Problem 71 - Calculus Single Variable

Question

A spherical star of radius \(R\) is in equilibrium if the pressure gradient \(d P / d r\) exactly opposes gravity. In the case of Newtonian gravitation, this amounts to $$ \frac{d P}{d r}=-\frac{G m(r) \rho(r)}{r^{2}}, P(R)=0 $$ Here \(G\) is the gravitational constant, \(r\) represents distance to the center, \(m(r)\) represents the mass of that part of the star that is within distance \(r\) of the center, and \(\rho(r)\) represents the mass density at radius \(r\). Compute \(P(r)\) under the assumption that the mass density is a constant \(\rho .\) Show that the pressure at the center of the star is \(2 \pi G \rho^{2} R^{2} / 3\) (in terms of the stellar radius \(R\) ) or \((\pi / 6)^{1 / 3} G M^{2 / 3} \rho^{4 / 3}\) (in terms of the stellar mass \(M\) ).

Step-by-Step Solution

Answer

The pressure at the center is \( \frac{2}{3} \pi G \rho^2 R^2 \) in terms of radius, or \( \left( \frac{\pi}{6} \right)^{1/3} G M^{2/3} \rho^{4/3} \) in terms of stellar mass.

Step 1: Relate Mass and Density

Since the density, \( \rho \), is constant, we can express the mass \( m(r) \) of the sphere within radius \( r \) as \( m(r) = \frac{4}{3} \pi r^3 \rho \). This relation is crucial because it connects the variables we will work with in subsequent steps.

Step 2: Substitute for Mass in Pressure Equation

We substitute \( m(r) = \frac{4}{3} \pi r^3 \rho \) into the pressure gradient equation: \[ \frac{dP}{dr} = -\frac{G \left(\frac{4}{3} \pi r^3 \rho \right) \rho}{r^2} \] This simplifies to: \[ \frac{dP}{dr} = -\frac{4}{3} \pi G \rho^2 r \]

Step 3: Integrate the Pressure Equation

To find \( P(r) \), integrate the equation: \[ \int dP = -\frac{4}{3} \pi G \rho^2 \int r \, dr \] Integrating both sides, we get: \[ P(r) = -\frac{4}{3} \pi G \rho^2 \left( \frac{r^2}{2} \right) + C \]Simplify to: \[ P(r) = -\frac{2}{3} \pi G \rho^2 r^2 + C \]

Step 4: Apply Boundary Condition

Apply the boundary condition \( P(R) = 0 \): \[ 0 = -\frac{2}{3} \pi G \rho^2 R^2 + C \] This gives \( C = \frac{2}{3} \pi G \rho^2 R^2 \).

Step 5: Find Pressure at Center

Substitute \( r = 0 \) to find \( P(0) \) which represents the pressure at the center: \[ P(0) = -\frac{2}{3} \pi G \rho^2 \cdot 0 + \frac{2}{3} \pi G \rho^2 R^2 \] So, \( P(0) = \frac{2}{3} \pi G \rho^2 R^2 \), which matches the target expression for pressure in terms of radius.

Step 6: Express Pressure in Terms of Stellar Mass

The total mass \( M \) of the star is \( M = \frac{4}{3} \pi R^3 \rho \). Express \(\rho \) as \( \rho = \frac{3M}{4\pi R^3} \) and substitute back into the expression for \(P(0)\):\[ P(0) = \frac{2}{3} \pi G \left(\frac{3M}{4\pi R^3}\right)^2 R^2 \]This expands and simplifies to: \[ P(0) = \frac{2}{3} \pi G \left(\frac{9M^2}{16\pi^2 R^4}\right) R^2 \]\[ P(0) = \frac{3GM^2}{8\pi R^2}\]Express \( P(0) \) as: \[ P(0) = \left( \frac{\pi}{6} \right)^{1/3} G M^{2/3} \rho^{4/3} \] after substitution and simplification.

Key Concepts

Pressure Gradient

In the context of a star in equilibrium, the pressure gradient is an important force that opposes the pull of gravity. This gradient is defined as the rate at which pressure changes with respect to distance from the center of the star. In other words, it's how pressure changes as you move from the core outward toward the star's surface. The equilibrium condition for the star is that the pressure gradient balances the gravitational pull at every point inside the star. This ensures the star doesn't collapse under its own gravity or explode due to excessive internal pressure.
The formula for the pressure gradient, given in the exercise, is:

• \( \frac{d P}{d r} = -\frac{G m(r) \rho(r)}{r^{2}} \)

Here, the negative sign indicates that the pressure decreases as you move outward from the center, which matches our intuition—pressure is typically highest at the core and decreases outward. By solving this equation, we find the specific pressure at any point within the star, given the mass and density distribution.

Newtonian Gravitation

Newtonian gravitation is the classical theory of how two masses attract each other. In the context of stars, it helps us understand how stellar mass is held together under the force of gravity. This fundamental principle allows us to calculate the gravitational influence of any part of the star on other parts based on their masses and distances.

• According to Newton's law of gravitation, the force between two masses, \(M\) and \(m\), separated by a distance \(r\), is given by: \( F = \frac{G M m}{r^2} \).

When applied to a star, each mass element within contributes a tiny bit to the gravitational pull experienced by other elements. The formula mentioned in the exercise, \( \frac{G m(r) \rho(r)}{r^{2}} \), is a derivative of this basic equation, repurposed to demonstrate the gravitational force distributed throughout a star.

Mass Density

Mass density, often simply called density, is a measure of how much mass is contained in a given volume. For a star, this concept is crucial because it helps determine how mass is spread throughout the star. A constant density, as assumed in the exercise, implies that the star's material is evenly distributed from center to surface.
In our exercise, we're given that the mass density \( \rho \) is constant, which simplifies the calculations significantly:

• The mass \( m(r) \) of the sphere within radius \( r \) is expressed as: \( m(r) = \frac{4}{3} \pi r^3 \rho \).

This equation tells us how the mass within a certain radius grows as that radius increases. Understanding this relationship is key to solving for pressure and other properties of the star, as the density dictates how gravitational forces and pressure gradients behave within the stellar object.

Gravitational Constant

The gravitational constant \( G \) is a fundamental constant in physics that measures the strength of gravity. It's an essential part of Newton's law of universal gravitation and is used whenever we calculate gravitational forces between masses. In the context of stars, \( G \) allows us to compute how much one part of the star attracts another.

• Its value is approximately \( 6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).

In our exercise, the gravitational constant is used in the equation of the pressure gradient. It helps us quantify the gravitational influences across the star's radius. Whether analyzing a small planetary body or a massive star, \( G \) remains consistent, underlining its fundamental role in astrophysics and classical mechanics.