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Chapter 4: Problem 41

A mass \(m\) moves in a circular orbit (centered 贸n the origin) in the field of an attractive central force with potential energy \(U=k r^{n} .\) Prove the virial theorem that \(T=n U / 2\).

Short Answer

Expert verified
Kinetic energy \(T\) is \(\frac{n}{2}\) times the potential energy \(U\).

Step by step solution

01

Understand the Problem

We are given a system where a mass \(m\) moves in a circular orbit around a central force with potential energy \(U = k r^n\). We are tasked with proving that the kinetic energy \(T\) is equal to \(\frac{n}{2} U\), which is a form of the virial theorem in this context.
02

Write Down Known Equations

In a central force problem, the force \( F \) is given by the negative gradient of the potential energy, \( F = - \frac{dU}{dr} \). The potential energy is given as \( U = k r^n \). The virial theorem for circular orbits states \( 2T + nU = 0 \). Here, we need to show \( T = \frac{n}{2} U \).
03

Apply the Condition of Circular Orbit

For a circular orbit, the centripetal force required (\( F_c = \frac{mv^2}{r} \)) is provided by the force from the potential energy, \( F = -\frac{dU}{dr} \). Hence, \[ \frac{mv^2}{r} = -\frac{d}{dr}(kr^n) \].
04

Calculate the Force

Calculate \( \frac{dU}{dr} \): \( \frac{d}{dr}(kr^n) = knr^{n-1} \). The force \( F \) is then \( F = -knr^{n-1} \).
05

Equate Forces for Circular Motion

Set the centripetal force equal to the negative potential energy force (since it's attractive): \[ \frac{mv^2}{r} = knr^{n-1} \]. Simplifying gives \[ mv^2 = knr^n \].
06

Express Kinetic Energy

The kinetic energy \( T \) is \( \frac{1}{2}mv^2 \). Substitute \( mv^2 = knr^n \) into the kinetic energy equation: \[ T = \frac{1}{2}knr^n \].
07

Relate Kinetic and Potential Energy

Using \( U = kr^n \), the kinetic energy can be expressed in terms of \( U \): \( T = \frac{1}{2}knr^n = \frac{1}{2}n(kr^n) = \frac{n}{2}U \). This establishes that \( T = \frac{n}{2}U \), proving the statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Central Force
A central force is a force that is directed along the line joining two bodies and depends solely on the distance between them. In physics, this type of force is crucial when studying objects that move under the influence of gravity, electric fields, or other similar forces. Some important characteristics of central forces include:
  • They always act along a line joining the center of one object to the center of the other.
  • Central forces are conservative, meaning that the work done by or against them depends only on the initial and final positions.
  • The force can be attractive (like gravity) or repulsive (like the force between similar charges).
In the context of the exercise, the central force is derived from the potential energy formula, \( U = k r^n \). The force itself is calculated by taking the negative gradient of the potential energy, which determines how the object will move in response to that force.
Circular Orbit
A circular orbit occurs when an object moves in a perfect circle around another object, with the force of attraction (like gravity) acting as the centripetal force keeping it in motion. This is a special case of the more general elliptical orbit, where the centripetal force is supplied by the central force derived from potential energy.
  • For a stable circular orbit, the centripetal force required to keep the object moving in a circle is provided entirely by the central force.
  • In our exercise, this relationship is shown by equating the centripetal force to the derivative of the potential energy.
  • Mathematically, the necessary centripetal force for a circular orbit is given by: \( F_c = \frac{mv^2}{r} \).
This compelling balance of forces ensures that the object continues to move along its circular path, neither falling into the center nor flinging off into space.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, given by the formula: \( T = \frac{1}{2} mv^2 \). It depends on both the mass of the object and its velocity squared. In our problem, kinetic energy plays a pivotal role in understanding the relation established by the virial theorem.
  • Kinetic energy is crucial in the context of orbital mechanics because it helps determine how fast the object is moving along its path.
  • In this scenario, the kinetic energy is directly tied to the potential energy through the virial theorem.
  • By analyzing the forces and motions involved, we proved that the kinetic energy can be expressed as a fraction of the potential energy: \( T = \frac{n}{2} U \), where \( n \) is derived from the potential energy relation.
This relationship is significant in many areas of physics, especially when studying systems that are bound by forces, like planets around a star.
Potential Energy
Potential energy is the stored energy in a system due to its position or configuration. In the case of central forces like gravity, potential energy can change as the relative positions of interacting bodies change. For this exercise, potential energy is defined by the expression \( U = k r^n \).
  • With central forces, potential energy is usually a function of distance between two objects.
  • The potential energy in the exercise plays a key role in determining the dynamics of the system, influencing how objects move in response to each other.
  • The virial theorem connects potential energy and kinetic energy in stable orbits, showing how energy balances within such systems: \( 2T + nU = 0 \).
Understanding potential energy and its implications helps us predict how the system behaves over time, particularly when we can express it in terms of other quantities of interest like kinetic energy.

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Most popular questions from this chapter

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.

Prove that if \(f(\mathbf{r})\) and \(g(\mathbf{r})\) are any two scalar functions of \(\mathbf{r},\) then \(\nabla(f g)=f \nabla g+g \nabla f\)

An interesting one-dimensional system is the simple pendulum, consisting of a point mass \(m\), fixed to the end of a massless rod (length \(l\) ), whose other end is pivoted from the ceiling to let it swing freely in a vertical plane, as shown in Figure \(4.26 .\) The pendulum's position can be specified by its angle \(\phi\) from the equilibrium position. (It could equally be specified by its distance \(s\) from equilibrium \(-\) indeed \(s=l \phi-\) but the angle is a little more convenient.) (a) Prove that the pendulum's potential energy (measured from the equilibrium level) is \(U(\phi)=m g l(1-\cos \phi)\). Write down the total energy \(E\) as a function of \(\phi\) and \(\dot{\phi}\). (b) Show that by differentiating your expression for \(E\) with respect to \(t\) you can get the equation of motion for \(\phi\) and that the equation of motion is just the familiar \(\Gamma=I \alpha\) (where \(\Gamma\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration \(\ddot{\phi}\) ). (c) Assuming that the angle \(\phi\) remains small throughout the motion, solve for \(\phi(t)\) and show that the motion is periodic with period \(\tau_{\mathrm{o}}=2 \pi \sqrt{l / g}\).

Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]
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