Problem 81 - University Physics with Modern Physics

Question

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be \(15.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the center and \(2.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) at the surface. What is the acceleration due to gravity at the surface of this planet?

Step-by-Step Solution

Answer

The gravitational acceleration at the surface is approximately 9.44 m/s².

Step 1: Define the density function

Since the density decreases linearly from the center to the surface, we can define the density function as \( \rho(r) = \rho_c - k r \), where \( \rho_c = 15.0 \times 10^3 \text{ kg/m}^3 \) is the density at the center, and \( \rho_s = 2.0 \times 10^3 \text{ kg/m}^3 \) is the density at the surface. The planet's radius \( R = 6371 \text{ km} = 6.371 \times 10^6 \text{ m} \).

Step 2: Find the value of k

The density function \( \rho(r) = \rho_c - k r \) gives us \( \rho(R) = \rho_s = 2.0 \times 10^3 \text{ kg/m}^3 \). Substituting, we find: \( 15.0 \times 10^3 - k \times 6.371 \times 10^6 = 2.0 \times 10^3 \). Solving for \( k \), we get \( k = \frac{15.0 \times 10^3 - 2.0 \times 10^3}{6.371 \times 10^6} \approx 2.040 \times 10^{-3} \text{ kg/m}^4 \).

Step 3: Calculate mass of the planet

The mass \( M \) of a spherically symmetric object is given by integrating the density over its volume: \( M = \int_0^R \rho(r) 4\pi r^2 dr \). Substituting \( \rho(r) = 15.0 \times 10^3 - 2.040 \times 10^{-3} r \) into this integral, we get:\[ M = \int_0^{6.371 \times 10^6} \left( 15.0 \times 10^3 - 2.040 \times 10^{-3} r \right) 4 \pi r^2 \, dr \]This integral evaluates to approximately \( M \approx 6.12 \times 10^{24} \text{ kg} \).

Step 4: Calculate gravity at the surface

The gravitational acceleration at the surface \( g \) is given by \( g = \frac{GM}{R^2} \), where \( G = 6.674 \times 10^{-11} \text{ N(m/kg)}^2 \). Substitute \( M = 6.12 \times 10^{24} \text{ kg} \) and \( R = 6.371 \times 10^6 \text{ m} \):\[ g = \frac{6.674 \times 10^{-11} \times 6.12 \times 10^{24}}{(6.371 \times 10^6)^2} \approx 9.44 \text{ m/s}^2 \].

Key Concepts

Spherical Symmetry

Imagine a perfect ball, smooth and uniform in all directions; this is the principle of spherical symmetry. When dealing with cosmic bodies, such as planets, spherical symmetry implies that the planet’s properties only depend on the distance from its center, not on direction. This symmetry simplifies many calculations, as it assumes the planet has a smooth and consistent density variation, depending solely on the radial distance.

• In our exercise, the density at the center is maximum and decreases gradually as we move toward the surface, representing this symmetry mathematically.
• This concept allows us to describe the entire mass and gravitational fields of planets in straightforward equations, especially when applying Newton’s law of universal gravitation.

Through spherical symmetry, we can comprehend complex celestial phenomena in a more manageable way.

Density Function

The density function describes how matter is distributed within an object. In this problem, the planet's density decreases linearly from the center to its surface. This is represented as a linear function: \( \rho(r) = \rho_c - kr \), where

• \( \rho_c \) is the density at the center, and
• \( k \) is a constant derived from the change in density over the radius.

Understanding this function is key to modeling the planet's mass distribution.

Here’s why it’s significant:

• Linear functions are simple, relying on basic arithmetic operations, making them easy to work with mathematically.
• They provide a very realistic description of the physical situation within spherically symmetric bodies like planets.

The linearly decreasing density helps predict not just how much matter is at each point but also calculate overall mass and gravitational effects.

Integration

In the context of physics, integration helps us find the total amount of a quantity that is distributed throughout a space, such as mass in this case. To calculate the planet’s total mass, we need to integrate the density function over its entire volume.
The integral involved is: \[ M = \int_0^R \rho(r) 4\pi r^2 \, dr \] This equation calculates the planet's mass by summing up all tiny spherical shells of mass from the center to the radius \( R \).

• \( 4\pi r^2 \) describes the surface area of each shell, while \( \rho(r) \) denotes how dense the shell is at a given radius \( r \).
• This requires the concept of the integral as an “infinite sum” of infinitesimal parts of the planet.

By understanding integration, we effectively combine the density and spherical symmetry concepts into a mathematical framework to model physical systems like planets.

Gravitational Acceleration

Gravitational acceleration signifies the rate at which an object accelerates due to gravity’s pull. At a planet’s surface, this attraction is described by the formula:\[ g = \frac{GM}{R^2} \] where:

• \( G \) is the gravitational constant, a universal factor ensuring the right unit balance and strength of force.
• \( M \) is the total mass calculated by integrating the density function.
• \( R \) is the radius of the planet.

This acceleration is crucial in determining phenomena such as free-fall speeds and the weight of objects at the surface.
With this knowledge, one can understand how varying density profiles affect gravitational forces. It explains why different planets have different gravity strengths, which influences everything from atmosphere retention to potential human colonization viability.