Exercise 3

Now the potential has the form: \[ V(x, y) = \frac{k}{2(x^2 + y^2)} \] \[ F_x = - \frac{\partial V}{\partial x} = - \frac{k}{2} (-1) (x^2 + y^2)^{-2} \, 2x = \frac{kx}{(x^2 + y^2)^2}\] \[ F_y = - \frac{\partial V}{\partial y} = \frac{ky}{(x^2 + y^2)^2}\]

The equations of motion are: \[ m \, \ddot{x}(t) = \frac{kx}{(x^2 + y^2)^2}\] \[ m \, \ddot{y}(t) = \frac{ky}{(x^2 + y^2)^2}\]

These are non-linear ordinary differential equations of 2nd order. Solving those equations is far beyond our scope. But even without having an analytical solution, we can make the following statements:

Now it is clear that there are no closed orbits. Independet of the initial conditions, the particle will depart from the origin, until it disappears in infinity.

The energy is conserved, according to the assumption that the force can be derived as the gradient of a potential. (These force fields are callled conservative.)

If you like a proof, here it is:

\[ E = T + V = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{k}{2(x^2 +y^2)} \]

For energy conservation, we have to show that the following applies: \[ \frac{\mathrm{d} E}{\mathrm{d}t } = 0 \] \[ \frac{\mathrm{d} E}{\mathrm{d}t } = \frac{1}{2}m (2 \dot{x}\ddot{x}+ 2 \dot{y}\ddot{y}) - \frac{k (2 x \dot{x} + 2 y \dot{y})}{2 (x^2+y^2)^2} \] \[ \frac{\mathrm{d} E}{\mathrm{d}t } = \dot{x} \left(m \ddot{x}- \frac{kx}{(x^2+y^2)^2} \right) + \dot{y} \left(m \ddot{y}- \frac{ky}{(x^2+y^2)^2} \right) = 0 \]

Since the laws of motion apply, the expressions in parentheses cancel out.