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A mass \(m\) moves in a circular orbit (centered ón 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

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a y^{2}+2 b y z+c z^{2},(\mathbf{b}) g(x, y, z)=\cos \left(a x y^{2} z^{3}\right),(\mathbf{c}) h(x, y, z)=a r,\) where \(a, b,\) and \(c\) are constants and \(r=\sqrt{x^{2}+y^{2}+z^{2}} .\) Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

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