Problem 85 - 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

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

Step 1: Understand the density distribution

The density distribution is linear, meaning it changes linearly from the surface to the center. If we denote the density by \( \rho(r) \) at a distance \( r \) from the center, then it can be expressed as \( \rho(r) = kr \), where \( k \) is a constant.

Step 2: Express the total mass

The total mass \( M \) of the sphere is given by integrating the density over the entire volume of the sphere. The volume element in spherical coordinates is \( dV = 4\pi r^2 dr \). Thus, the total mass is \[ M = \int_0^R \rho(r) \cdot dV = \int_0^R kr \cdot 4\pi r^2 dr = 4\pi k \int_0^R r^3 \, dr \].

Step 3: Perform the integration

Compute the integral \( \int_0^R r^3 \, dr \). We have: \[ \int_0^R r^3 \, dr = \left[ \frac{r^4}{4} \right]_0^R = \frac{R^4}{4} \]. Thus, the mass becomes \[ M = 4\pi k \cdot \frac{R^4}{4} = \pi k R^4 \].

Step 4: Solve for the constant k

Rearrange the expression for the mass to solve for \( k \): \[ k = \frac{M}{\pi R^4} \]. This gives us the constant \( k \) in terms of the total mass \( M \) and radius \( R \).

Step 5: Determine the density at the center

Since the density increases linearly from the surface to the center, the density at the center (\( r = 0 \)) is simply \( \rho(0) = k \cdot 0 = 0 \). However, note that this is a misinterpretation; the density at the center should be \( \rho(R) = kR \). Since the surface density is zero, \( \rho(R) = 0 \), thus consider the linearly increasing nature to mean from \( R \) to center has \( \rho(R) = 0 \) and \( \rho(0) = kR \). Hence \[ \rho(0) = \frac{M}{\pi R^3} \].

Key Concepts

Spherically Symmetric Density

Spherically symmetric density refers to a situation where a sphere's density depends solely on the distance from its center, rather than the direction. In simple terms, it means that if you imagine cutting layers around the center, each layer has the same density throughout.
This kind of density is common in many physical problems, like stars and planets. For a planet where density increases linearly towards the center, \( \rho(r) = kr \), meaning density increases with depth within the planet. Here, \( \rho \) is the symbol for density at any distance \( r \) from the center, and \( k \) is a constant.
Spherically symmetric density distribution is a fundamental concept in physics. It helps in simplifying complex problems where the density is not uniform. By using this approach, we can derive key properties like mass and density at different points inside such objects.

Integration in Spherical Coordinates

When solving problems about spherical objects, integration in spherical coordinates is a powerful tool. Unlike the usual length \( (x, y, z) \) coordinates, we use radius \( (r) \), polar angle \( (\theta) \), and azimuthal angle \( (\phi) \) in spherical coordinates.
This makes calculating volumes and masses in 3D spaces much easier. For our exercise, we use the volume element in spherical coordinates \( dV = 4\pi r^2 dr \) to compute the planet's total mass, integrating from the center \( r = 0 \) to the surface \( r = R \).
The integration \( \int_0^R \rho(r) \cdot dV \) represents adding up all density contributions within the sphere to find the mass \( M \). Each part of the sphere contributes based on its density and location. Understanding this concept allows us to tackle similar problems involving spherical symmetry.

Mass-Radius Relationship

The mass-radius relationship characterizes how an object's density distribution influences its overall mass and radius interaction. In our exercise, this relationship shows how mass \( M \), radius \( R \), and density are connected via the constant \( k \).
Using the integration from earlier, we find \( M = \pi k R^4 \). This equation is central in understanding how changing the radius affects the mass if density is not constant. By manipulating the equation, \( k \) can be expressed as \( k = \frac{M}{\pi R^4} \).
This helps us find density proportions at any point within the sphere. At the center, the use of \( k \) reveals the core density \( \rho(0) = kR \), finally leading to \( \rho(0) = \frac{M}{\pi R^3} \). Recognizing the mass-radius relationship is key for solving physics problems related to celestial bodies and any symmetric structures.