Question

7.21. Repulsive potential Consider an "anti-gravitational" potential (or more mundanely, the elec- trostatic potential between two like charges), V(r) = α/r, where α > 0. What is the basic change in the analysis of Section 7.4? Show that circular, elliptical, and parabolic orbits do not exist. Draw the figure analogous to Fig. 7.9 for the hyperbolic orbit.

Answer 1

Let's begin by collecting the formulas 1/r = m alpha/L^2 (1 + epsilon cos theta) where epsilon = squareroot 1 + 2 E l/m alpha^2 r_max = {k/1 - epsilon = L^2/m alpha (1 - epsilon) (if epsilon < 1) if epsilon greaterthanorequalto 1 r_min = k/1 + epsilon = L^2/m alpha(1 + epsilon) k congruent L^2/m alpha = r(1 + epsilon cos theta) = r + epsilon x for alpha < 0, we get a for value of 1/r if epsilon cos theta < -1 implies cos theta < -1/epsilon this is possible only for epsilon > 1 which means circular and parabolic orbits are impossible. The hyperbolic orbit is plotted below the easiest way to make the connection with is to recognize that the other branch of hyperbola for (the unphysical branch with x < 0 -plotted) to the orbit for attractive force with but inverted through origin because r is negative note that the dashed black curve is a plot of 1/r = -P_1/L_3 (1 + epsilon cos theta)