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Chapter 2: Problem 25

For which values of \(\alpha\) is motion along a circular orbit in the field with potential energy \(U=r^{\alpha},-2 \leq \alpha<\infty\), Liapunov stable?

Short Answer

Expert verified
Answer: The motion along a circular orbit is Liapunov stable for \(\alpha > 2\).

Step by step solution

01

Find the equation of motion in the potential field

We start by finding the force acting on a particle in the potential field. The force can be found by taking the negative gradient of the potential energy function: \(F = -\frac{\partial U}{\partial r} = -\alpha r^{\alpha - 1}\). Next, we use Newton's second law, \(F = m*r''\), where \(m\) is the mass of the particle and \(r''\) is the radial acceleration. Plugging in the force: \(m * r'' = -\alpha r^{\alpha-1}\). Hence the equation of motion is: \(r'' = -\frac{\alpha}{m}r^{\alpha - 1}\).
02

Find the conditions for a circular orbit

For a circular orbit, the centripetal acceleration should balance the radial acceleration: \(m \omega^2*r = m r''\), where \(\omega\) is the angular velocity. Therefore, the condition for a circular orbit is: \(\omega^2 * r = -\frac{\alpha}{m}r^{\alpha - 1}\). Simplifying and rearranging gives: \(r^{\alpha - 2} = -\frac{\alpha}{\omega^2 * m}\).
03

Determine the Liapunov stability criterion

To check for Liapunov stability, let's consider small perturbations \(\delta r\) in the radial distance. The general equation of motion for the perturbed distance is given by: \((r + \delta r)'' = -\frac{\alpha}{m}(r + \delta r)^{\alpha - 1}\). For small perturbations, we can approximate \((r + \delta r)^{\alpha - 1} \approx r^{\alpha - 1} + (\alpha - 1) r^{\alpha - 2}\delta r\). Therefore, the equation of motion becomes: \((r + \delta r)'' = -\frac{\alpha}{m} (r^{\alpha - 1} + (\alpha - 1) r^{\alpha - 2} \delta r)\). Now, let's substitute the condition for a circular orbit, \(r^{\alpha - 2} = -\frac{\alpha}{\omega^2 * m}\): \((r + \delta r)'' = -\omega^2 (r * (\alpha - 2) \delta r)\).
04

Apply the Liapunov stability criterion

For Liapunov stability, the perturbation \(\delta r\) must remain bounded, which means that the perturbed motion should have oscillatory behavior. Thus, we need the coefficient of \(\delta r\) to be positive: \(\alpha - 2 > 0 \Rightarrow \alpha > 2\). So, for motion along a circular orbit to be Liapunov stable, we must have \(\alpha > 2\). In conclusion, the motion along a circular orbit in the field with potential energy \(U = r^\alpha, -2 \leq \alpha < \infty\), is Liapunov stable for \(\alpha > 2\).

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