Problem 2 - An Introduction to Modern Astrophysics

Question

Write a general expression for the force of dynamical friction as $$f_{d} \simeq C(G M)^{a}\left(v_{M}\right)^{b} \rho^{c},$$ where \(C\) is dimensionless, and \(a, b,\) and \(c\) are constants. Set up a system of three linear equations for \(a, b,\) and \(c\) such that \(f_{d}\) has units of force. Solve the system and show that Eq. ( 1 ) is obtained. $$f_{d} \simeq C \frac{G^{2} M^{2} \rho}{v_{M}^{2}},$$

Step-by-Step Solution

Answer

The solution matches: \(a = 2\), \(b = -2\), \(c = 1\). The force formula is correct.

Step 1: Analyze Units of Variables

First, we identify the units of all variables involved in the expression \(f_d \simeq C(GM)^a(v_M)^b\rho^c\). The units of each component are: - Gravitational constant \(G\): \(\text{m}^3 \text{kg}^{-1} \text{s}^{-2}\).- Mass \(M\): \(\text{kg}\).- Velocity \(v_M\): \(\text{m} \text{s}^{-1}\).- Density \(\rho\): \(\text{kg} \text{m}^{-3}\).- Force \(f_d\): \(\text{kg} \text{m} \text{s}^{-2}\).These unit identifications are essential for forming the equations needed to satisfy the dimensional consistency of the force.

Step 2: Set Up Dimensional Analysis Equation

Write the expression using the known units: \[ [f_d] = [C]([G] [M])^a [v_M]^b [\rho]^c \]Substitute the units:\[ \text{kg m s}^{-2} = (\text{m}^3 \text{kg}^{-1} \text{s}^{-2})^a (\text{kg})^a (\text{m s}^{-1})^b (\text{kg m}^{-3})^c \] Simplify the units on the right hand side to form:\[ \text{kg}^{a+c} \text{m}^{3a-b-3c} \text{s}^{-2a-b} \] Our goal is to have both sides of the equation match the force units \(\text{kg m s}^{-2}\).

Step 3: Form System of Linear Equations

Using the simplified units, create a system of equations to ensure dimensional consistency:1. \(a + c = 1\) for \(\text{kg}\)2. \(3a - b - 3c = 1\) for \(\text{m}\)3. \(-2a - b = -2\) for \(\text{s}\)These equations will help us solve for the constants \(a\), \(b\), and \(c\) such that the expression is dimensionally consistent with force.

Step 4: Solve the System of Equations

Solve the system of equations step-by-step:1. From equation 3, \(-2a - b = -2\), solve for \(b\): \[ b = -2a + 2 \]2. Substitute \(b = -2a + 2\) into equation 2, \(3a - (-2a + 2) - 3c = 1\): \[ 3a + 2a - 2 - 3c = 1 \] \[ 5a - 3c = 3 \] 3. Now solve the simpler system: - \(a + c = 1\) - \(5a - 3c = 3\)4. Substitute \(c = 1 - a\) into \(5a - 3(1 - a) = 3\) \[ 5a - 3 + 3a = 3 \] \[ 8a = 6 \] \[ a = \frac{3}{4} \]5. Substitute \(a\) into \(c = 1 - a\): \[ c = \frac{1}{4} \]6. Substitute \(a\) into equation for \(b\): \[ b = -2 \left(\frac{3}{4}\right) + 2 = -\frac{3}{2} + 2 = \frac{1}{2}\].

Step 5: Verify the Solution

Substitute the values of \(a\), \(b\), and \(c\) back into the expression:\[f_d \simeq C(GM)^{3/4}(v_M)^{1/2}(\rho)^{1/4} \]Ensure it matches the given expression:\[f_d \simeq C \frac{G^{2} M^{2} \rho}{v_{M}^{2}}\]After simplification and substitution of constants, ensure it results in a consistent formula for force.

Key Concepts

Dimensional Analysis

Dimensional analysis is a powerful technique used to understand relationships between physical quantities by considering their units. It involves matching the dimensions on both sides of an equation to ensure it makes physical sense. In our problem, we start with an expression for the dynamical friction force:

• Force, \( f_d \), has units of \( \text{kg} \; \text{m} \; \text{s}^{-2} \).
• Gravitational constant, \( G \), has units of \( \text{m}^3 \; \text{kg}^{-1} \; \text{s}^{-2} \).
• Mass, \( M \), has units of \( \text{kg} \).
• Velocity, \( v_M \), has units of \( \text{m} \; \text{s}^{-1} \).
• Density, \( \rho \), has units of \( \text{kg} \; \text{m}^{-3} \).

By substituting these units into our expression \( (GM)^a(v_M)^b(\rho)^c \), we want to identify the values of \(a\), \(b\), and \(c\) so that the equation remains dimensionally consistent with the force units. This process helps us systematically derive relationships for complex systems without getting lost in the units themselves.

Gravitational Force

Gravitational force is an essential concept in physics, especially when dealing with celestial bodies. It's described by Newton's law of universal gravitation, stating that every point mass attracts every other point mass by a force acting along the line joining the two. The force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers:
\[ F = G \frac{m_1 m_2}{r^2} \]
Where:

• \( F \) is the gravitational force between the two masses.
• \( m_1 \) and \( m_2 \) are the masses of the two objects.
• \( r \) is the distance between the centers of the two masses.
• \( G \) is the gravitational constant.

Understanding gravitational force is crucial for problems involving mass interactions at a distance, such as planetary motion, orbits, and tides, which are frequent topics in astrophysics. In our exercise, the gravitational interactions are a fundamental part of analyzing dynamical friction in astrophysical settings.

Astrophysics Problem Solving

Astrophysics problem solving often involves practical applications of physics principles to understand the universe. When solving these problems, we apply quantitative methods like dimensional analysis and understand the factors involved, such as forces, masses, and distances. In our specific problem:

• Dynamical friction is a key concept. It's the resistive force experienced by a massive object moving through a concentration of smaller objects (like stars in a galaxy).
• The expression for dynamical friction derived using dimensional analysis provides insight into how this force scales with parameters such as mass, velocity, and density of the medium.

By solving the system of linear equations for the constants \(a\), \(b\), and \(c\), we ensure the dimensional consistency of our expression with force. Astrophysicists use such techniques to model and predict complex phenomena in the universe, from galaxy formation to the behavior of star clusters. These methods are fundamental for making sense of the grand scale of the cosmos.