Question

Prove that the potential energy of a central force \(\mathbf{F}=-k r^{n} \hat{\mathbf{r}}(\text { with } n \neq-1)\) is \(U=k r^{n+1} /(n+1)\). In particular, if \(n=1,\) then \(\mathbf{F}=-k \mathbf{r}\) and \(U=\frac{1}{2} k r^{2}\).

Short answer

Potential energy is \( U = \frac{k r^{n+1}}{n+1} \); for \( n=1 \), \( U = \frac{1}{2} kr^2 \).

Step-by-step solution

Define Potential Energy

Potential energy, denoted as \( U \), is defined for a central force \( \mathbf{F} = -abla U \). This means the force is the negative gradient of the potential energy.

Relate Force and Potential Energy

The force \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \) is acting radially, so we can write \(-abla U = -k r^n \hat{\mathbf{r}} \). We assume \( U = U(r) \) depends solely on \( r \), leading to \(-\frac{dU}{dr} = -k r^n \).

Integrate to Find U(r)

Integrate \(-\frac{dU}{dr} = -k r^n \) with respect to \( r \) to find the potential energy function. This gives us \( U(r) = \int k r^n \, dr = k \frac{r^{n+1}}{n+1} + C \), where \( C \) is an integration constant.

Apply Specific Case (n=1)

Substitute \( n=1 \) into the derived formula \( U(r) = \frac{k r^{n+1}}{n+1} \). This yields \( U(r) = \frac{k r^2}{2} \). The specific case of \( \mathbf{F} = -k \mathbf{r} \) corresponds to this expression for potential energy.

Key concepts

Central Force

When we talk about central forces, we are referring to forces that act along a line that passes through a central point. This point is often the origin in a coordinate system. Such forces are always directed towards or away from this central point, and their magnitude depends only on the distance from the center.
A central force can be represented mathematically as \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), where:- \( k \) is a constant that determines the strength of the force.- \( r \) is the radial distance from the origin.- \( n \) is the power through which this distance is raised, affecting how the force changes with distance.
This kind of force is quite common in physics. Examples include gravitational and electrostatic forces, where the force strength depends on the inverse square of the distance (when \( n = -2 \)). Understanding central forces is key to grasping concepts like orbiting planets and charged particle interactions.

Gradient of Potential Energy

The concept of the gradient comes from calculus and is crucial to understanding potential energy in the context of forces. The gradient essentially provides the rate and direction of change of a quantity. In the context of physics, it's used to derive the force exerted by a potential energy field.
The force associated with potential energy is given by the negative gradient of that potential energy \( U \), and is expressed as \( \mathbf{F} = -abla U \). This means:

• The force points in the direction where potential energy decreases the fastest.
• The magnitude of the force indicates how steeply the potential energy decreases.

This concept helps in finding the equilibrium positions and understanding stability, as areas where the gradient (or force) is zero correspond to potential minima or maxima. It provides a bridge between energy landscapes and the forces that drive physical systems.

Integration

Integration is a mathematical tool that is used to determine the whole from its parts. In the context of this exercise, it's the key to finding the potential energy function from a given force. When we say we integrate a force function, we mean we are calculating the accumulated effect of this force over a path or distance.
For the central force \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), following the expression \( -\frac{dU}{dr} = -k r^n \), the integration process yields the potential energy function \( U(r) \).

• Integrating \( -k r^n \) with respect to \( r \), we compute: \[ U(r) = \int k r^n \, dr = \frac{k r^{n+1}}{n+1} + C \]
• Here, \( C \) is the constant of integration, representing potential energy at a reference position.

Through integration, we transform force data into an understandable energy framework, facilitating the analysis of mechanical systems and predictions of motion dynamics.

Radial Force

Radial force is a term used to describe any force that acts along a radius from a central point, either towards or away from it. This kind of force is fundamental in systems with spherical symmetry, like planets orbiting stars or electrons circling nuclei.
Radial forces simplify many analysis and calculation processes because they maintain symmetry in relation to the center point. In our central force example \( \mathbf{F} = -k r^n \hat{\mathbf{r}} \), the term \( \hat{\mathbf{r}} \) is a unit vector pointing radially outward from the center. This notation indicates that the force vector points along the radius:

• Radial symmetry ensures that the force's effect only depends on the distance from the center (\( r \)) and not the direction in space.
• This makes it easier to calculate potential energies and predict the behavior of objects under such forces.

Radial forces are crucial in simplifying the mathematics of many physical systems, allowing us to focus only on how force and potential change with distance.