The total mechanical energy \( E \) of the particle is the sum of its Kinetic Energy (\( K \)) and Potential Energy (\( U \)).

Potential Energy: Since the force is attractive \( F = -k/r^2 \), the potential energy is: \[ U(r) = -\frac{k}{r} \]

Kinetic Energy: In polar coordinates, \( v^2 = (\dot{r})^2 + (r\dot{\theta})^2 \). \[ K = \frac{1}{2}m \left[ \left(\frac{dr}{dt}\right)^2 + r^2 \left(\frac{d\theta}{dt}\right)^2 \right] \]

The angular momentum \( L = m r^2 \dot{\theta} \) is conserved. We can rewrite the angular velocity as: \[ \frac{d\theta}{dt} = \frac{L}{m r^2} \] Substituting this into the Kinetic Energy term: \[ \frac{1}{2}m r^2 \left(\frac{L}{m r^2}\right)^2 = \frac{1}{2}m r^2 \frac{L^2}{m^2 r^4} = \frac{L^2}{2m r^2} \]

Substituting the terms back into the Total Energy equation: \[ E = \underbrace{\frac{1}{2}m \left(\frac{dr}{dt}\right)^2}_{\text{Radial KE}} + \underbrace{\frac{L^2}{2m r^2}}_{\text{Rotational KE}} – \underbrace{\frac{k}{r}}_{\text{Potential Energy}} \] Since the force is conservative, the Total Energy \( E \) is constant. Therefore: \[ \Delta \left( \frac{1}{2}m \left(\frac{dr}{dt}\right)^2 + \frac{L^2}{2m r^2} – \frac{k}{r} \right) = 0 \]