Let \(U(r)=-k r^{-\beta}, 0<\beta<2 .\) Find \(\Phi_{0}=\lim _{E \rightarrow-0} \Phi\).

Given the potential energy function \(U(r) = -kr^{-\beta}\), we can find the kinetic energy function \(T(v)\) as \(T(v) = \frac{1}{2}mv^2\) (we're assuming the system has only one mass m). The total energy of the system is given by the sum of kinetic and potential energy: \(E = T(v) + U(r)\). Now, we will express \(\Phi\) in terms of \(k\), \(r\), \(\beta\), and \(E\). The phase volume \(\Phi\) is the volume of the region in phase space enclosed by the surface of constant energy \(E\). To find \(\Phi\), we will integrate the kinetic energy \(T(v)\) over the allowed volume for each value of \(r\):

Integrating \(T(v)\) over the allowed volume for each value of \(r\), we obtain the following expression for \(\Phi\): $$ \Phi = \oint T(v)dr = \int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{\pi}\frac{1}{2}mv^2(1 - \frac{kr^{-\beta}}{E}) r^2\sin{\theta}d\theta d\phi dr $$ Where we have used the expression for the allowed volume in spherical coordinates.

The integral can be simplified into: $$ \Phi = \frac{1}{2}mV\int_{0}^{\infty}v^2(1 - \frac{kr^{-\beta}}{E}) r^2dr $$ Where \(V = 4\pi\) is the total solid angle.

To find \(\Phi_0 = \lim_{E\rightarrow -0} \Phi\), we can substitute the expression for \(\Phi\) into the limit: $$ \Phi_0 = \lim_{E\rightarrow -0} \frac{1}{2}mV\int_{0}^{\infty}v^2(1 - \frac{kr^{-\beta}}{E}) r^2dr $$ Now, we perform the limit as \(E\rightarrow -0\). The potential energy U(r) should be negative (since the particle is bound), so as \(E\rightarrow -0\) we get \(1-\frac{kr^{-\beta}}{E} \rightarrow \infty\). This means that the limit of the integral will be dominated by the term with the smallest power of \(r\). As \(0<\beta<2\), the term \(1-\frac{kr^{-\beta}}{E}\) will dominate the integral, and hence, the integral goes to infinity when \(E\rightarrow -0\). Therefore, $$ \Phi_0 = \lim_{E\rightarrow -0} \Phi = \infty $$ Thus, the phase volume \(\Phi_0\) diverges to infinity as the total energy \(E\) approaches 0.