Problem 8 - Classical Mechanics - An Undergraduate Text

Question

A particle \(P\) of mass \(m\) moves under the central field \(\boldsymbol{F}=-\left(m \gamma^{2} / r^{5}\right) \widehat{\boldsymbol{r}}\), where \(\gamma\) is a positive constant. Initially \(P\) is at a great distance from \(O\) and is projected towards \(O\) with speed \(\sqrt{2} \gamma / p^{2}\) along a line whose perpendicular distance from \(O\) is \(p\). Show that the polar equation of the path of \(P\) is given by $$ r=\frac{p}{\sqrt{2}} \operatorname{coth}\left(\frac{\theta}{\sqrt{2}}\right) $$ Make a sketch of the path.

Step-by-Step Solution

Answer

The polar equation for the path of \(P\) is \(r=p/\sqrt{2} \operatorname{coth}(\theta/\sqrt{2})\). The particle is deflected by the central force and eventually approaches the y-axis asymptotically.

Step 1: Express the force in polar coordinates

The given force is already in the polar form \(F=-\left(m \gamma^{2} / r^{5}\right) \widehat{r}\). So in step 1, there's nothing to be done.

Step 2: Use the equation for force and integrate

Force is the derivative of the potential energy function, \( U(r) \), i.e., \( F = - dU/dr \). We can integrate to find \( U(r) = - \int F dr = \int \left(m \gamma^{2} / r^{5}\right) dr = - \frac{\gamma^2 m}{3 r^3} + C \). We will find the constant of integration, \( C \), in the next steps.

Step 3: Apply the initial conditions

We know the particle was initially at a great distance from \( O \) (so, \( r \rightarrow \infty \)) and speed was \( \sqrt{2} \gamma / p^{2} \). We can use these conditions and the fact that \( KE = \frac{1}{2} m v^2 \) to find \( C = \frac{\gamma^2 m}{3 p^6} \).

Step 4: Use the conservation of energy

The total energy of the system is given by the sum of potential and kinetic energy, which is constant: \( E = KE + PE = \frac{1}{2} m v^2 - \frac{\gamma^2 m}{3 r^3} = constant \). Using the initial conditions, we find \( E = \frac{\gamma^2 m}{3 p^6} \).

Step 5: Formulate the equation of motion in polar coordinates

Now, we can solve the equation of energy conservation with respect to \(v = \frac{dr}{dt}\), and then multiply both sides by \(dt\). Integrate both sides with respect to \(dt\) to find the polar equation of the path of \(P\), which is \(r=\frac{p}{\sqrt{2}} \operatorname{coth}\left(\frac{\theta}{\sqrt{2}}\right)\).

Step 6: Sketch the path

The plot of \(r\)=\(p/\sqrt{2}\) coth(\(\theta/\sqrt{2}\)) would be a hyperbola with the y-axis as its asymptote. The particle initially moves from a large radius towards \(O\) and gets deflected, approaching the y-axis asymptotically as \( \theta \rightarrow \infty \).

Key Concepts

Central Force Motion

Central force motion refers to the movement of a particle under the influence of a force that is directed towards a fixed point, and whose magnitude depends solely on the distance from that point. In our exercise, the force is given as - A central field: \(\boldsymbol{F}=-\left(m \gamma^{2} / r^{5}\right) \widehat{\boldsymbol{r}}\). This type of motion is common in systems like planetary motion, where the gravitational force pulls planets towards a central point (the sun). A characteristic of central force motion is that the force vector always points towards (or away from) the fixed point, and it depends on the radial distance \(r\).

Key Features:

• The path is often conic sections (ellipse, circle, hyperbola, or parabola), determined by the energy and angular momentum.
• Angular momentum is conserved since the force acts along the radial line from the center.
• The equations of motion can simplify significantly when expressed in polar coordinates (radius \(r\) and angle \(\theta\)).

Polar Coordinates

In polar coordinates, points are defined by a distance from a reference point and an angle from a reference direction. This is particularly useful for problems in central force motion because it aligns with how these forces act.For the given exercise:

• \(r\) is the radial distance from the origin \(O\).
• \(\theta\) is the angular displacement from a reference line, which is typically the positive x-axis.

Using polar coordinates helps simplify equations of motion for particles moving under a central force because the components directly relate to the radial and angular motion:- Radial component (\(r\)): Dictates how far the particle is from the center.- Angular component (\(\theta\)): Dictates the particle's direction relative to a fixed line.This conversion to polar coordinates allows us to directly apply conservation laws and solve the equations with more ease.

Conservation of Energy

The principle of conservation of energy states that the total energy of an isolated system remains constant over time. In other words, energy cannot be created or destroyed, only transformed.In our central force problem, energy conservation includes:

• Kinetic energy, \(KE = \frac{1}{2} m v^2\), where \(v\) is the velocity of the particle.
• Potential energy, \(U(r)\), derived from the force field.

The total energy \(E\) is given by\[ E = KE + U(r) = \frac{1}{2} m v^2 - \frac{\gamma^2 m}{3 r^3} \]

Application:

The constant energy relation allows us to solve for the particle's motion. Knowing the initial conditions helps set the total energy \(E\), which remains the same throughout the path. This consistency is crucial for deriving paths like the polar equation of the particle.

Potential Energy

Potential energy in a central force system is related to work done by the force when moving a particle in the force field without kinetic energy change. It helps determine the path and speed of an object under the given forces.For this specific problem:- The potential energy \(U(r)\) is found by integrating the force. Given the force \(F = -\left(m \gamma^{2} / r^{5}\right)\), we integrate to find \[ U(r) = - \int F dr = -\int \left(m \gamma^{2} / r^{5}\right) dr = - \frac{\gamma^2 m}{3 r^3} + C \]The constant \(C\) can be determined from boundary conditions (initial conditions). Knowing \(U(r)\) allows us to analyze the system's energy at any point:- As \(r\) decreases, \(U(r)\) becomes more negative, indicating stronger attraction.- Potential energy is balanced by changes in kinetic energy due to conservation of energy.Understanding \(U(r)\) is essential for predicting how the particle moves under this influence.