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

We can obtain the equation of motion for a particle moving in a potential field by applying the principle of conservation of energy. The conservation of energy equation is given by: \[E = T + U\] Where \(E\) is the total energy, \(T\) is the kinetic energy, and \(U\) is the potential energy. For a particle in central potential, the kinetic energy is given by: \[T = \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \frac{1}{2} m r^2 \left(\frac{d\theta}{dt}\right)^2\] Now, we can write the conservation of energy equation as: \[E = \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \frac{1}{2} m r^2 \left(\frac{d\theta}{dt}\right)^2 - kr^{-\beta}\]

To express the equation of motion in terms of \(\Phi\), we need to rewrite it in terms of polar coordinates \((r, \theta)\). First, let's find the relation between \(\frac{dr}{dt}\) and \(\frac{d\theta}{dt}\). We have: \[\Phi = -\int \frac{d\theta}{dr} \frac{dr}{dt} dt = -\int \frac{d\theta}{dt} dr = -\int \frac{d\theta}{dt} \frac{dr}{d\theta} d\theta\] Now, we can express \(\frac{dr}{dt}\) in terms of \(\frac{d\theta}{dt}\) and \(\Phi\) as follows: \[\frac{dr}{dt} = -\frac{d\Phi}{d\theta} \frac{d\theta}{dt}\] Substituting this expression back into the equation of motion, we get: \[E = \frac{1}{2} m \left(-\frac{d\Phi}{d\theta} \frac{d\theta}{dt}\right)^2 + \frac{1}{2} m r^2 \left(\frac{d\theta}{dt}\right)^2 - kr^{-\beta}\]

We can now solve the equation of motion for \(\Phi\). First, divide the equation by \(m\): \[E/m = \frac{1}{2} \left(-\frac{d\Phi}{d\theta} \frac{d\theta}{dt}\right)^2 + \frac{1}{2} r^2 \left(\frac{d\theta}{dt}\right)^2 - \frac{k}{m} r^{-\beta}\] Now, we can eliminate \(\frac{d\theta}{dt}\) to find the expression for \(\Phi\): \[\frac{d\Phi}{d\theta} = -\sqrt{2 m \left[E/m - \frac{1}{2} r^2 \left(\frac{d\theta}{dt}\right)^2 + \frac{k}{m} r^{-\beta}\right]}\] Integrating both sides with respect to \(\theta\), we get: \[\Phi(\theta) = -\int \sqrt{2 m \left[E/m - \frac{1}{2} r^2 \left(\frac{d\theta}{dt}\right)^2 + \frac{k}{m} r^{-\beta}\right]} d\theta\]

To find \(\Phi_0\), we need to take the limit of \(\Phi(\theta)\) as \(E\) approaches \(0\): \[\Phi_0 = \lim_{E \rightarrow -0} \Phi(\theta) = -\int \sqrt{2 m \left[-\frac{1}{2} r^2 \left(\frac{d\theta}{dt}\right)^2 + \frac{k}{m} r^{-\beta}\right]} d\theta\] Notice that \(\Phi_0\) depends on the potential energy function \(U(r)\) and the mass of the particle \(m\). Since \(0 < \beta < 2\), we can conclude that \(\Phi_0\) exists and is finite, but the exact value depends on the specific values of \(k\), \(m\), and \(\beta\).