Question

Consider a particle of reduced mass \(\mu\) orbiting in a central force with \(U=k r^{n}\) where \(k n>0\). (a) Explain what the condition \(k n>0\) tells us about the force. Sketch the effective potential energy \(U_{\mathrm{eff}}\) for the cases that \(n=2,-1,\) and \(-3 .\) (b) Find the radius at which the particle (with given angular momentum \(\ell\) ) can orbit at a fixed radius. For what values of \(n\) is this circular orbit stable? Do your sketches confirm this conclusion? (c) For the stable case, show that the period of small oscillations about the circular orbit is \(\tau_{\mathrm{osc}}=\tau_{\mathrm{orb}} / \sqrt{n+2} .\) Argue that if \(\sqrt{n+2}\) is a rational number, these orbits are closed. Sketch them for the cases that \(n=2,-1,\) and 7.

Short answer

Stable circular orbits exist for \( n > -2 \). Closed orbits occur when \( \sqrt{n+2} \) is rational.

Step-by-step solution

Understanding the Force Condition

The condition \( k n > 0 \) implies that both the constant \( k \) and the power \( n \) in the potential energy function \( U = k r^n \) have the same sign. If \( k > 0 \) and \( n > 0 \), the force is repulsive when acting in a radial direction. Conversely, if \( k < 0 \) and \( n < 0 \), the force is attractive.

Sketching Effective Potential Energy

The effective potential energy \( U_{ ext{eff}} \) is given by \( U_{ ext{eff}} = U(r) + \frac{\ell^2}{2\mu r^2} \). Sketches for \( n=2, -1, \) and \( -3 \) are as follows:- **For \( n=2 \):** The potential \( U(r) = k r^2 \) creates a parabolic rise, while \( \frac{\ell^2}{2\mu r^2} \) term contributes a barrier at small \( r \).- **For \( n=-1 \):** \( U(r) = k r^{-1} \) describes a well with a steep drop-off, combined with a centrifugal barrier.- **For \( n=-3 \):** \( U(r) = k r^{-3} \) has a sharper drop and a strong repulsive force at small \( r \) due to the centrifugal term.

Finding Fixed Orbital Radius

For a stable circular orbit at radius \( r_0 \), the condition \( \frac{dU_{ ext{eff}}}{dr}|_{r=r_0} = 0 \) must be satisfied. This leads to the equation:\[kn r_0^{n-1} + \frac{\ell^2}{\mu r_0^3} = 0.\]Solving gives the equilibrium radius \( r_0 \) in terms of \( \ell, \mu, k, \) and \( n \). For stability, the second derivative \( \frac{d^2U_{ ext{eff}}}{dr^2} \mid_{r=r_0} > 0 \) should be positive.

Analyzing Stability of Circular Orbit

The orbit stability condition \( \frac{d^2U_{ ext{eff}}}{dr^2} > 0 \) after substituting and simplifying leads to \( n > -2 \), meaning orbits are stable if \( n > -2 \). This is consistent with sketches, where for \( n = 2 \) and \( n = -1 \), stability holds but for \( n = -3 \), it does not.

Calculating Period of Small Oscillations

The frequency \( \omega \) of small oscillations about \( r_0 \) is derived using the effective potential:\[\omega^2 = \frac{1}{\mu} \frac{d^2U_{ ext{eff}}}{dr^2} \Big|_{r=r_0} = \frac{\ell^2 (n+2)}{\mu r_0^4}.\]The oscillation period \( \tau_{osc} \) is related to the orbital period \( \tau_{orb} \) by \( \tau_{osc} = \tau_{orb} / \sqrt{n+2} \). Orbits are closed when \( \sqrt{n+2} \) is rational.

Confirming Closed Orbits and Sketching

For \( n=2, -1, \) and \( 7 \), the closed orbits are examined:- **For \( n=2 \):** \( \sqrt{4} = 2 \), a rational number, confirms closed orbits.- **For \( n=-1 \):** \( \sqrt{1} = 1 \), also leads to closed orbits.- **For \( n=7 \):** \( \sqrt{9} = 3 \), implying closed orbits.Sketches show circular and elliptical paths depending on \( n \).

Key concepts

Effective Potential Energy

In central force motion, understanding the effective potential energy is crucial. The effective potential energy equation is expressed as:

• \( U_{\text{eff}}(r) = U(r) + \frac{\ell^2}{2\mu r^2} \)

Here, \( U(r) = k r^n \) represents the potential energy due to the central force, while the term \( \frac{\ell^2}{2\mu r^2} \) arises from the particle's angular momentum \( \ell \). This additional term accounts for the centrifugal barrier, which tends to repel the particle as it gets too close to the center.

The condition \( k n > 0 \) ensures that the force is attractive when both \( k < 0 \) and \( n < 0 \), or repulsive when both are positive. Sketching the effective potential energy for different \( n \) values helps to understand the behavior:

• **For** \( n = 2 \) : The potential energy is parabolic (rising with \( r \)), combined with an increasing centrifugal barrier as \( r \) decreases.
• **For** \( n = -1 \) : The potential energy shows a steep drop, moderated by the centrifugal barrier creating a well-like profile. This is a classic case of an attractive force.
• **For** \( n = -3 \) : A sharp decrease in potential energy denotes a more pronounced attractive force, heavily influenced by the centrifugal term at small radii.

Effective potential sketches aid in visualizing how a particle would navigate these energy landscapes.

Stability of Circular Orbits

Determining the stability of circular orbits is central to comprehending particle motion in a force field. A stable circular orbit exists when small perturbations or deviations don't lead to the particle spiraling away.

Mathematically, this means the effective potential energy curve must have a local minimum at the orbit radius \( r_0 \). This criterion is expressed as:

• \( \frac{d^2U_{\text{eff}}}{dr^2} \Big|_{r=r_0} > 0 \)

For the attraction to be stable, calculations reveal that the power \( n \) must satisfy \( n > -2 \). This condition ensures that the effective potential has a stable minimum point.

When sketching effective potential energy:

• \( n = 2 \) and \( n = -1 \) show stability since the second derivative indicates a potential well.
• Conversely, when \( n = -3 \), the potential is more likely to have no stable point, making the orbit unstable.

These sketches confirm analytical conclusions about orbit stability.

Oscillation Period

The oscillation period about a stable circular orbit provides insight into the dynamics of particle motion around the orbit. If the orbit is slightly disturbed, the time it takes for the particle to complete small oscillations around the stable radius is given by the oscillation period \( \tau_{\text{osc}} \).

This period is determined by:

• \( \tau_{\text{osc}} = \frac{\tau_{\text{orb}}}{\sqrt{n+2}} \)

Here, \( \tau_{\text{orb}} \) is the period of the actual orbit, while \( \sqrt{n+2} \) adjusts for the stability factor of the potential.

For an orbit to be closed, \( \sqrt{n+2} \) must be a rational number, which is verification that the orbit will repeat itself after completing a cycle. Examples include:

• **For** \( n = 2 \) : \( \sqrt{4} = 2 \), resulting in a closed orbit.
• **For** \( n = -1 \) : \( \sqrt{1} = 1 \), also indicates a closed orbit since it cycles perfectly.
• **For** \( n = 7 \) : \( \sqrt{9} = 3 \), which ensures the orbit cycles back completely after a loop.

Understanding the oscillation period and its determiners helps in predicting the stability and periodicity of the particle orbits.