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Mobile App Chapter 9: Problem 12
Suppose the density of a star is given by.
$$\rho(r)=\left\\{\begin{array}{ll}
\rho_{0} & r
Short Answer
Expert verified
The expression for \(M(r)\) is the sum of the masses of the two regions. The value of \(\rho_{0}\) can be found using the given mass and radius.
Step by step solution
01
- Understand the density function
The density function \(\rho(r)\) has three regions: \(\rho(r) = \rho_{0}\) for \(r < r_{0} \), \(\rho(r) = \rho_{0} (r_{0} / r)^{2}\) for \(r_{0} < r < R\), and \(\rho(r) = 0\) for \(r > R\).
02
- Calculate mass for region 1 (\(r < r_{0}\))
The mass \(M_{1}\) contained within radius \(r_{0}\) can be found by integrating the density over the volume: \[M_{1} = \int_{0}^{r_{0}} 4 \pi r^{2} \rho_{0} \, dr = 4 \pi \rho_{0} \int_{0}^{r_{0}} r^{2} \, dr = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} \right]_{0}^{r_{0}} = \frac{4 \pi \rho_{0} r_{0}^{3}}{3} \]
03
- Calculate mass for region 2 (\(r_{0} < r < R\))
The mass \(M_{2}\) for \(r_{0} < r < R\) can be found by integrating the density over the volume: \[M_{2} = \int_{r_{0}}^{R} 4 \pi r^{2} \rho_{0} \(\frac{r_{0}}{r}\)^{2} \, dr = 4 \pi \rho_{0} r_{0}^{2} \int_{r_{0}}^{R} \frac{1}{r} \, dr = 4 \pi \rho_{0} r_{0}^{2} \left[ \ln(r) \right]_{r_{0}}^{R} = 4 \pi \rho_{0} r_{0}^{2} \ln \left( \frac{R}{r_{0}} \right) \]
04
- Find the total mass \(M\)
The total mass \(M\) of the star is the sum of \(M_{1}\) and \(M_{2}\): \[M = \frac{4 \pi \rho_{0} r_{0}^{3}}{3} + 4 \pi \rho_{0} r_{0}^{2} \ln \left( \frac{R}{r_{0}} \right) \]
05
- Plug in the given values
Using \(M = M_{\odot}\), \(R = R_{\odot}\), and \(r_{0} = 0.1 R_{\odot}\), substitute into the expression for \(M\) yields: \[M_{\odot} = 4 \pi \rho_{0} \left( \frac{(0.1 R_{\odot})^{3}}{3} + (0.1 R_{\odot})^{2} \ln \left( \frac{R_{\odot}}{0.1 R_{\odot}} \right) \right) \]
06
- Simplify the expression
Simplify the expression to solve for \(\rho_{0}\): \[M_{\odot} = 4 \pi \rho_{0} \left( \frac{(0.1 R_{\odot})^{3}}{3} + (0.1 R_{\odot})^{2} \ln (10) \right) \]i.e. \[M_{\odot} = 4 \pi \rho_{0} \left( \frac{0.001 R_{\odot}^{3}}{3} + 0.01 R_{\odot}^{2} \ln (10) \right) \]
07
- Solve for \(\rho_{0}\)
Re-arranging the expression to isolate \(\rho_{0}\): \[\rho_{0} = \frac{M_{\odot}}{4 \pi \left( \frac{0.001 R_{\odot}^{3}}{3} + 0.01 R_{\odot}^{2} \ln (10) \right)} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stellar Density Profile
The stellar density profile describes how the density of a star varies with distance from its center. In the given problem, the density function \(\rho(r)\) is composed of three distinct regions. For distances less than \(r_0\), the density is constant at \(\rho_0\). Between \(r_0\) and \(R\), the density decreases with the square of the distance. Beyond \(R\), the density is zero. This type of varying density is common in astrophysical models as stars are not homogenous spheres. Understanding how density changes within a star is crucial for calculating its mass and understanding its structure.
Mass Integration
To find the mass of the star, we need to integrate the density function over the star's volume. This process, known as mass integration, involves adding up the small amounts of mass contributed by each tiny volume within the star. For \(r < r_0\), the integration yields \(M_1 = \frac{4\pi \rho_0 r_0^3}{3}\). For the region \(r_0 < r < R\), the mass is found by integrating the density function: \(M_2 = 4 \pi \rho_0 r_0^2 \ln (\frac{R}{r_0})\). The total mass is then the sum of these two integrals: \(M = M_1 + M_2\). Accurate integration is essential to determine the mass distribution within the star.
Polytropic Models
Polytropic models are a useful way of describing stellar structures. In these models, the pressure and density of a star are related by a simple power law. Although the problem does not explicitly mention polytropes, the density profile resembles polytropic behavior, often represented as \(\rho \propto r^{-2}\). Polytropic models simplify the complex physics of stars due to their straightforward mathematical relations, making them valuable in stellar astrophysics. They help in solving and approximating the internal structures and mass distributions of stars.
Astrophysical Constants
Astrophysical constants are crucial for accurate calculations in astrophysics. In this problem, we encounter constants like the solar mass \(M_{\odot}\) and the solar radius \(R_{\odot}\). These constants are standard benchmarks in astrophysics. The problem's final step involves using these constants to derive \rho_0\. By plugging the known values into the models, we can calculate the specific density at the core of the hypothetical star. Hence, constants ensure that theoretical models can be applied to real-world scenarios effectively, providing meaningful and comparable results.
