by breaking them down into infinitesimally small parts and summing them. In this problem, we use integral calculus to determine the total mass of a spherical object using its density distribution.
Performing the integral \( \int_0^R (R - r) r^2 \ dr \) involves finding an antiderivative, which, when evaluated, gives the net "sum" of all the infinitesimal mass contributions within the sphere. For our density problem, integrating the function \( \rho(r) = \alpha (R - r) \) with respect to the radius involves solving the integral of a polynomial. The resulting expression plays a crucial role in further calculations to deduce physical quantities such as the constant \( \alpha \) and the central density.